Class:9-10( অনুশীলনী:৩.১,৩.২,৩.৩,৩.৪- এর বহুনির্বাচনি(2)

Class:9-10_math_img.

1) \displaystyle a+b=\sqrt{7} এবং \displaystyle a-b=\sqrt{3} হলে, \displaystyle ab=কত?
দেওয়া আছে,
\displaystyle a+b=\sqrt{7} এবং ,
\displaystyle a-b=\sqrt{3}
এখন, \displaystyle ab=\left(\frac{a+b}{2}\right)^{2} -\left(\frac{a-b}{2}\right)^{2}
\displaystyle =\left(\frac{\sqrt{7}}{2}\right)^{2} -\left(\frac{\sqrt{3}}{2}\right)^{2}
\displaystyle =\frac{7}{4} -\frac{3}{4}
\displaystyle =\frac{7-3}{4}
\displaystyle =\frac{4}{4}
\displaystyle =1

Class:9-10_math_img.

2) \displaystyle a^{2} -\sqrt{3} a+1=0 হলে, \displaystyle a+\frac{1}{a}এর মান কত?
দেওয়া আছে,,
\displaystyle a^{2} -\sqrt{3} a+1=0
\displaystyle \Longrightarrow a^{2} +1=\sqrt{3} a
\displaystyle \Longrightarrow a\left( a+\frac{1}{a}\right) =\sqrt{3} a
\displaystyle \Longrightarrow \ a+\frac{1}{a} =\sqrt{3}

Class:9-10_math_img.

3) \displaystyle x^{4} +x^{2} +1=0 হলে, \displaystyle x^{2} +\frac{1}{x^{2}} =কত?
দেওয়া আছে,,
\displaystyle x^{4} +x^{2} +1=0
\displaystyle \Longrightarrow x^{4} +1=-x^{2}
\displaystyle \Longrightarrow x^{2}\left( x^{2} +\frac{1}{x^{2}}\right) =-x^{2}
\displaystyle \Longrightarrow \left( x^{2} +\frac{1}{x^{2}}\right) =-1

Class:9-10_math_img.

4) \displaystyle a=\sqrt{3} +\sqrt{2} হলে, \displaystyle a+\frac{1}{a} =কত?
দেওয়া আছে,
\displaystyle a=\sqrt{3} +\sqrt{2}
\displaystyle \Longrightarrow \frac{1}{a} =\frac{1}{\sqrt{3} +\sqrt{2}}
\displaystyle \Longrightarrow \frac{1}{a} =\frac{\left(\sqrt{3} -\sqrt{2}\right)}{\left(\sqrt{3} +\sqrt{2}\right) .\left(\sqrt{3} -\sqrt{2}\right)}

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

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

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

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

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

Class:9-10_math_img.

5) \displaystyle a+\frac{1}{a} =5 হলে, \displaystyle \frac{6a}{a^{2} +a+1} =কত?
দেওয়া আছে,
\displaystyle a+\frac{1}{a} =5
\displaystyle \Longrightarrow \frac{a^{2} +1}{a} =5
\displaystyle \Longrightarrow a^{2} +1=5a

এখন,
\displaystyle \frac{6a}{a^{2} +a+1}

\displaystyle =\frac{6a}{5a+a}

\displaystyle =\frac{6a}{6a}
\displaystyle =1

Class:9-10_math_img.

6) \displaystyle x=7+4\sqrt{3} হলে,, \displaystyle x^{2}এর মান কত?
দেওয়া আছে,,
\displaystyle x=7+4\sqrt{3}
এখন,
\displaystyle x^{2} =\left( 7+4\sqrt{3}\right)^{2}
\displaystyle =( 7)^{2} +2.7.4\sqrt{3} +\left(\sqrt{3}\right)^{2}
\displaystyle =49+56\sqrt{3} +3
\displaystyle =52+56\sqrt{3}

Class:9-10_math_img.

7) \displaystyle x^{4} -x^{2} +1=0 হলে, \displaystyle \left( x+\frac{1}{x}\right)^{2} =কত?
দেওয়া আছে,
\displaystyle \Longrightarrow x^{4} -x^{2} +1=0
\displaystyle \Longrightarrow x^{4} +1=x^{2}
\displaystyle \Longrightarrow x^{2}\left( x^{2} +\frac{1}{x^{2}}\right) =x^{2}
\displaystyle \Longrightarrow x^{2} +\frac{1}{x^{2}} =1
\displaystyle \Longrightarrow \left( x+\frac{1}{x}\right)^{2} -2.x.\frac{1}{x} =1
\displaystyle \Longrightarrow \left( x+\frac{1}{x}\right)^{2} -2=1
\displaystyle \Longrightarrow \left( x+\frac{1}{x}\right)^{2} =2+1
\displaystyle \Longrightarrow \left( x+\frac{1}{x}\right)^{2} =3

Class:9-10_math_img.

8) \displaystyle 2x+\frac{2}{ও} =3 হলে, \displaystyle x^{2} +\frac{1}{x^{2}}এর মান কত?
দেওয়া আছে,
\displaystyle 2x+\frac{2}{x} =3
\displaystyle \Longrightarrow 2\left( x+\frac{1}{x}\right) =3
\displaystyle \Longrightarrow \left( x+\frac{1}{x}\right) =\frac{3}{2}
এথন,
\displaystyle x^{2} +\frac{1}{x^{2}}
\displaystyle =\left( x+\frac{1}{x}\right)^{2} -2.x.\frac{1}{x}
\displaystyle =\left(\frac{3}{2}\right)^{2} -2
\displaystyle =\frac{9}{4} -2
\displaystyle =\frac{9-8}{4}
\displaystyle =\frac{1}{4}

Class:9-10_math_img.

9) \displaystyle a^{2} +b^{2} +c^{2} +2ab-2bc-2ca=কত?
এখন,
\displaystyle a^{2} +b^{2} +c^{2} +2ab-2bc-2ca
\displaystyle =( -a)^{2} +( -b)^{2} +c^{2} +2( -a)( -b) +2( -b) .c+2c.( -a)
\displaystyle =( -a-b+c)^{2}
\displaystyle =( c-a-b)^{2}

Class:9-10_math_img.

10) \displaystyle a+b=5 এবং \displaystyle ab=0 হলে, \displaystyle ( a-b)^{2} =?
দেওয়া আছে,,
\displaystyle a+b=5
এবং, \displaystyle ab=0
এখন,
\displaystyle ( a-b)^{2} =( a+b)^{2} -4ab
\displaystyle =( 5)^{2} -4.0
\displaystyle =25

Class:9-10_math_img

11) \displaystyle x+\frac{1}{x} =10 হলে, \displaystyle \sqrt{x} -\frac{1}{\sqrt{x}}এর মান কত?
দেওয়া আছে,,
\displaystyle x+\frac{1}{x} =10
এখন,
\displaystyle \left(\sqrt{x} -\frac{1}{\sqrt{x}}\right)^{2} =\left(\sqrt{x}\right)^{2} -2.\sqrt{x} .\frac{1}{\sqrt{x}} +\left(\frac{1}{\sqrt{x}}\right)^{2}
\displaystyle \Longrightarrow \left(\sqrt{x} -\frac{1}{\sqrt{x}}\right)^{2} =x+\frac{1}{x} -2
\displaystyle \Longrightarrow \left(\sqrt{x} -\frac{1}{\sqrt{x}}\right)^{2} =10-2
\displaystyle \Longrightarrow \left(\sqrt{x} -\frac{1}{\sqrt{x}}\right)^{2} =8
\displaystyle \Longrightarrow \sqrt{\left(\sqrt{x} -\frac{1}{\sqrt{x}}\right)^{2}} =\pm \sqrt{8}
\displaystyle \Longrightarrow \sqrt{x} -\frac{1}{\sqrt{x}} =\pm \sqrt{4.2}
\displaystyle \Longrightarrow \sqrt{x} -\frac{1}{\sqrt{x}} =\pm 2\sqrt{2}

Class:9-10_math_img.

12) \displaystyle x=\sqrt{7} -\sqrt{6} হয়, তবে \displaystyle \frac{1}{x} =কত?
দেওয়া আছে,
\displaystyle x=\sqrt{7} -\sqrt{6}

\displaystyle \Longrightarrow \frac{1}{x} =\frac{1}{\sqrt{7} -\sqrt{6}}
\displaystyle \Longrightarrow \frac{1}{x} =\frac{\left(\sqrt{7} +\sqrt{6}\right)}{\left(\sqrt{7} -\sqrt{6}\right)\left(\sqrt{7} +\sqrt{6}\right)}

\displaystyle \Longrightarrow \frac{1}{x} =\frac{\left(\sqrt{7} +\sqrt{6}\right)}{\left(\sqrt{7}\right)^{2} -\left(\sqrt{6}\right)^{2}}

\displaystyle \Longrightarrow \frac{1}{x} =\frac{\left(\sqrt{7} +\sqrt{6}\right)}{7-6}
\displaystyle \frac{1}{x} =\sqrt{7} +\sqrt{6}

Class:9-10_math_img.

13) \displaystyle x+y=\sqrt{5} এবং, \displaystyle x-y=\sqrt{3} হলে \displaystyle xy=কত?
দেওয়া আছে,
\displaystyle x+y=\sqrt{5}
এবং, \displaystyle x-y=\sqrt{3}
এখন,
\displaystyle xy=\left(\frac{x+y}{2}\right)^{2} -\left(\frac{x-y}{2}\right)^{2}

\displaystyle =\left(\frac{\sqrt{5}}{2}\right)^{2} -\left(\frac{\sqrt{3}}{2}\right)^{2}

\displaystyle =\frac{5}{4} -\frac{3}{4}

\displaystyle =\frac{5-3}{4}

\displaystyle =\frac{2}{4}

=\displaystyle \frac{1}{2} =0.5

Class:9-10_math_img.

14) \displaystyle x^{4} -x^{2} +1=0 হলে, \displaystyle x^{2} +\frac{1}{x^{2}} এর মান নির্নয় কর।
দেওয়া আছে,,
\displaystyle x^{4} -x^{2} +1=0
\displaystyle \Longrightarrow x^{4} +1=x^{2}
\displaystyle \Longrightarrow x^{2}\left( x^{2} +\frac{1}{x^{2}}\right) =x^{2}
\displaystyle =x^{2} +\frac{1}{x^{2}} =1

Class:9-10_math_img.

15) \displaystyle 2x+\frac{2}{x} =3 হলে, \displaystyle 4\left( x^{2} +\frac{1}{x^{2}}\right) =কত?
দেওয়া আছে,,
\displaystyle 2x+\frac{2}{x} =3
এখন,
\displaystyle 4\left( x^{2} +\frac{1}{x^{2}}\right)
\displaystyle =4x^{2} +\frac{4}{x^{2}}
\displaystyle =\left( 2x+\frac{2}{x}\right)^{2} -2.2x.\frac{2}{x}
\displaystyle =( 3)^{2} -2.2.2
\displaystyle =9-8
\displaystyle =1

Class:9-10_math_img

16) \displaystyle a+\frac{1}{a} =2 হলে, \displaystyle \sqrt{a} +\frac{1}{\sqrt{a}} এর মান কত?
দেওয়া আছে,
\displaystyle a+\frac{1}{a} =2
এখন,
\displaystyle \left(\sqrt{a} +\frac{1}{\sqrt{a}}\right)^{2} =\left(\sqrt{a}\right)^{2} +2.\sqrt{a} .\frac{1}{\sqrt{a}} +\left(\frac{1}{\sqrt{a}}\right)^{2}

\displaystyle \Longrightarrow \left(\sqrt{a} +\frac{1}{\sqrt{a}}\right)^{2} =a+\frac{1}{a} +2
\displaystyle \Longrightarrow \left(\sqrt{a} +\frac{1}{\sqrt{a}}\right)^{2} =2+2
\displaystyle \Longrightarrow \left(\sqrt{a} +\frac{1}{\sqrt{a}}\right)^{2} =4
\displaystyle \Longrightarrow \sqrt{\left(\sqrt{a} +\frac{1}{\sqrt{a}}\right)^{2}} =\pm \sqrt{4}
\displaystyle \Longrightarrow \sqrt{a} +\frac{1}{\sqrt{a}} =\pm 2

Class:9-10_math_img.

17) \displaystyle y=\sqrt{6} +\sqrt{5} হলে, \displaystyle \frac{1}{y} এর মান কত?
দেওয়া আছে,
\displaystyle y=\sqrt{6} +\sqrt{5}
\displaystyle \Longrightarrow \frac{1}{y} =\frac{1}{\sqrt{6} +\sqrt{5}}
\displaystyle \Longrightarrow \frac{1}{y} =\frac{(\sqrt{6} -\sqrt{5)}}{\left(\sqrt{6} +\sqrt{5}\right)\left(\sqrt{6} -\sqrt{5}\right)}
\displaystyle \Longrightarrow \frac{1}{y} =\frac{(\sqrt{6} -\sqrt{5)}}{\left(\sqrt{6}\right)^{2} -\left(\sqrt{5}\right)^{2}}
\displaystyle \Longrightarrow \frac{1}{y} =\frac{(\sqrt{6} -\sqrt{5)}}{6-5}
\displaystyle \frac{1}{y} =\sqrt{6} -\sqrt{5}

Class:9-10_math_img.

18) \displaystyle x+y=4 এবং \displaystyle xy=1 হলে \displaystyle ( x-y)এর মান নির্নয় কর।
দেওয়া আছে,,
\displaystyle x+y=4 এবং \displaystyle xy=1
আমরা জানি,
\displaystyle ( x-y)^{2} =( x+y)^{2} -4xy
\displaystyle \Longrightarrow ( x-y)^{2} =( 4)^{2} -4.1
\displaystyle \Longrightarrow ( x-y)^{2} =16-4
\displaystyle \Longrightarrow ( x-y)^{2} =12
\displaystyle \Longrightarrow \sqrt{( x-y)^{2}} =\pm \sqrt{4.3}
\displaystyle \Longrightarrow x-y=\pm 2\sqrt{3}

Class:9-10_math_img.

20) \displaystyle a+b=7 এবং \displaystyle ab=12 হলে, \displaystyle a-b=কত?
দেওয়া আছে,,
\displaystyle a+b=7 এবং \displaystyle ab=12
আমরা জানি,
\displaystyle ( a-b\ )^{2} =( a+b)^{2} -4ab
\displaystyle =( 7)^{2} -4.12
\displaystyle =49-48
\displaystyle =1

Post Author: showrob

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