Class:9-10(অনুশীলনি:৩.১,৩.২,৩.৩,৩.৪-সৃজনশীল:5-এর বীজগানিতিক রাশি)

1) যদি \displaystyle p^{2} =5+2\sqrt{6} ,\ a^{3} +a^{-3} =18\sqrt{3} হয়, তাহলে-

ক) \displaystyle p-\frac{1}{p} এর মান বের কর।

খ) দেখাও যে, \displaystyle a=\sqrt{3} +\sqrt{2} যখন,\displaystyle a^{3} -a^{-3}  >0.

গ) প্রমান কর যে, \displaystyle \frac{p^{10} +1}{p^{5}} =178\sqrt{3}

ক)

দেওয়া আছে,,

\displaystyle p^{2} =5+2\sqrt{6}

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

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

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

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

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

সুতরাং, \displaystyle \frac{1}{p} =\frac{1}{\sqrt{3} +\sqrt{2}}

\displaystyle =\frac{1.\left(\sqrt{3} -\sqrt{2}\right)}{\left(\sqrt{3} +\sqrt{2}\right) .\left(\sqrt{3} -\sqrt{2}\right)} [উভয়পক্ষ \displaystyle \sqrt{3} -\sqrt{2} দ্বারা ভাগ করে পাই ]

\displaystyle =\frac{\left(\sqrt{3} -\sqrt{2}\right)}{\left(\sqrt{3}\right)^{2} -\left(\sqrt{2}\right)^{2}} [ আমরা জানি, \displaystyle ( a+b) .( a-b) সূত্র অনুযায়ী ]

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

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

\displaystyle \frac{1}{p} =\sqrt{3} -\sqrt{2}

এখন, \displaystyle p-\frac{1}{p} =\left(\sqrt{3} +\sqrt{2}\right) -\left(\sqrt{3} -\sqrt{2}\right)

\displaystyle =\sqrt{3} +\sqrt{2} -\sqrt{3} +\sqrt{2}

\displaystyle =2\sqrt{2}

নির্নয় মান: \displaystyle 2\sqrt{2}

খ)

দেওয়া আছে,

\displaystyle a^{3} +a^{-3} =18\sqrt{3}

বা \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

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

\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} [ বর্গমূল করে ]

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

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

\displaystyle =3\sqrt{3} +6\sqrt{3} +9\sqrt{2} +2\sqrt{2}

\displaystyle =3\sqrt{3} +9\sqrt{2} +6\sqrt{3} ++2\sqrt{2}

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

গ)

(ক) হতে পাই,

\displaystyle p=\sqrt{3} +\sqrt{2}

এবং \displaystyle \frac{1}{p} =\sqrt{3} -\sqrt{2}

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

\displaystyle =2\sqrt{3}

\displaystyle \Longrightarrow \left( p+\frac{1}{p}\right)^{2} =\left( 2\sqrt{3}\right)^{2} [ বর্গ করে পাই ]

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

\displaystyle \Longrightarrow p^{2} +\frac{1}{p^{2}} +2=4.3

\displaystyle \Longrightarrow p^{2} +\frac{1}{p^{2}} =12-2

\displaystyle \Longrightarrow p^{2} +\frac{1}{p^{2}} =10------( i) \ নং

আবার, \displaystyle p+\frac{1}{p} =2\sqrt{3}

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

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

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

\displaystyle \Longrightarrow p^{3} +\frac{1}{p^{3}} +6\sqrt{3} =8.3\sqrt{3}

\displaystyle \Longrightarrow p^{3} +\frac{1}{p^{3}} =24\sqrt{3} -6\sqrt{3}

\displaystyle \Longrightarrow p^{3} +\frac{1}{p^{3}} =18\sqrt{3} ------( ii) নং\

(i) ও (ii) গুন করে পাই,

\displaystyle \left( p^{2} +\frac{1}{p^{2}}\right)\left( p^{3} +\frac{1}{p^{3}}\right) =10.18\sqrt{3} [ মান বসাই ]

\displaystyle \Longrightarrow p^{5} +\frac{1}{p^{5}} +p+\frac{1}{p} =180\sqrt{3}

\displaystyle \Longrightarrow p^{5} +\frac{1}{p^{5}} +2\sqrt{3} =180\sqrt{3} [ মান বসাই ]

\displaystyle \Longrightarrow p^{5} +\frac{1}{p^{5}} =180\sqrt{3} -2\sqrt{3}

\displaystyle \Longrightarrow p^{5} +\frac{1}{p^{5}} =178\sqrt{3}

\displaystyle \Longrightarrow \frac{p^{10} +1}{p^{5}} =178\sqrt{3} ( প্রমানিত )

2) যদি \displaystyle p^{2} =7+4\sqrt{3} হলে,

ক) \displaystyle p এর মান নির্নয় কর ।

খ) \displaystyle \frac{p^{6} -1}{p^{3}} এর মান নির্নয় কর ।

গ) প্রমান কর যে, \displaystyle p^{5} +\frac{1}{p^{5}} =724

ক)

দেওয়া আছে,

\displaystyle p^{2} =7+4\sqrt{3}

\displaystyle \Longrightarrow p^{2} =4+4\sqrt{3} +3

\displaystyle \Longrightarrow p^{2} =2^{2} +2.2.\sqrt{3} +\left(\sqrt{3}\right)^{2}

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

\displaystyle \Longrightarrow \sqrt{p^{2}} =\sqrt{\left( 2+\sqrt{3}\right)^{2}} [ বর্গমূল করে পাই ]

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

খ)

(ক) হতে পাই,

\displaystyle p=2+\sqrt{3}

\displaystyle \frac{1}{p} =\frac{1}{2+\sqrt{3}}

\displaystyle =\frac{1.\left( 2-\sqrt{3}\right)}{\left( 2+\sqrt{3}\right) .\left( 2-\sqrt{3}\right)} [উভয়পক্ষ \displaystyle 2-\sqrt{3} দ্বারা ভাগ করে পাই ]

\displaystyle =\frac{\left( 2-\sqrt{3}\right)}{( 2)^{2} -\left(\sqrt{3}\right)^{2}} [ \displaystyle ( a+b)( a-b) <strong>=a^{2} -b^{2} সূত্র প্রয়োগ করে পাই ]

\displaystyle =\frac{\left( 2-\sqrt{3}\right)}{4-3}

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

\displaystyle \frac{1}{p} =2-\sqrt{3}

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

\displaystyle =2\sqrt{3}

এখন, \displaystyle \frac{p^{6} -1}{p^{3}}

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

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

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

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

\displaystyle =( 2)^{3} .\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 =30\sqrt{3}

নির্নয় মান: \displaystyle 30\sqrt{3}

গ)

(ক) হতে পাই,

\displaystyle p=2+\sqrt{3}

(খ) হতে পাই,

\displaystyle \frac{1}{p} =2-\sqrt{3}

এখন, \displaystyle p+\frac{1}{p} =2+\sqrt{3} +2-\sqrt{3}

\displaystyle =4

বামপক্ষ\displaystyle =p^{5} +\frac{1}{p^{5}}

\displaystyle =\left( p^{3} +\frac{1}{p^{5}}\right) .\left( p^{2} +\frac{1}{p^{2}}\right) -\left( p+\frac{1}{p}\right)

\displaystyle =\left{\left( p+\frac{1}{p}\right)^{3} -3.p.\frac{1}{p}\left( p+\frac{1}{p}\right)\right} .\left{\left( p+\frac{1}{p}\right)^{2} -2.p.\frac{1}{p}\right} -\left( p+\frac{1}{p}\right)

\displaystyle =\left( 4^{3} -3.4\right)\left( 4^{2} -2\right) -4

\displaystyle =( 64-12) .( 16-2) -4

\displaystyle =( 52.14-4)

\displaystyle =728-4

\displaystyle =724

সুতরাং \displaystyle p^{5} +\frac{1}{p^{5}} =724 ( প্রমানিত )

3) যদি \displaystyle x+\frac{1}{x} =6 হয়,

ক) \displaystyle \left( x-\frac{1}{x}\right)^{2}এর মান কত?

খ) দেখাও যে, \displaystyle x^{3} +\frac{1}{x^{3}} =198

গ) প্রমান কর যে, \displaystyle x^{5} +\frac{1}{x^{5}} =6726

ক)

দেওয়া আছে,

\displaystyle x+\frac{1}{x} =6

আমরা জানি,

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

\displaystyle =( 6)^{2} -4

\displaystyle =36-4

\displaystyle =32
নির্নয় মান: 32

খ)

দেওয়া আছে,

\displaystyle x+\frac{1}{x} =6

বামপক্ষ\displaystyle =x^{3} +\frac{1}{x^{3}}

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

\displaystyle =( 6)^{3} -3.6

\displaystyle =6.6.6-18

\displaystyle =216-18

\displaystyle =198

\displaystyle =ডানপক্ষ

সুতরাং, \displaystyle x^{3} +\frac{1}{x^{3}} =198 ( দেখানো হলো )

গ)

দেওয়া আছে,

\displaystyle x+\frac{1}{x} =6

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

\displaystyle \Longrightarrow x^{2} +2.x.\frac{1}{x} +\frac{1}{x^{2}} =36

\displaystyle \Longrightarrow x^{2} +2+\frac{1}{x^{2}} =36

\displaystyle \Longrightarrow x^{2} +\frac{1}{x^{2}} =36-2

\displaystyle \Longrightarrow x^{2} +\frac{1}{x^{2}} =34

(খ) হতে পাই,
\displaystyle x^{3} +\frac{1}{x^{3}} =198

এখন, \displaystyle \left( x^{3} +\frac{1}{x^{3}}\right) .\left( x^{2} +\frac{1}{x^{2}}\right) =198.34

\displaystyle \Longrightarrow x^{5} +x+\frac{1}{x} +\frac{1}{x^{5}} =6732

\displaystyle \Longrightarrow x^{5} +\frac{1}{x^{5}} +\left( x+\frac{1}{x}\right) =6732

\displaystyle \Longrightarrow x^{5} +\frac{1}{x^{5}} +6=6732 [ মান বসাই ]

\displaystyle \Longrightarrow x^{5} +\frac{1}{x^{5}} =6732-6

\displaystyle \Longrightarrow x^{5} +\frac{1}{x^{5}} =6726 ( দেখানো হল )

4) \displaystyle x+y=\sqrt{3} এবং \displaystyle x^{2} -y^{2} =\sqrt{6}

খ) \displaystyle xy এর মান নির্নয় কর ।

খ) দেখাও যে, \displaystyle x^{3} +y^{3} +\frac{\sqrt{27}}{4} =3\sqrt{3}

গ) \displaystyle 16xy\left( x^{2} +y^{2}\right) এর মান নির্নয় কর ।

ক)

দেওয়া আছে,,

\displaystyle x+y=\sqrt{3}

এবং, \displaystyle x^{2} -y^{2} =\sqrt{6}

\displaystyle \Longrightarrow ( x+y)( x-y) =\sqrt{6} [ \displaystyle a^{2} -b^{2} =( a+b)( a-b) সূত্র অনুসারে ]

\displaystyle \Longrightarrow x-y=\left(\frac{\sqrt{6}}{x+y}\right)

\displaystyle \Longrightarrow x-y=\left(\frac{\sqrt{6}}{\sqrt{3}}\right)

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

\displaystyle \Longrightarrow x-y=\sqrt{2}

আমরা জানি,

\displaystyle xy=\left(\frac{x+y}{2}\right)^{2} -\left(\frac{x-y}{2}\right)^{2}

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

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

\displaystyle =\frac{3}{4} -\frac{2}{4}

\displaystyle =\frac{3-2}{4}

\displaystyle =\frac{1}{4}

নির্নয় মান: \displaystyle \frac{1}{4}

খ)

দেওয়া আছে,

\displaystyle x+y=\sqrt{3}

(ক) হতে পাই,

\displaystyle xy=\frac{1}{4}
বামপক্ষ\displaystyle =x^{3} +y^{3} +\frac{\sqrt{27}}{4}

\displaystyle =( x+y)^{3} -3xy( x+y) +\frac{\sqrt{27}}{4}

\displaystyle =\left(\sqrt{3}\right)^{3} -3.\frac{1}{4} .\sqrt{3} +\frac{\sqrt{9.3}}{4}

\displaystyle =\left(\sqrt{3}\right)^{2} .\sqrt{3} -\frac{\sqrt{3} .\sqrt{3} .\sqrt{3}}{4} +\frac{3\sqrt{3}}{4}

\displaystyle =3\sqrt{3} -\frac{3\sqrt{3}}{4} +\frac{3\sqrt{3}}{4}

\displaystyle =3\sqrt{3}

\displaystyle =ডানপক্ষ

সুতরাং, \displaystyle x^{3} +y^{3} +\frac{\sqrt{27}}{4} =3\sqrt{3} ( দেখানো হলো )

গ)

দেওয়া আছে,,

\displaystyle x+y=\sqrt{3}

(ক) হতে পাই,

\displaystyle x-y=\sqrt{2}

\displaystyle xy=\frac{1}{4}

প্রদওরাশি\displaystyle =16xy\left( x^{2} +y^{2}\right)

\displaystyle =16.\frac{1}{4} .\frac{( x+y)^{2} +( x-y)^{2}}{2} [ আমরা জানি, \displaystyle 2\left( a^{2} +b^{2}\right) =( a+b)^{2} +( a-b)^{2} ]

\displaystyle =4.\frac{( x+y)^{2} +( x-y)^{2}}{2}

\displaystyle =4.\frac{\left(\sqrt{3}\right)^{2} +\left(\sqrt{2}\right)^{2}}{2} [ মান বসাই ]

\displaystyle =2.( 3+2)

\displaystyle =2.5

\displaystyle =10

সুতরাং, নির্নয় মান: 10

5) \displaystyle a+b+c,\ a^{2} +b^{2} +c^{2} দুইটি বীজগানিতিক রাশি ।

ক) ১ম রাশি\displaystyle =0 হলে, প্রমান কর যে, \displaystyle a^{3} +b^{3} +c^{3} =3abc

খ) ১ম রাশি\displaystyle =10, ২য় রাশি\displaystyle =34 হলে,\displaystyle ( a-b)^{2} +( b-c)^{2} +( c-a)^{2}এর মান কত?

গ) ১ম রাশি\displaystyle =0 হলে,প্রমান কর যে,\displaystyle \frac{( b+c)^{2}}{6bc} +\frac{( c+a)^{2}}{6ca} +\frac{( a+b)^{2}}{6ab} =\frac{1}{2}

ক)

দেওয়া আছে,

\displaystyle a+b+c=0

\displaystyle \Longrightarrow a+b=-c

\displaystyle \Longrightarrow ( a+b)^{3} =( -c)^{3} [ উভয় পক্ষ ঘন করে পাই ]

\displaystyle \Longrightarrow a^{3} +b^{3} +3ab( a+b) =-c^{3}

\displaystyle \Longrightarrow a^{3} +b^{3} +3ab.( -c) =-c^{3} [ \displaystyle a+b=-c ] মান বসাই

\displaystyle \Longrightarrow a^{3} +b^{3} -3abc=-c^{3}

\displaystyle \Longrightarrow a^{3} +b^{3} +c^{3} =3abc ( প্রমানিত )

খ)

১ম শর্তমতে, \displaystyle a+b+c=10

২য় শর্তমতে, \displaystyle a^{2} +b^{2} +c^{2} =38

এখানে, \displaystyle a+b+c=10

\displaystyle \Longrightarrow ( a+b+c)^{2} =( 10)^{2} [ বর্গ করে ]

\displaystyle \Longrightarrow a^{2} +b^{2} +c^{2} +2ab+2bc+2ca=100 [ সূত্র প্রয়োগ করে পাই ]

\displaystyle \Longrightarrow 38+2( ab+bc+ca) =100

\displaystyle \Longrightarrow 2( ab+bc+ca) =100-38

\displaystyle \Longrightarrow 2( ab+bc+ca) =62

প্রদওরাশি\displaystyle =( a-b)^{2} +( b-c)^{2} +( c-a)^{2}

\displaystyle =a^{2} -2ab+b^{2} +b^{2} -2bc+c^{2} +c^{2} -2ca+a^{2}

\displaystyle =2a^{2} +2b^{2} +2c^{2} -2ab-2bc-2ca

\displaystyle =2\left( a^{2} +b^{2} +c^{2}\right) -2( ab+bc+ca)

\displaystyle =2.38-62

\displaystyle =76-62

\displaystyle =14

নির্নয় মান: \displaystyle 14

গ)

দেওয়া আছে,

১ম রাশি\displaystyle =a+b+c

১ম রাশি\displaystyle =0 হলে,\displaystyle a+b+c=0

\displaystyle a+b=-c

এবং,\displaystyle b+c=-a

আবার, \displaystyle c+a=-b

ক) হতে পাই, \displaystyle a^{3} +b^{3} +c^{3} =3abc

এখন, \displaystyle \frac{( b+c)^{2}}{6bc} +\frac{( c+a)^{2}}{6ca} +\frac{( a+b)^{2}}{6ab}

\displaystyle =\frac{( -a)^{2}}{6bc} +\frac{( -b)^{2}}{\ 6ca} +\frac{( -c)^{2}}{6ab}

\displaystyle =\frac{a^{2}}{6ab} +\frac{b^{2}}{6ca} +\frac{c^{2}}{6ab}

\displaystyle =\frac{a^{3} +b^{3} +c^{3}}{6abc}

\displaystyle =\frac{3abc}{6abc}

\displaystyle =\frac{1}{2} ( প্রমানিত )

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