Class:9-10(বীজগানিতিক রাশি- অনুশীলনী:৩.১,৩.২,৩.৩,৩.৪- এর সৃজনশীল:3)

1) \displaystyle x^{2} -3=2\sqrt{2} হলে,

ক) \displaystyle x এর মান নির্নয় কর।

খ)\displaystyle x^{4} +\frac{1}{x^{4}} এর মান নির্নয় কর।

গ) প্রমান কর যে, \displaystyle x^{5} +\frac{1}{x^{5}} =58\sqrt{2}

ক)

দেওয়া আছে,
\displaystyle x^{2} -3=2\sqrt{2}

\displaystyle \Longrightarrow x^{2} =3+2\sqrt{2}

\displaystyle \Longrightarrow x^{2} =2+2\sqrt{2} +1

\displaystyle \Longrightarrow x^{2} =\left(\sqrt{2}\right)^{2} +2.\sqrt{2} .1+( 1)^{2}

\displaystyle \Longrightarrow x^{2} =\left(\sqrt{2} +1\right)^{2}

\displaystyle \Longrightarrow \sqrt{x^{2}} =\sqrt{\left(\sqrt{2} +1\right)^{2}}

\displaystyle \Longrightarrow x=\sqrt{2} +1

খ)

(ক) হতে পাই,
\displaystyle x=\sqrt{2} +1

\displaystyle \Longrightarrow \frac{1}{x} =\frac{1}{\sqrt{2} +1}

\displaystyle \Longrightarrow \frac{1}{x} =\frac{1.\left(\sqrt{2} -1\right)}{\left(\sqrt{2} +1\right) .\left(\sqrt{2} -1\right)} [ লব ও হর কে \displaystyle \sqrt{2} -1 দ্বারা গুন করে পাই ]

\displaystyle \Longrightarrow \frac{1}{x} =\frac{\left(\sqrt{2} -1\right)}{\left(\sqrt{2}\right)^{2} -( 1)^{2}}

\displaystyle \Longrightarrow \frac{1}{x} =\frac{\left(\sqrt{2} -1\right)}{2-1}

\displaystyle \Longrightarrow \frac{1}{x} =\frac{\left(\sqrt{2} -1\right)}{1}

\displaystyle \Longrightarrow \frac{1}{x} =\sqrt{2} -1

এখন,

\displaystyle x+\frac{1}{x} =\sqrt{2} +1+\sqrt{2} -1

\displaystyle =2\sqrt{2}

প্রদও রাশি\displaystyle =x^{4} +\frac{1}{x^{4}}

\displaystyle =\left( x^{2}\right)^{2} +\left(\frac{1}{x^{2}}\right)^{2}

\displaystyle =\left( x^{2} +\frac{1}{x^{2}}\right)^{2} -2.x^{2} .\frac{1}{x^{2}}

\displaystyle =\left{\left( x+\frac{1}{x}\right)^{2} -2.x.\frac{1}{x}\right}^{2} -2

\displaystyle =\left{\left( 2\sqrt{2}\right)^{2} -2\right}^{2} -2

\displaystyle =\left{( 2)^{2} .\left(\sqrt{2}\right)^{2} -2\right}^{2} -2

\displaystyle =( 4.2-2)^{2} -2

\displaystyle =( 8-2)^{2} -2

\displaystyle =( 6)^{2} -2

\displaystyle =6.6-2

\displaystyle =36-2

\displaystyle =34

গ)

(খ) হতে পাই,
\displaystyle x+\frac{1}{x} =2\sqrt{2} এবং \displaystyle x^{4} +\frac{1}{x^{4}} =34

সুতরাং, \displaystyle \left( x+\frac{1}{x}\right)\left( x^{4} +\frac{1}{x^{4}}\right) =2\sqrt{2} .34

\displaystyle \Longrightarrow x^{5} +x^{3} +\frac{1}{x^{3}} +\frac{1}{x^{5}} =68\sqrt{2}

\displaystyle \Longrightarrow x^{5} +\frac{1}{x^{5}} =68\sqrt{2} -\left( x^{3} +\frac{1}{x^{3}}\right)

\displaystyle \Longrightarrow x^{5} +\frac{1}{x^{5}} =68\sqrt{2} -\left{\left( x+\frac{1}{x}\right)^{3} -3.x.\frac{1}{x}\left( x+\frac{1}{x}\right)\right}

\displaystyle \Longrightarrow x^{5} +\frac{1}{x^{5}} =68\sqrt{2} -\left{\left( 2\sqrt{2}\right)^{3} -3.2\sqrt{2}\right}

\displaystyle \Longrightarrow x^{5} +\frac{1}{x^{5}} =\ 68\sqrt{2} -\left{( 2)^{3} .\left(\sqrt{2}\right)^{3} -6\sqrt{2}\right}

\displaystyle \Longrightarrow x^{5} +\frac{1}{x^{5}} =68\sqrt{2} -\left{\left( 8.\sqrt{2} .\sqrt{2} .\sqrt{2}\right) -6\sqrt{2}\right}

\displaystyle \Longrightarrow x^{5} +\frac{1}{x^{5}} =68\sqrt{2} -\left{8.\left(\sqrt{2}\right)^{2} .\sqrt{2} -6\sqrt{2}\right}

\displaystyle \Longrightarrow x^{5} +\frac{1}{x^{5}} =68\sqrt{2} -\left( 8.2\sqrt{2} -6\sqrt{2}\right)

\displaystyle \Longrightarrow x^{5} +\frac{1}{x^{5}} =68\sqrt{2} -\left( 16\sqrt{2} -6\sqrt{2}\right)

\displaystyle \Longrightarrow x^{5} +\frac{1}{x^{5}} =68\sqrt{2} -10\sqrt{2}

\displaystyle \Longrightarrow x^{5} +\frac{1}{x^{5}} =58\sqrt{2}

2) কোন সংখ্যা ও ঐ সংখ্যার গুনাক্তক বিপরীত সংখ্যার যোগফল \displaystyle 2\sqrt{3}

ক) সংখ্যাটিকে \displaystyle a চলকে প্রকাশ করে উপরের তথ্যকেএকটি সমীকরনের মাধ্যমে প্রকাশ কর ।

খ) \displaystyle a^{3} +\frac{1}{a^{3}} এর মান নির্নয় কর।

গ) প্রমান কর যে, \displaystyle a=\sqrt{3} +\sqrt{2}

ক)

ধরি, সংখ্যাটি \displaystyle =a

সুতরং, সংখ্যাটির গুনাক্তক বিপরীত সংখ্যা \displaystyle =\frac{1}{a}

প্রশ্ননুসারে, \displaystyle a+\frac{1}{a} =2\sqrt{3}

খ)

(ক) হতে পাই,
\displaystyle a+\frac{1}{a} =2\sqrt{3}

প্রদওরাশি\displaystyle =a^{3} +\frac{1}{a^{3}}

\displaystyle =\left( a+\frac{1}{a}\right)^{3} -3.a.\frac{1}{a}\left( a+\frac{1}{a}\right)

\displaystyle =\left( 2\sqrt{3}\right)^{3} -3.2\sqrt{3}

\displaystyle =( 2)^{3} .\left(\sqrt{3}\right)^{3} -6\sqrt{3}

\displaystyle =8.\sqrt{3} .\sqrt{3} .\sqrt{3} -6\sqrt{3}

\displaystyle =8.\left(\sqrt{3}\right)^{2} .\sqrt{3} -6\sqrt{3}

\displaystyle =8.3\sqrt{3} -6\sqrt{3}

\displaystyle =24\sqrt{3} -6\sqrt{3}

\displaystyle =18\sqrt{3}

গ)

(খ) হতে পাই,
\displaystyle a^{3} +\frac{1}{a^{3}} =18\sqrt{3}

\displaystyle \Longrightarrow \frac{a^{6} +1}{a^{3}} =18\sqrt{3}

\displaystyle \Longrightarrow a^{6} +1=18\sqrt{3} a^{3}

\displaystyle \Longrightarrow a^{6} -18\sqrt{3} a^{3} +1=0

\displaystyle \Longrightarrow \left( a^{3}\right)^{2} -2.a^{3} .9\sqrt{3} +\left( 9\sqrt{3}\right)^{2} -\left( 9\sqrt{3}\right)^{2} +1=0

\displaystyle \Longrightarrow \left( a^{3} -9\sqrt{3}\right)^{2} -( 9)^{2} .\left(\sqrt{3}\right)^{2} +1=0

\displaystyle \Longrightarrow \left( a^{3} -9\sqrt{3}\right)^{2} -81.3+1=0

\displaystyle \Longrightarrow \left( a^{3} -9\sqrt{3}\right)^{2} =243-1

\displaystyle \Longrightarrow \left( a^{3} -9\sqrt{3}\right)^{2} =242

\displaystyle \Longrightarrow \left( a^{3} -9\sqrt{3}\right)^{2} =121.2

\displaystyle \Longrightarrow \sqrt{\left( a^{3} -9\sqrt{3}\right)^{2}} =\sqrt{121} .\sqrt{2}

\displaystyle \Longrightarrow a^{3} -9\sqrt{3} =11\sqrt{2}

\displaystyle \Longrightarrow a^{3} =9\sqrt{3} +11\sqrt{2}

\displaystyle \Longrightarrow a^{3} =3\sqrt{3} +6\sqrt{3} +9\sqrt{2} +2\sqrt{2}

\displaystyle \Longrightarrow a^{3} =3\sqrt{3} +9\sqrt{2} +6\sqrt{3}+\displaystyle 2\sqrt{2}

\displaystyle \Longrightarrow a^{3} =\left(\sqrt{3}\right)^{3} +3.\left(\sqrt{3}\right)^{2} .\sqrt{2} +3.\sqrt{3} .\left(\sqrt{2}\right)^{2} +\left(\sqrt{2}\right)^{3}

\displaystyle \Longrightarrow a^{3} =\left(\sqrt{3} +\sqrt{2}\right)^{3}

\displaystyle \Longrightarrow a=\sqrt{3} +\sqrt{2} ( প্রমানিত )

3) \displaystyle b^{2} -2\sqrt{6} b+1=0

ক) দেখাও যে, \displaystyle b+\frac{1}{b} =2\sqrt{6}

খ) \displaystyle \frac{1}{b^{3}}\left( b^{6} -1\right) এর মান নির্নয় কর।

গ) প্রমান কর যে, \displaystyle b^{5} +\frac{1}{b^{5}} =922\sqrt{6}

ক)

দেওয়া আছে,
\displaystyle b^{2} -2\sqrt{6} b+1=0

\displaystyle \Longrightarrow b^{2} +1=2\sqrt{6} b

\displaystyle \Longrightarrow \frac{b^{2}}{b} +\frac{1}{b} +\frac{2\sqrt{6} b}{b} [ উভয়পক্ষ \displaystyle b দ্বারা ভাগ করে পাই ]

\displaystyle \Longrightarrow b+\frac{1}{b} =2\sqrt{6}

খ)

ক) হইতে পাই,
\displaystyle b+\frac{1}{b} =2\sqrt{6}

\displaystyle \left( b-\frac{1}{b}\right)^{2} =\left( b+\frac{1}{b}\right)^{2} -4.b.\frac{1}{b}

\displaystyle \Longrightarrow \left( b-\frac{1}{b}\right)^{2} =\left( 2\sqrt{6}\right)^{2} -4

\displaystyle \Longrightarrow \left( b-\frac{1}{b}\right)^{2} =( 2)^{2} .\left(\sqrt{6}\right)^{2} -4

\displaystyle \Longrightarrow \left( b-\frac{1}{b}\right)^{2} =( 4.6-4)

\displaystyle \Longrightarrow \left( b-\frac{1}{b}\right)^{2} =24-4

\displaystyle \Longrightarrow \left( b-\frac{1}{b}\right)^{2} =20

\displaystyle \Longrightarrow \sqrt{\left( b-\frac{1}{b}\right)^{2}} =\sqrt{4.}\sqrt{5}

\displaystyle \Longrightarrow b-\frac{1}{b} =2\sqrt{5}

প্রদওরাশি\displaystyle =\frac{1}{b^{3}}\left( b^{6} -1\right)

\displaystyle =\frac{b^{6}}{b^{3}} -\frac{1}{b^{3}}

\displaystyle =\left( b^{6-3}\right) -\frac{1}{b^{3}}

\displaystyle =b^{3} -\frac{1}{b^{3}}

\displaystyle =\left( b-\frac{1}{b}\right)^{3} +3b.\frac{1}{b}\left( b-\frac{1}{b}\right)

\displaystyle =\left( 2\sqrt{5}\right)^{3} +3.2\sqrt{5}

\displaystyle =( 2)^{3} .\left(\sqrt{5}\right)^{3} +6\sqrt{5}

\displaystyle =2.2.2.\sqrt{5} .\sqrt{5} .\sqrt{5} +6\sqrt{5}

\displaystyle =8.\left(\sqrt{5}\right)^{2} .\sqrt{5} +6\sqrt{5}

\displaystyle =8.5\sqrt{5} +6\sqrt{5}

\displaystyle =40\sqrt{5} +6\sqrt{5}

\displaystyle =46\sqrt{5}

গ)

(ক) হতে পাই,
\displaystyle b+\frac{1}{b} =2\sqrt{6}

\displaystyle \Longrightarrow \left( b+\frac{1}{b}\right)^{2} =\left( 2\sqrt{6}\right)^{2} [ বর্গ করে পাই ]

\displaystyle \Longrightarrow b^{2} +2b.\frac{1}{b} +\frac{1}{b^{2}} =( 2)^{2} .\left(\sqrt{6}\right)^{2}

\displaystyle \Longrightarrow b^{2} +\frac{1}{b^{2}} +2=4.6

\displaystyle \Longrightarrow b^{2} +\frac{1}{b^{2}} +2=24

\displaystyle \Longrightarrow b^{2} +\frac{1}{b^{2}} =24-2

\displaystyle \Longrightarrow b^{2} +\frac{1}{b^{2}} =22
আবার,

\displaystyle b+\frac{1}{b} =2\sqrt{6}

\displaystyle \Longrightarrow \left( b+\frac{1}{b}\right)^{3} =\left( 2\sqrt{6}\right)^{3} [ ঘন করে ]

\displaystyle \Longrightarrow b^{3} +\frac{1}{b^{3}} +3.b.\frac{1}{b}\left( b+\frac{1}{b}\right) =( 2)^{3} .\left(\sqrt{6}\right)^{3}

\displaystyle \Longrightarrow b^{3} +\frac{1}{b^{3}} +3.2\sqrt{6} =2.2.2.\sqrt{6} .\sqrt{6} .\sqrt{6}

\displaystyle \Longrightarrow b^{3} +\frac{1}{b^{3}} +6\sqrt{6} =8.\left(\sqrt{6}\right)^{2} .\sqrt{6}

\displaystyle \Longrightarrow b^{3} +\frac{1}{b^{3}} +6\sqrt{6} =8.6.\sqrt{6}

\displaystyle \Longrightarrow b^{3} +\frac{1}{b^{3}} =48\sqrt{6} -6\sqrt{6}

\displaystyle \Longrightarrow b^{3} +\frac{1}{b^{3}} =42\sqrt{6}

এখন, \displaystyle \left( b^{2} +\frac{1}{b^{2}}\right)\left( b^{3} +\frac{1}{b^{3}}\right) =22.42\sqrt{6}

\displaystyle \Longrightarrow b^{5} +\frac{1}{b} +b+\frac{1}{b^{5}} =924\sqrt{6}

\displaystyle \Longrightarrow b^{5} +\frac{1}{b^{5}} +b+\frac{1}{b} =924\sqrt{6}

\displaystyle \Longrightarrow b^{5} +\frac{1}{b^{5}} +2\sqrt{6} =924\sqrt{6}

\displaystyle \Longrightarrow b^{5} +\frac{1}{b^{5}} =924\sqrt{6} -2\sqrt{6}

\displaystyle \Longrightarrow b^{5} +\frac{1}{b^{5}} =922\sqrt{6}

4) \displaystyle x^{2} =5+2\sqrt{6} ,a+b+c=m,a^{2} +b^{2} +c^{2} =n,a^{3} +b^{3} =p^{3}

ক) \displaystyle x এর মান নির্নয় কর।

খ)প্রমান কর যে, \displaystyle \frac{x^{8} +1}{x^{4}} =98

গ) যদি\displaystyle c=0 হয়, দেখাও যে,\displaystyle m^{3} +2p^{3} =3mn

ক)

দেওয়া আছে,
\displaystyle x^{2} =5+2\sqrt{6}

\displaystyle \Longrightarrow x^{2} =3+2\sqrt{6} +2

\displaystyle \Longrightarrow x^{2} =\left(\sqrt{3}\right)^{2} +2.\sqrt{3} .\sqrt{2} +\left(\sqrt{2}\right)^{2}

\displaystyle \Longrightarrow x^{2} =\left(\sqrt{3} +\sqrt{2}\right)^{2}

\displaystyle \Longrightarrow \sqrt{x^{2}} =\sqrt{\left(\sqrt{3} +\sqrt{2}\right)^{2}}

\displaystyle \Longrightarrow x=\sqrt{3} +\sqrt{2}

খ)

(ক) হতে পাই,
\displaystyle x=\sqrt{3} +\sqrt{2}

\displaystyle \Longrightarrow \frac{1}{x} =\frac{1}{\sqrt{3} +\sqrt{2}}

\displaystyle \Longrightarrow \frac{1}{x} =\frac{1.\left(\sqrt{3} -\sqrt{2}\right)}{\left(\sqrt{3} +\sqrt{2}\right)\left(\sqrt{3} -\sqrt{2}\right)} [ হর ও লবকে \displaystyle \sqrt{3} -\sqrt{2} দ্বারা ভাগ করে পাই ]

\displaystyle \Longrightarrow \frac{1}{x} =\frac{\left(\sqrt{3} -\sqrt{2}\right)}{\left(\sqrt{3}\right)^{2} -\left(\sqrt{2}\right)^{2}}

\displaystyle \Longrightarrow \frac{1}{x} =\frac{\left(\sqrt{3} -\sqrt{2}\right)}{3-2}

\displaystyle \Longrightarrow \frac{1}{x} =\frac{\left(\sqrt{3} -\sqrt{2}\right)}{1}

\displaystyle \Longrightarrow \frac{1}{x} =\sqrt{3} -\sqrt{2}

সুতরাং \displaystyle x+\frac{1}{x} =\sqrt{3} +\sqrt{2} +\sqrt{3} -\sqrt{2}

\displaystyle =2\sqrt{3}

বামপক্ষ\displaystyle =\frac{x^{8} +1}{x^{4}}

\displaystyle =\frac{x^{8}}{x^{4}} +\frac{1}{x^{4}}

\displaystyle =\left( x^{8-4}\right) +\frac{1}{x^{4}}

\displaystyle =x^{4} +\frac{1}{x^{4}}

\displaystyle =\left( x^{2}\right)^{2} +\left(\frac{1}{x^{2}}\right)^{2}

\displaystyle =\left( x^{2} +\frac{1}{x^{2}}\right)^{2} -2.x^{2} .\frac{1}{x^{2}}

\displaystyle =\left(\left( x+\frac{1}{x}\right)^{2} -2.x.\frac{1}{x}\right)^{2} -2 [ ২য় ব্রাকেট ও দিতে পার ]

\displaystyle =\left(\left( 2\sqrt{3}\right)^{2} -2\right)^{2} -2

\displaystyle =\left(( 2)^{2} .\left(\sqrt{3}\right)^{2} -2\right)^{2} -2

\displaystyle =(( 4.3) -2)^{2} -2

\displaystyle =( 12-2)^{2} -2

\displaystyle =( 10)^{2} -2

\displaystyle =10.10-2

\displaystyle =100-2

\displaystyle =98

= ডানপক্ষ

সুতরাং
\displaystyle \frac{x^{8} +1}{x^{4}} =98

গ)

দেওয়া আছে,,

\displaystyle c=0

\displaystyle a+b+c=m

বা,\displaystyle a+b+0=m

বা, \displaystyle a+b=m

এবং,

\displaystyle a^{2} +b^{2} +c^{2} =n

বা, \displaystyle a^{2} +b^{2} +0^{2} =n

বা,\displaystyle a^{2} +b^{2} =n

\displaystyle a^{3} +b^{3} =p^{3}

বামপক্ষ\displaystyle =m^{3} +2p^{3}

\displaystyle =( a+b)^{3} +2\left( a^{3} +b^{3}\right) [ মান বসাই ]

\displaystyle =a^{3} +3a^{2} b+3ab^{2} +b^{3} +2a^{3} +2b^{3} [ সূত্র প্রয়োগ করে পাই ]

\displaystyle =3a^{3} +3a^{2} b+3ab^{2} +3b^{3}

\displaystyle =3\left( a^{3} +a^{2} b+ab^{2} +b^{3}\right)

\displaystyle =3\left( \ a^{2}( a+b) +b^{2}( a+b)\right) [ ২য় ব্রাকেট ও দিতে পার ]

\displaystyle =3( a+b)\left( a^{2} +b^{2}\right)

\displaystyle =3mn [ মান বসাই ]

=ডানপক্ষ

সুতরাং, \displaystyle m^{3} +2p^{3} =3mn\ \ ( দেখানো হলো )

5) \displaystyle x^{2} +\frac{1}{x^{2}} =10

ক) \displaystyle \left( x+\frac{1}{x}\right) এর মান কত ?

খ) প্রমান কর যে, \displaystyle \frac{x^{8} -1}{x^{4}} =40\sqrt{6}

গ) \displaystyle x^{5} -\frac{1}{x^{5}} এর মান নির্নয় কর।

ক)

দেওয়া আছে,
\displaystyle x^{2} +\frac{1}{x^{2}} =10

\displaystyle \Longrightarrow \left( x+\frac{1}{x}\right)^{2} -2.x.\frac{1}{x} =10

\displaystyle \Longrightarrow \left( x+\frac{1}{x}\right)^{2} -2=10

\displaystyle \Longrightarrow \left( x+\frac{1}{x}\right)^{2} =10+2

\displaystyle \Longrightarrow \left( x+\frac{1}{x}\right)^{2} =12

\displaystyle \Longrightarrow \left( x+\frac{1}{x}\right)^{2} =4.3

\displaystyle \Longrightarrow \sqrt{\left( x+\frac{1}{x}\right)^{2}} =\sqrt{4} .\sqrt{3} [ উভয় পক্ষ বর্গমূল করে পাই ]

\displaystyle \Longrightarrow \left( x+\frac{1}{x}\right) =2\sqrt{3}

খ)

দেওয়া আছে,
\displaystyle x^{2} +\frac{1}{x^{2}} =10

\displaystyle \Longrightarrow \left( x-\frac{1}{x}\right)^{2} +2.x.\frac{1}{x} =10

\displaystyle \Longrightarrow \left( x-\frac{1}{x}\right)^{2} +2=10

\displaystyle \Longrightarrow \left( x-\frac{1}{x}\right)^{2} =10-2

\displaystyle \Longrightarrow \left( x-\frac{1}{x}\right)^{2} =8

\displaystyle \Longrightarrow \left( x-\frac{1}{x}\right)^{2} =4.2

\displaystyle \Longrightarrow \sqrt{\left( x-\frac{1}{x}\right)^{2}} =\sqrt{4} .\sqrt{2} [ উভয় পক্ষ বর্গমূল করে পাই ]

\displaystyle \Longrightarrow \left( x-\frac{1}{x}\right) =2\sqrt{2}

(ক) হতে পাই,

\displaystyle \left( x+\frac{1}{x}\right) =2\sqrt{3}

বামপক্ষ\displaystyle =\frac{x^{8} -1}{x^{4}}

\displaystyle =\frac{x^{8}}{x^{4}} -\frac{1}{x^{4}}

\displaystyle =x^{4} -\frac{1}{x^{4}}

\displaystyle =\left( x^{2}\right)^{2} -\left(\frac{1}{x^{2}}\right)^{2}

\displaystyle =\left( x^{2} +\frac{1}{x^{2}}\right)\left( x^{2} -\frac{1}{x^{2}}\right) [ \displaystyle a^{2} -b^{2} =( a+b)( a-b) \] সূত্র প্রয়োগ করে পাই

\displaystyle =10.\left( x+\frac{1}{x}\right) .\left( x-\frac{1}{x}\right)

\displaystyle =10.2\sqrt{3} .2\sqrt{2} [ মান বসাই ]

\displaystyle =10.2.2.\sqrt{3} .\sqrt{2}

\displaystyle =40\sqrt{6}
\displaystyle =ডানপক্ষ

সুতরাং, \displaystyle \frac{x^{8} -1}{x^{4}} =40\sqrt{6} ( প্রমানিত )

Post Author: showrob

1 thought on “Class:9-10(বীজগানিতিক রাশি- অনুশীলনী:৩.১,৩.২,৩.৩,৩.৪- এর সৃজনশীল:3)

    soikot

    (April 8, 2020 - 8:33 pm)

    Very nice post?

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