May 2026

Proof of Given Expression = 28√2 Question \[ \frac{3^{-3} \times 6^2 \times \sqrt{98}} {5^2 \times \sqrt[3]{\frac{1}{25}} \times 15^{-4/3} \times 3^{1/3}} \] Solution \[ 6^2 = 2^2 \times 3^2,\quad \sqrt{98} = 7\sqrt{2},\quad 15 = 3 \times 5 \] \[ \sqrt[3]{\frac{1}{25}} = \frac{1}{5^{2/3}} \] \[ = \frac{3^{-3} \times 2^2 \times 3^2 \times 7\sqrt{2}} {5^2 \times \frac{1}{5^{2/3}} \times […]

Proof Prove: \[ \frac{(0.6)^0 – (0.1)^{-1}}{(3/8)^{-1}(3/2)^3 – (1/3)^{-1}} = -\frac{3}{2} \] Solution \[ = \frac{1 – 10}{\frac{8}{3}\cdot \frac{27}{8} – 3} \] \[ = \frac{-9}{9 – 3} \] \[ = \frac{-9}{6} \] \[ = -\frac{3}{2} \] Hence Proved Next Question / Full Exercise

Proof Prove: \[ \left(\frac{64}{125}\right)^{-2/3} + \frac{1}{\left(\frac{256}{625}\right)^{1/4}} + \left(\frac{\sqrt{25}}{\sqrt[3]{64}}\right)^0 = \frac{61}{16} \] Proof \[ \left(\frac{64}{125}\right)^{-2/3} = \left(\frac{4^3}{5^3}\right)^{-2/3} = \left(\frac{4}{5}\right)^{-2} = \left(\frac{5}{4}\right)^2 = \frac{25}{16} \] \[ \frac{1}{\left(\frac{256}{625}\right)^{1/4}} = \frac{1}{\left(\frac{4^4}{5^4}\right)^{1/4}} = \frac{1}{\frac{4}{5}} = \frac{5}{4} \] \[ \left(\frac{\sqrt{25}}{\sqrt[3]{64}}\right)^0 = 1 \] \[ = \frac{25}{16} + \frac{5}{4} + 1 \] \[ = \frac{25}{16} + \frac{20}{16} + \frac{16}{16} \] \[ =

Proof Prove: \[ \frac{2^n + 2^{n-1}}{2^{n+1} – 2^n} = \frac{3}{2} \] Proof \[ = \frac{2^n + 2^{n-1}}{2^{n+1} – 2^n} \] \[ = \frac{2^{n-1}(2 + 1)}{2^n(2 – 1)} \] \[ = \frac{2^{n-1} \cdot 3}{2^n \cdot 1} \] \[ = \frac{3}{2} \] Hence Proved Next Question / Full Exercise

Proof Prove: \[ \sqrt{\frac{1}{4}} + (0.01)^{-1/2} – 27^{2/3} = \frac{3}{2} \] Solution \[ \sqrt{\frac{1}{4}} = \frac{1}{2} \] \[ 0.01 = 10^{-2} \] \[ (0.01)^{-1/2} = (10^{-2})^{-1/2} = 10^1 = 10 \] \[ 27^{2/3} = (3^3)^{2/3} = 3^2 = 9 \] \[ = \frac{1}{2} + 10 – 9 \] \[ = \frac{1}{2} + 1 \] \[

Proof of exponential identity equals 10 Prove: \[ \frac{2^{1/2}\cdot 3^{1/2}\cdot 4^{1/4}}{10^{-1/5}\cdot 5^{3/5}} \div \frac{3^{4/3}\cdot 5^{-7/5}}{4^{-3/5}\cdot 6} = 10 \] Solution \[ = \frac{2^{1/2}\cdot 3^{1/2}\cdot (2^2)^{1/4}}{(2\cdot5)^{-1/5}\cdot 5^{3/5}} \div \frac{3^{4/3}\cdot 5^{-7/5}}{(2^2)^{-3/5}\cdot 2\cdot 3} \] \[ = \frac{2^{1/2}\cdot 3^{1/2}\cdot 2^{1/2}}{2^{-1/5}\cdot 5^{2/5}} \times \frac{2^{-6/5}\cdot 2\cdot 3}{3^{4/3}\cdot 5^{-7/5}} \] \[ = 2^{1+1/5-6/5+1} \cdot 3^{1/2+1-4/3} \cdot 5^{-2/5+7/5} \] \[ = 2^1

Proof Prove: \[ \left(\frac{1}{4}\right)^{-2} – 3\times 8^{2/3} \times 4^0 + \left(\frac{9}{16}\right)^{-1/2} = \frac{16}{3} \] Solution \[ \left(\frac{1}{4}\right)^{-2} – 3\times 8^{2/3} \times 4^0 + \left(\frac{9}{16}\right)^{-1/2} \] \[ = 4^2 – 3\times (2^3)^{2/3} \times 1 + \left(\frac{3^2}{4^2}\right)^{-1/2} \] \[ = 16 – 3\times 2^2 + \left(\frac{3}{4}\right)^{-1} \] \[ = 16 – 12 + \frac{4}{3} \] \[ =

Proof Prove: \[ 9^{3/2} – 3\times 5^0 – (1/81)^{-1/2} = 15 \] Solution \[ 9^{3/2} = (3^2)^{3/2} = 3^3 = 27 \] \[ 5^0 = 1 \] \[ (1/81)^{-1/2} = (3^{-4})^{-1/2} = 3^2 = 9 \] \[ = 27 – 3\times 1 – 9 \] \[ = 27 – 3 – 9 \] \[ =

Proof Prove: \[ \left(\frac{\sqrt{3\cdot 5^{-3}}}{\sqrt[3]{3^{-1}\cdot 5}}\right)\times \sqrt[6]{3\cdot 5^6} = \frac{3}{5} \] Proof \[ = \frac{(3\cdot 5^{-3})^{1/2}}{(3^{-1}\cdot 5)^{1/3}} \times (3\cdot 5^6)^{1/6} \] \[ = \frac{3^{1/2} \cdot 5^{-3/2}}{3^{-1/3}\cdot 5^{1/3}} \times 3^{1/6}\cdot 5 \] \[ = 3^{1/2 + 1/3 + 1/6} \cdot 5^{-3/2 – 1/3 + 1} \] \[ = 3^{\frac{3+2+1}{6}} \cdot 5^{\frac{-9-2+6}{6}} \] \[ = 3^1 \cdot

Simplify (√2/√3)^5 (6/7)^2 Simplify: \(\left(\frac{\sqrt{2}}{\sqrt{3}}\right)^5 \left(\frac{6}{7}\right)^2\) Solution \[ \left(\frac{\sqrt{2}}{\sqrt{3}}\right)^5 \left(\frac{6}{7}\right)^2 \] \[ = \frac{(\sqrt{2})^5}{(\sqrt{3})^5} \cdot \frac{6^2}{7^2} \] \[ = \frac{2^{5/2}}{3^{5/2}} \cdot \frac{36}{49} \] \[ = \frac{4\sqrt{2}}{9\sqrt{3}} \cdot \frac{36}{49} \] \[ = \frac{144\sqrt{2}}{441\sqrt{3}} \] \[ = \frac{48\sqrt{2}}{147\sqrt{3}} \] \[ = \frac{48\sqrt{6}}{441} \] Final Answer: \[ \boxed{\frac{48\sqrt{6}}{441}} \] Next Question / Full Exercise