x2-4x+1=0 এবং a2+1a2=5 দুইটি বীজগাণিতিক সমীকরণ।n ^ 2 + 13n - 48

p4+p2+1=(p2)2+2p2.1+12-p2=(p2+1)2-p2=(p2+1+p)(p2+1-p)=(p2+p+1)(p2-p+1) দেওয়া আছে, a2+1a2=5বা, (a-1a)2+2.a.1a=5বা,(a-1a)2=5-2বা, (a-1a)2=3a-1a=3এখন, a-1a3=(a-1a)3+3.a.1a(a-1a)=33+3.3=33+33=63 দেওয়া আছে, x2-4x+1=0x2+1=4xx2+1x=4x + 1x = 4বামপক্ষ ==26x2+1x2=26 (x+1x

$x^{2} - 4 x + 1 = 0$ এবং $a^{2} + \frac{1}{a^{2}} = 5$ দুইটি বীজগাণিতিক সমীকরণ।n ^ 2 + 13n - 48

Created: 9 months ago | Updated: 9 months ago

Updated: 9 months ago

136

(ক)

উৎপাদকে বিশ্লেষণ কর :

p^{4} + p^{2} + 1

উত্তরঃ

$p^{4} + p^{2} + 1$

$= (p^{2})^{2} + 2 p^{2} . 1 + 1^{2} - p^{2}$

$= (p^{2} + 1)^{2} - p^{2}$

$= (p^{2} + 1 + p) (p^{2} + 1 - p)$

$= (p^{2} + p + 1) (p^{2} - p + 1)$

LivemcQ

9 months ago

(খ)

a^{3} - \frac{1}{a^{3}}

এর মান নির্ণয় কর।

উত্তরঃ

দেওয়া আছে, $a^{2} + \frac{1}{a^{2}} = 5$

বা, $(a - \frac{1}{a})^{2} + 2 . a . \frac{1}{a} = 5$

বা, $(a - \frac{1}{a})^{2} = 5 - 2$

বা, $(a - \frac{1}{a})^{2} = 3$

$a - \frac{1}{a} = \sqrt{3}$

এখন, $a - \frac{1}{a^{3}} = (a - \frac{1}{a})^{3} + 3 . a . \frac{1}{a} (a - \frac{1}{a})$

$= {(\sqrt{3})}^{3} + 3 . \sqrt{3} = 3 \sqrt{3} + 3 \sqrt{3} = 6 \sqrt{3}$

LivemcQ

9 months ago

(গ)

প্রমাণ কর যে,

26 (x^{2} + \frac{1}{x^{2}}) = 7 (x^{3} + \frac{1}{x^{3}})

উত্তরঃ

দেওয়া আছে,

$x^{2} - 4 x + 1 = 0 x^{2} + 1 = 4 x \frac{x^{2} + 1}{x} = 4 x + \frac{1}{x} = 4$

বামপক্ষ =

$= 26 (x^{2} + \frac{1}{x^{2}}) = 26 \{(x + \frac{1}{x})^{2} - 2 . x . \frac{1}{x}\} = 26 (4^{2} - 2) = 26 (16 - 2) = 26 \times 14 = 364$

ডানপক্ষ =

$= 7 (x^{3} + \frac{1}{x^{3}}) = 7 \{(x + \frac{1}{x})^{3} - 3 . x . \frac{1}{x} (x + \frac{1}{x})\} = 7 (4^{3} - 3 \times 4) = 7 (64 - 12) = 7 x 52 = 364$