class:9-10 (বীজগানিতিক রাশি: অনুশীলনী:৩.১,৩.২,৩.৩,৩.৪- এর সৃজনশীল(1)।

1) \displaystyle a^{4} +a^{2} b^{2} +b^{4} =21 এবং \displaystyle a^{2} +ab+b^{2} =7
ক) \displaystyle \ a^{2} -ab+b^{2} এর মান নির্নয় কর।
খ) প্রমান কর যে,\displaystyle \ 8ab( a^{2} +b^{2} =80
গ) \displaystyle \ a,b এর মান বের কর।

ক)

দেওয়া আছে,
\displaystyle a^{4} +a^{2} b^{2} +b^{4} =21,এবং \displaystyle a^{2} +ab+b^{2} =7
এখন,

\displaystyle a^{4} +a^{2} b^{2} +b^{4} =21

\displaystyle \Longrightarrow \left( a^{2}\right)^{2} +2a^{2} b^{2} +\left( b^{2}\right)^{2} -a^{2} b^{2} =21

\displaystyle \Longrightarrow \left( a^{2} +b^{2}\right)^{2} -( ab)^{2} =21

\displaystyle \Longrightarrow \left( a^{2} +b^{2} +ab\right)\left( a^{2} +b^{2} -ab\right) =21

\displaystyle \Longrightarrow \left( a^{2} +ab+b^{2}\right)\left( a^{2} -ab+b^{2}\right) =21

\displaystyle \Longrightarrow 7.\left( a^{2} -ab+b^{2}\right) =21

\displaystyle \Longrightarrow a^{2} -ab+b^{2} =\frac{21}{7}

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

খ)

দেওয়া আছে,

\displaystyle a^{2} +ab+b^{2} =7———-(i) নং

ক) হতে পাই, \displaystyle a^{2} +ab+b^{2} =3 ——(ii) নং

সমীকরন (i) ও (ii) যোগ করে পাই,,

\displaystyle \Longrightarrow \left( a^{2} +ab+b^{2}\right) +\left( a^{2} -ab+b^{2}\right) =7+3

\displaystyle \Longrightarrow a^{2} +ab+b^{2} +a^{2} -ab+b^{2} =10

\displaystyle \Longrightarrow 2a^{2} +2b^{2} =10

\displaystyle \Longrightarrow 2\left( a^{2} +b^{2}\right) =10

\displaystyle \Longrightarrow a^{2} +b^{2} =\frac{10}{2}

\displaystyle \Longrightarrow a^{2} +b^{2} =5

সমীকরন (i) ও (ii) বিয়োগ করে পাই,,

\displaystyle \Longrightarrow \left( a^{2} +ab+b^{2}\right) -\left( a^{2} -ab+b^{2}\right) =7-3

\displaystyle \Longrightarrow a^{2} +ab+b^{2} -a^{2} +ab-b^{2} =4

\displaystyle \Longrightarrow 2ab=4

\displaystyle \Longrightarrow ab=\frac{4}{2}

\displaystyle \Longrightarrow ab=2

বামপক্ষ\displaystyle =8ab\left( a^{2} +b^{2}\right)

\displaystyle =8.2.5

\displaystyle =80

= ডানপক্ষ ( প্রমানিত)

(গ)

খ) হতে পাই,

\displaystyle a^{2} +b^{2} =5 —–(i) নং

\displaystyle ab=2 —–(ii) নং

আমরা জানি,

\displaystyle \left( a^{2} -b^{2}\right)^{2} =\left( a^{2} +b^{2}\right)^{2} -4a^{2} b^{2}

\displaystyle \Longrightarrow \left( a^{2} -b^{2}\right)^{2} =\left( a^{2} +b^{2}\right)^{2} -4( ab)^{2}

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

\displaystyle \Longrightarrow \left( a^{2} -b^{2}\right)^{2} =25-4.4

\displaystyle \Longrightarrow \left( a^{2} -b^{2}\right)^{2} =25-16

\displaystyle \Longrightarrow \left( a^{2} -b^{2}\right)^{2} =9

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

\displaystyle \Longrightarrow a^{2} -b^{2} =3 —-(iii) নং

সমীকরন(i) ও (iii) যোগ করে পাই,

\displaystyle \left( a^{2} +b^{2}\right) +\left( a^{2} -b^{2}\right) =5+3

\displaystyle \Longrightarrow \left( a^{2} +b^{2} +a^{2} -b^{2}\right) =8

\displaystyle \Longrightarrow 2a^{2} =8

\displaystyle \Longrightarrow a^{2} =\frac{8}{2}

\displaystyle \Longrightarrow a^{2} =4

\displaystyle \Longrightarrow \sqrt{a^{2}} =\sqrt{4}

\displaystyle \Longrightarrow a=2

\displaystyle a এর মান (ii) নং সমীকরনে বসাই,,

\displaystyle ab=2

\displaystyle \Longrightarrow 2.b=2

\displaystyle \Longrightarrow b=\frac{2}{2}

\displaystyle b=1

সুতরং, \displaystyle a=2 এবং \displaystyle b=1

2)\displaystyle x-\frac{1}{x} =m একটি বীজগানিতিক রাশি।
ক) \displaystyle x^{2} +\frac{1}{x^{2}} এর মান নির্নয় কর।
খ) দেখাও যে, \displaystyle \frac{x^{8} +1}{x^{4}} =m^{4} +4m^{2} +2
গ) যদি \displaystyle x^{4} =119-\frac{1}{x^{4}} হয়, তাহলে প্রমান কর যে, m=±3

ক)


দেওয়া আছে,
\displaystyle x-\frac{1}{x} =m
\displaystyle \Longrightarrow \left( x-\frac{1}{x}\right)^{2} =m^{2} [ বর্গ করে পাই ]
\displaystyle \Longrightarrow x^{2} -2.x.\frac{1}{x} +\frac{1}{x^{2}} =m^{2}
\displaystyle \Longrightarrow x^{2} -2+\frac{1}{x^{2}} =m^{2}
\displaystyle \Longrightarrow x^{2} +\frac{1}{x^{2}} =m^{2} +2

খ)

ক) হইতে পাই,
\displaystyle x^{2} +\frac{1}{x^{2}} =m^{2} +2

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

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

\displaystyle \Longrightarrow x^{4} +2+\frac{1}{x^{4}} =m^{4} +4m^{2} +4

\displaystyle \Longrightarrow x^{4} +\frac{1}{x^{4}} =m^{4} +4m^{2} +4-2

\displaystyle \Longrightarrow x^{4} +\frac{1}{x^{4}} =m^{4} +4m^{2} +2

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

\displaystyle \Longrightarrow \frac{x^{8} +1}{x^{4}} =m^{4} +4m^{2} +2 ( দেখানো হলো )

গ)

দেওয়া আছে,
\displaystyle x^{4} =119-\frac{1}{x^{4}}

\displaystyle \Longrightarrow x^{4} +\frac{1}{x^{4}} =119

\displaystyle \Longrightarrow \left( x^{2}\right)^{2} +\left(\frac{1}{x^{2}}\right)^{2} =119

\displaystyle \Longrightarrow \left( x^{2} +\frac{1}{x^{2}}\right)^{2} -2.x^{2} .\frac{1}{x^{2}} =119

\displaystyle \Longrightarrow \left( x^{2} +\frac{1}{x^{2}}\right)^{2} =119+2

\displaystyle \Longrightarrow \left( x^{2} +\frac{1}{x^{2}}\right)^{2} =121

\displaystyle \Longrightarrow \sqrt{\left( x^{2} +\frac{1}{x^{2}}\right)^{2}} =\sqrt{121} [ বর্গমূল করে পাই ]

\displaystyle \Longrightarrow x^{2} +\frac{1}{x^{2}} =11

\displaystyle \Longrightarrow \left( x-\frac{1}{x}\right)^{2} +2.x.\frac{1}{x} =11

\displaystyle \Longrightarrow \left( x-\frac{1}{x}\right)^{2} =11-2

\displaystyle \Longrightarrow \left( x-\frac{1}{x}\right)^{2} =9

\displaystyle \Longrightarrow \sqrt{\left( x-\frac{1}{x}\right)^{2}} =\pm \sqrt{9} [ বর্গমূল করে পাই ]

\displaystyle \Longrightarrow x-\frac{1}{x} =\pm 3 [ দেওয়া আছে, \displaystyle x-\frac{1}{x} =m ]

\displaystyle \Longrightarrow m=\pm 3 ( প্রমানিত )

3) \displaystyle p+q+r=8,p^{2} +q^{2} +r^{2} =42,x+y+3z=0 হলে,

ক) \displaystyle 2p-q+r এর বর্গ নির্নয় কর।

খ) সরল কর: \displaystyle \frac{( x+y)^{2}}{3xy} +\frac{( y+3z)^{2}}{9yz} +\frac{( 3z+x)^{2}}{9zx}

গ) \displaystyle ( p-q)^{2} +( q-r)^{2} +( r-p)^{2} এর মান নির্নয় কর।

ক)

\displaystyle ( 2p-q+r) এর বর্গ:\displaystyle ( 2p-q+r)^{2}

ধরি,\displaystyle 2p-q=m

\displaystyle \Longrightarrow ( m+r)^{2}

\displaystyle \Longrightarrow m^{2} +2.m.r+r^{2}

\displaystyle \Longrightarrow ( 2p-q)^{2} +2.( 2p-q) .r+r^{2}

\displaystyle \Longrightarrow ( 2p)^{2} -2.2p.q+q^{2} +4pr-2qr+r^{2}

\displaystyle \Longrightarrow 4p^{2} -4pq+q^{2} +4pr-2qr+r^{2}

\displaystyle \Longrightarrow 4p^{2} +q^{2} +r^{2} -4pq-2qr+4pr

খ)

দেওয়া আছে,

\displaystyle x+y+3z=0

\displaystyle \Longrightarrow x+y=-3z —— (i) নং

\displaystyle \Longrightarrow x+3z=-y ——(ii) নং

\displaystyle \Longrightarrow y+3z=-x —— (iii) নং

এখন,
\displaystyle \frac{( x+y)^{2}}{3xy} +\frac{( y+3z)^{2}}{9yz} +\frac{( 3z+x)^{2}}{9zx}

\displaystyle =\frac{( -3z)^{2}}{3xy} +\frac{( -x)^{2}}{9yz} +\frac{( -y)^{2}}{9zx} [(i),(ii) ও (iii) নং থেকে পাই ]

\displaystyle =\frac{9z^{2}}{3xy} +\frac{x^{2}}{9yz} +\frac{y^{2}}{9zx}

\displaystyle =\frac{27z^{2} +x^{2} +y^{2}}{9xyz}

\displaystyle =\frac{27z^{2} +( x+y)^{3} -3xy( x+y)}{9xyz}

\displaystyle =\frac{27z^{2} +( -3z)^{3} -3xy.( -3z)}{9xyz}

\displaystyle =\frac{27z^{2} -27z^{2} +9xyz}{9xyz}

\displaystyle =\frac{9xyz}{9xyz}

\displaystyle =1

গ)

দেওয়া আছে,

\displaystyle p+q+r=8

এবং, \displaystyle p^{2} +q^{2} +r^{2} =42

প্রদও রাশি \displaystyle =( p-q)^{2} +( q-r)^{2} +( r-p)^{2}

\displaystyle =p^{2} -2pq+q^{2} +q^{2} -2qr+r^{2} +r^{2} -2rp+p^{2}

\displaystyle =2p^{2} +2q^{2} +2r^{2} -2pq-2qr-2pr

\displaystyle =2\left( p^{2} +q^{2} +r^{2}\right) -2( pq+qr+pr)

\displaystyle =2\left( p^{2} +q^{2} +r^{2}\right) -\left{( p+q+r)^{2} -\left( p^{2} +q^{2} +r^{2}\right)\right}

\displaystyle =2.42-\left{( 8)^{2} -42\right}

\displaystyle =84-( 64-42)

\displaystyle =84-64+42

\displaystyle =62

4) \displaystyle A=15x^{2} -2xy-8y^{2} ,B=5x-4y,P=x^{4} +x^{2} y^{2} +y^{4} ,Q=x^{2} -xy+y^{2}

ক) \displaystyle u-v=4 এবং \displaystyle uv=21 হলে,\displaystyle u+v এর মান বের কর।

খ) \displaystyle A কে দুইটি বর্গের বিয়োগফলরূপে প্রকাশ কর।

গ) \displaystyle P=35 এবং \displaystyle Q=7 হলে দেখাও যে, \displaystyle x-y=\pm 2\sqrt{2}

ক)

দেওয়া আছে,

\displaystyle u-v=4 এবং \displaystyle uv=21

আমরা জানি,

\displaystyle \Longrightarrow ( u+v)^{2} =( u-v)^{2} +4uv

\displaystyle \Longrightarrow ( u+v)^{2} =( 4)^{2} +4.21

\displaystyle \Longrightarrow ( u+v)^{2} =16+84

\displaystyle \Longrightarrow ( u+v)^{2} =100

\displaystyle \Longrightarrow \sqrt{( u+v)^{2}} =\pm \sqrt{100}

\displaystyle \Longrightarrow u+v=\pm 10

সুতরং নির্নয় মান: ±10

খ)

দেওয়া আছে,

\displaystyle A=15x^{2} -2xy-8y^{2}

\displaystyle =15x^{2} -12xy+10xy-8y^{2}

\displaystyle =3x( 5x-4y) +2y( 5x-4y)

\displaystyle =( 5x-4y)( 3x+2y)

ধরি,

\displaystyle ( 5x-4y) =a

\displaystyle ( 3x+2y) =b

আমরা জানি,

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

\displaystyle =\left{\frac{( 5x-4y) +( 3x+2y)}{2}\right}^{2} -\left{\frac{( 5x-4y) -( 3x+2y)}{2}\right}^{2}

\displaystyle =\left(\frac{5x-4y+3x+2y}{2}\right)^{2} -\left(\frac{5x-4y-3x-2y}{2}\right)^{2}

\displaystyle =\left(\frac{8x-2y}{2}\right)^{2} -\left(\frac{2x-6y}{2}\right)^{2}

\displaystyle =\left{\frac{2( 4x-y)}{2}\right}^{2} -\left{\frac{2( x-3y)}{2}\right}^{2}

\displaystyle =( 4x-y)^{2} -( x-3y)^{2}

গ)

দেওয়া আছে,

\displaystyle Q=7,P=35

তাহলে,

\displaystyle x^{2} -xy+y^{2} =7 ——- (i) নং

আবার,

\displaystyle x^{4} +x^{2} y^{2} +y^{4} =35

\displaystyle \Longrightarrow \left( x^{2}\right)^{2} +2x^{2} y^{2} +\left( y^{2}\right)^{2} -x^{2} y^{2} =35

\displaystyle \Longrightarrow \left( x^{2} +y^{2}\right)^{2} -( xy)^{2} =35

\displaystyle \Longrightarrow \left( x^{2} +y^{2} +xy\right)\left( x^{2} +y^{2} -xy\right) =35

\displaystyle \Longrightarrow \left( x^{2} +xy+y^{2}\right)\left( x^{2} -xy+y^{2}\right) =35

\displaystyle \Longrightarrow \left( x^{2} +xy+y^{2}\right) .7=35

\displaystyle \Longrightarrow \left( x^{2} +xy+y^{2}\right) =\frac{35}{7}

\displaystyle \Longrightarrow \left( x^{2} +xy+y^{2}\right) =5 ——-(ii) নং

(i) ও (ii) নং বিয়োগ করে পাই,

\displaystyle \left( x^{2} -xy+y^{2}\right) -\left( x^{2} +xy+y^{2}\right) =7-5

\displaystyle \Longrightarrow x^{2} -xy+y^{2} -x^{2} -xy-y^{2} =2

\displaystyle \Longrightarrow -2xy=2

\displaystyle \Longrightarrow xy=-\frac{2}{2}

\displaystyle \Longrightarrow xy=-1

(i) ও (ii) নং যোগ করে পাই,

\displaystyle \left( x^{2} -xy+y^{2}\right) +\left( x^{2} +xy+y^{2}\right) =7+5

\displaystyle \Longrightarrow x^{2} -xy+y^{2} +x^{2} +xy+y^{2}) =12

\displaystyle \Longrightarrow 2x^{2} +2y^{2} =12

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

\displaystyle \Longrightarrow x^{2} +y^{2} =\frac{12}{2}

\displaystyle \Longrightarrow x^{2} +y^{2} =6

\displaystyle \Longrightarrow ( x-y)^{2} +2xy=6

\displaystyle \Longrightarrow ( x-y)^{2} +2.( -1) =6

\displaystyle \Longrightarrow ( x-y)^{2} -2=6

\displaystyle \Longrightarrow ( x-y)^{2} =6+2

\displaystyle \Longrightarrow ( x-y)^{2} =8

\displaystyle \Longrightarrow \sqrt{( x-y)^{2}} =\pm \sqrt{8}

\displaystyle \Longrightarrow x-y=\pm \sqrt{4.2}

\displaystyle \Longrightarrow x-y=\pm \sqrt{4} .\sqrt{2}

\displaystyle \Longrightarrow x-y=\pm 2\sqrt{2} ( দেখানো হলো )

5) \displaystyle p+q+r,p^{2} +q^{2} +r^{2} দুইটি বীজগানিতিক রাশি।

ক) ১ম রাশি\displaystyle =0 হলে, প্রমান কর যে, \displaystyle p^{2} +q^{2} +r^{2} =3pqr

খ) ১ম রাশি\displaystyle =10, 2য় রাশি\displaystyle =38 হলে, \displaystyle ( p-q)^{2} +( q-r)^{2} +( r-p)^{2} এর মান নির্নয় কর।

গ) ১ম রাশি\displaystyle =0 হলে, প্রমান কর যে, \displaystyle \frac{( q+r)^{2}}{6qr} +\frac{( r+p)^{2}}{6rp} +\frac{( p+q)^{2}}{6pq} =\frac{1}{2}

ক)

দেওয়া আছে,

১ম রাশি\displaystyle =p+q+r

শর্তমতে,

\displaystyle p+q+r=0

\displaystyle \Longrightarrow p+q=-r

\displaystyle \Longrightarrow ( p+q)^{3} =( -r)^{3} [ উভয় পক্ষ ঘন করে পাই ]

\displaystyle \Longrightarrow p^{3} +q^{3} +3pq( p+q) =-r^{3}

\displaystyle \Longrightarrow p^{3} +q^{3} +3pq( -r) =-r^{3}

\displaystyle \Longrightarrow p^{3} +q^{3} -3pqr=-r^{3}

\displaystyle \Longrightarrow p^{3} +q^{3} +r^{3} =3pqr ( প্রমানিত )

খ)

১ম শর্তমতে,
\displaystyle p+q+r=10

২য় শর্তমতে,
\displaystyle p^{2} +q^{2} +r^{2} =38

এখানে,

\displaystyle p+q+r=10

\displaystyle \Longrightarrow ( p+q+r)^{2} =( 10)^{2} [উভয় পক্ষ বর্গ করে পাই ]

\displaystyle \Longrightarrow p^{2} +q^{2} +r^{2} +2pq+2qr+2rp=100

\displaystyle \Longrightarrow 38+2( pq+qr+rp) =100

\displaystyle \Longrightarrow 2( pq+qr+rp) =100-38

\displaystyle \Longrightarrow 2( pq+qr+rp) =62

প্রদওরাশি\displaystyle =( p-q)^{2} +( q-r)^{2} +( r-p)^{2}

\displaystyle =p^{2} -2pq+q^{2} +q^{2} -2qr+r^{2} +r^{2} -2rp+p^{2}

\displaystyle =2p^{2} +2q^{2} +2r^{2} -2pq-2qr-2rp

\displaystyle =2\left( p^{2} +q^{2} +r^{2}\right) -2( pq+qr+rp)

\displaystyle =2.38-62

\displaystyle =76-62

\displaystyle =14

গ)

দেওয়া আছে,
১ম রাশি\displaystyle =p+q+r

সুতরাং ১ম রাশি\displaystyle =0 হলে,\displaystyle p+q+r=0

\displaystyle \Longrightarrow p+q=-r

\displaystyle \Longrightarrow q+r=-p

\displaystyle \Longrightarrow r+p=-q

এবং (ক) থেকে পাই,, \displaystyle p^{3} +q^{3} +r^{3} =3pqr

এখন, \displaystyle \frac{( p+r)^{2}}{6qr} +\frac{( r+p)^{2}}{6rp} +\frac{( p+q)^{2}}{6pq}

\displaystyle =\frac{( -p)^{2}}{6qr} +\frac{( -q)^{2}}{6rp} +\frac{( -r)^{2}}{6pq}

\displaystyle =\frac{p^{2}}{6qr} +\frac{q^{2}}{6rp} +\frac{r^{2}}{6pq}

\displaystyle =\frac{p^{2} .p+q^{2} .q+r^{2} .r}{6pqr}

\displaystyle =\frac{p^{3} +q^{3} +r^{3}}{6pqr}

\displaystyle =\frac{3pqr}{6pqr} [ ক) হতে পাই, \displaystyle p^{3} +q^{3} +r^{3} =3pqr ]

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

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