যদি,$x(2x - 5) =\frac{1}{2}$ হয় তবে, প্রমাণ কর যে, $8x^3-\frac{1}{27x^3}=135$

(সংক্ষিপ্ত প্রশ্ন)

দেওয়া আছে, $a^2=5+2\sqrt{6}$

বা, $a^2=3+2\sqrt{6}+2={\left(\sqrt{3}\right)}^2+2\sqrt{3}.\sqrt{2}={\left(\sqrt{3}+\sqrt{2}\right)}^2$

বা, $a =\sqrt{3}+\sqrt{2}$

বা, $\frac{1}{a}= \frac{1}{\sqrt{3}+\sqrt{2}}$

বা, $\frac{1}{a}=\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}=\frac{\sqrt{\mathbf{3}}\mathbf{-}\sqrt{\mathbf{2}}}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2}=\frac{\sqrt{3}-\sqrt{2}}{3-2}$

$\therefore \frac{1}{a}=\sqrt{3}-\sqrt{2.}$

ক-হতে প্রাপ্ত, $a=\sqrt{3}+\sqrt{2}$ এবং $\frac{1}{a}=\sqrt{3}-\sqrt{2}$

এখন, $a + \frac{1}{a}= \left(\sqrt{3}+\sqrt{2}\right) +\left(\sqrt{3}-\sqrt{2}\right)= \sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}=2\sqrt{3}$

$a^2+\frac{1}{a^2}={\left(a+\frac{1}{a}\right)}^2-2 .a.\frac{1}{a}={\left(2\sqrt{3}\right)}^2-2=12-2=10$

$a^3+\frac{1}{a^3}={\left(a+\frac{1}{a}\right)}^3-3 . a.\frac{1}{a} \left(a+\frac{1}{a}\right)$

$={\left(2\sqrt{3}\right)}^3-3\times 2\sqrt{3}=24\sqrt{3}-6\sqrt{3}=18\sqrt{3}$

বামপক্ষ $=\left(\frac{a^{10+1}}{a^5}\right)=a^5+\frac{1}{a^5}$

                                         $= \left(a^2+\frac{1}{a^2}\right) \left(a^3+\frac{1}{a^3}\right)-\left(a+\frac{1}{a}\right)=10\times 18\sqrt{3}-2\sqrt{3}$

                                          $= 180\sqrt{3}-2\sqrt{3}=178\sqrt{3}$ = ডানপক্ষ

$\frac{a^{10+1}}{a^5}=178\sqrt{3}$ (প্রমাণিত)

দেওয়া আছে, $x =\sqrt{3}+\sqrt{2}$

       বা, $\frac{1}{x}=\frac{\sqrt{3}-\sqrt{2}}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2}=\frac{\sqrt{3}-\sqrt{2}}{3-2}=\sqrt{3}-\sqrt{2}$

     $\therefore x+\frac{1}{x}=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}=2\sqrt{3}$

এখন, $P=x^3+\frac{1}{x^3}$

                $=(x{)}^3+{\left(\frac{1}{x}\right)}^3$

                 $={\left(x+\frac{1}{x}\right)}^3-3.x.\frac{1}{x}\left(x+\frac{1}{x}\right)$

                   $={\left(2\sqrt{3}\right)}^3-3\times 2\sqrt{3}$

                   $=24\sqrt{3}-6\sqrt{3}=18\sqrt{3}$

$\therefore P = 18\sqrt{3.}$ (প্রমাণিত)

প্রদত্ত রাশি $=m^3-3m^2+3m-2$

                   $=m^3-3m^2.l+3m.l^2-l^3-1$

                    $=(m-1{)}^3-1$

                      $=(m-1-1)\left\{\right(m-1{)}^2+(m-1)+1\}$

                        $=(m-2)(m^2-2m+1+m-1+1)$

                         $=(m-2)(m^2-m+1).$

দেওয়া আছে , $P=x^2-2\sqrt{30}-11$

$P = 0$ হলে, $x^2-2\sqrt{30}-11=0$

বা, $x^2=11+2\sqrt{30}$

বা, $x^2=6+2\sqrt{5}\times \sqrt{6}+5.={\left(\sqrt{6}\right)}^2+2\sqrt{6}\times \sqrt{5}+{\left(\sqrt{5}\right)}^2$

বা, $x^2={\left(\sqrt{6}+\sqrt{5}\right)}^2$

বা, $x =\sqrt{6}+\sqrt{5}$ [ বর্গমূল করে]

বা, $\frac{1}{x}$= $\frac{1}{\sqrt{6}+\sqrt{5}}$

$=\frac{\sqrt{6}-\sqrt{5}}{\left(\sqrt{6}+\sqrt{5}\right) \left(\sqrt{6}-\sqrt{5}\right)}$ [হর ও লবকে  $\left(\sqrt{6}-\sqrt{5}\right)$ দ্বারা গুণ করে ]

$= \frac{\sqrt{6}-\sqrt{5}}{{\left(\sqrt{6}\right)}^2-{\left(\sqrt{5}\right)}^2}=\frac{\sqrt{6}-\sqrt{5}}{6-5}=\sqrt{6}-\sqrt{5}$

$\therefore x+\frac{1}{x}=\sqrt{6}+\sqrt{5}+\sqrt{6}-\sqrt{5}=2\sqrt{6}$

এবং $x-\frac{1}{x}=\sqrt{6}+\sqrt{5}-\sqrt{6}+\sqrt{5}$ = $2\sqrt{5}$

$\therefore x^3+\frac{1}{x^3}={\left(x+\frac{1}{x}\right)}^3-3.x,\frac{1}{x}\left(x+\frac{1}{x}\right)$

$={\left(2\sqrt{6}\right)}^3-3\times 2\sqrt{6}$

$= 8 \times 6\sqrt{6}-6\sqrt{6}$

$=48\sqrt{6}-6\sqrt{6}=42\sqrt{6}$

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

$2\sqrt{6}\times 2\sqrt{5}=4\sqrt{6}\sqrt{5}$

এখন, $\left(x^2-\frac{1}{x^2}\right),\left(x^3+\frac{1}{x^3}\right)=x^5+\frac{1}{x}-x-\frac{1}{x^3}$

বা, $4\sqrt{6}\sqrt{5}\times 42\sqrt{6}+\left(x-\frac{1}{x}\right)=x^5-\frac{1}{x^5}$

বা, $1008\sqrt{5}+2\sqrt{5}=x^5-\frac{1}{x^5}$

বা, $1010\sqrt{5}=x^5-\frac{1}{x^5}$

$\therefore x^5-\frac{1}{x^5}=1010\sqrt{5}$

নির্ণেয় মান: $1010\sqrt{5}$

দেওয়া আছে, $Q= \left(\log \sqrt{27}-\log 10\sqrt{10}-\log \frac{1}{8}\right)÷\log \sqrt{1.44}$

                     $=\left\{\log {\left(3\right)}^{\frac{3}{2}}-\log {10}^{1+\frac{1}{2}}-\log 2^{-3}\right\}÷log\sqrt{\frac{144}{100}}$

                      $=\left\{\frac{3}{2}\log 3-\log {10}^{\frac{3}{2}}-\left(-3\right)\log 2\right\}÷\log \sqrt{{\left(\frac{10}{12}\right)}^2}$

                       $=\left\{\frac{3}{2}\log 3 -\frac{3}{2}\log 10+2\log 2\right\}÷\log \frac{12}{10}$

                       $=\left\{\frac{3}{2}\left(\log 3-\log 10\right)+\frac{3}{2}\times 2 \log 2 \right\}÷\log \frac{6}{5}$

                         $= \left(\frac{3}{2}\log \frac{3}{10}+\frac{3}{2}\log 4\right) ÷ \log \frac{6}{5}$

                           $=\frac{3}{2}\left\{\log \left(\frac{3}{10}\times 4\right)\right\}÷\log \frac{6}{5}=\frac{3}{2}\log \frac{6}{5}÷\log \frac{6}{5}=\frac{3}{2}$

$Q = \frac{3}{2}.$ (প্রমাণিত)