গণিত প্রশ্ন

উত্তর ও ব্যাখ্যা

1. ক)

   $x^2 + 1 = 5x$। $x \neq 0$ বলে $x$ দ্বারা ভাগ করে, $x + \frac{1}{x} = 5$

   $\left(x - \frac{1}{x}\right)^2 = \left(x + \frac{1}{x}\right)^2 - 4 = 25 - 4 = 21$

   $\therefore x - \frac{1}{x} = \mathbf{\pm\sqrt{21}}$

2. খ)

   দেওয়া আছে, $y^4 = 527 - \frac{1}{y^4} \Rightarrow y^4 + \frac{1}{y^4} = 527$

   $\left(y^2 + \frac{1}{y^2}\right)^2 = 527 + 2 = 529 \Rightarrow y^2 + \frac{1}{y^2} = 23$ ($y > 0$)

   $\left(y + \frac{1}{y}\right)^2 = 23 + 2 = 25 \Rightarrow y + \frac{1}{y} = 5$

   $y^3 + \frac{1}{y^3} = \left(y + \frac{1}{y}\right)^3 - 3\left(y + \frac{1}{y}\right) = 125 - 15 = 110$

   উভয়পক্ষকে $y^3$ দ্বারা গুণ করে, $y^6 + 1 = 110y^3 \Rightarrow y^6 - 110y^3 + 1 = 0$ (দেখানো হলো)

3. গ)

   দেওয়া আছে, $m^2 - 2\sqrt{30} = 11 \Rightarrow m^2 = 11 + 2\sqrt{30} = 6 + 5 + 2\sqrt{6\cdot 5} = (\sqrt{6}+\sqrt{5})^2$

   $m > 0$ বলে $m = \sqrt{6}+\sqrt{5}$ এবং $\frac{1}{m} = \sqrt{6}-\sqrt{5}$।

   $m^2 = 11 + 2\sqrt{30}$, $\frac{1}{m^2} = 11 - 2\sqrt{30}$

   $m^2 + \frac{1}{m^2} = 22$, $m^2 - \frac{1}{m^2} = 4\sqrt{30}$

   $\frac{m^8-1}{m^4} = m^4 - \frac{1}{m^4} = \left(m^2 + \frac{1}{m^2}\right)\left(m^2 - \frac{1}{m^2}\right) = 22 \times 4\sqrt{30} = \mathbf{88\sqrt{30}}$