যদি $A=\frac{\left(2^{x+4}\right)-4.2^{x+1}}{2^{x+2}+ 2},$ $B=\frac{2^{2p+1} . 3^{2p+q} . 5^{p-q} . 6^p}{3^{p-2 }. 6^{2p-2}. {10}^p.{15}^q}$ এবং $C=\frac{5^{m+1}}{{\left(5^m\right)}^{m-1}}+\frac{{25}^{m+1}}{{\left(5^{m-1}\right)}^{m+1}}$হয় তবে-

দেওয়া আছে,  $x^{x\sqrt{x}}= {\left(x\sqrt{x}\right)}^x$

বা, $x^{x\sqrt{x}}= {\left(x^{1+\frac{1}{2}}\right)}^x$

বা, $x^{x\sqrt{x}}= x^{\frac{3}{2}x}$

বা, $x\sqrt{x}= \frac{3}{2}x$

বা, $\sqrt{x}= \frac{3}{2}$

বা, ${\left(\sqrt{x}\right)}^2= {\left(\frac{3}{2}\right)}^2$

$\therefore$ $x = \frac{9}{4}$

দেওয়া আছে,

$A=\frac{2^{n+4}-4.2^{n+1}}{2^{n+2} ÷ 2}$

$=\frac{2^n.2^4- 2^2.2^n.2^1}{2^{n+2-1}}=\frac{2^n.2^4-2^{2+1}.2^n}{2^{n+1}}$

$=\frac{2^n\left(2^4-2^3\right)}{2^{n+1}}=\frac{2^n\left(16-8\right)}{2^n.2}=\frac{8}{2}=4$

এবং $C=\frac{5^{m+1}}{{\left(5^m\right)}^{m-1}}÷ \frac{{25}^{m+1}}{{\left(5^{m-1}\right)}^{m+1}}=\frac{5^{m+1}}{S^{m^2-m}}÷\frac{5^{m+1}}{5^{m-1}}$

$=5^{m+1-m^2+m} ÷ 5^{2m+2-m^2+1}$

$=5^{2m-m^2+1-2m+m^2-3}=5^{-2}=\frac{1}{5^2}=\frac{1}{25}$

প্রদত্ত রাশি $= A÷\sqrt{C}= 4 ÷ \sqrt{\frac{1}{25}}= 4 ÷ \frac{1}{5}= 4 \times \frac{5}{1} = 20$

নির্ণেয় মান : 20

দেওয়া আছে,

$B=\frac{2^{2p+1}. 3^{2p+q}.5^{p+q}. 6^p}{3^{p-2}.6^{2p-2} .{10}^p.{15}^q}$

= $\frac{2^{2p+1},3^{2p+q}. 5^{P+q}. {(2 \times 3)}^p}{3^{P-2}. {(2 \times 3)}^{2p+2}. {(5 \times 2)}^p. {(3\times 5)}^q}$

$=\frac{2^{2p+1}.3^{2p+q}.5^{p+q}.2^p.3^p}{3^{p-2 }.2^{2p+2}.3^{2p+2}.5^p.2^p.3^q.5^q}$

$=2^{2p+1+p-2p-2-p}. 3^{2p+q+p-p+2-2p-2-q}. 5^{P+4-P-4}$

$=2^{-1}\times 3^0\times 5^0=\frac{1}{2}\times 1=\frac{1}{2}$

বামপক্ষ $=\frac{A}{8}- B + 25C$

$= \frac{4}{8}-\frac{1}{2}+25\times \frac{1}{25}$

= $\frac{1}{2}-\frac{1}{2}$+ 1 = 1 = ডানপক্ষ

$\therefore$ $\frac{A}{8}- B + 25C$ = 1 (প্রমাণিত)