গণিত প্রশ্ন

উত্তর ও ব্যাখ্যা

1. ক)

   $p^3+3p+36$ এ $p=-3$ বসালে $-27-9+36=0$। সুতরাং $(p+3)$ একটি উৎপাদক।

   $p^3+3p+36 = p^3+3p^2-3p^2-9p+12p+36$
   $= p^2(p+3)-3p(p+3)+12(p+3) = \mathbf{(p+3)(p^2-3p+12)}$

2. খ)

   $m+n=\sqrt6,\ m-n=\sqrt5$

   $m^2+n^2 = \dfrac{(m+n)^2+(m-n)^2}{2} = \dfrac{6+5}{2}=\dfrac{11}{2}$

   $4mn = (m+n)^2-(m-n)^2 = 6-5=1 \Rightarrow mn=\dfrac14$

   সুতরাং $24mn(m^2+n^2) = 24\times\dfrac14\times\dfrac{11}{2} = \mathbf{33}$।

3. গ)

   $x=\sqrt5+2$ হলে $\dfrac1x = \dfrac{1}{\sqrt5+2} = \dfrac{\sqrt5-2}{5-4}=\sqrt5-2$

   $x+\dfrac1x = 2\sqrt5$

   $x^2+\dfrac1{x^2} = (2\sqrt5)^2-2 = 18$

   $x^3+\dfrac1{x^3} = (2\sqrt5)^3-3\times2\sqrt5 = 40\sqrt5-6\sqrt5=34\sqrt5$

   $\left(x^2+\dfrac1{x^2}\right)\left(x^3+\dfrac1{x^3}\right) = x^5+\dfrac1{x^5}+x+\dfrac1x$
   $\Rightarrow x^5+\dfrac1{x^5} = 18\times34\sqrt5-2\sqrt5 = 612\sqrt5-2\sqrt5 = 610\sqrt5$ (প্রমাণিত)।