Educational

Proof of Given Expression = 1 Question \[ \left[ \left(\frac{x^{a(a-b)}}{x^{a(a+b)}}\right) \div \left(\frac{x^{b(b-a)}}{x^{b(b+a)}}\right) \right]^{a+b} \] Solution \[ = \left[ \frac{x^{a(a-b)-a(a+b)}} {x^{b(b-a)-b(b+a)}} \right]^{a+b} \] \[ = \left[ \frac{x^{-2ab}}{x^{-2ab}} \right]^{a+b} \] \[ = (1)^{a+b} \] \[ = 1 \] Answer \[ \boxed{1} \] Next Question / Full Exercise

Proof of 1/(1+x^(a-b)) + 1/(1+x^(b-a)) = 1 Question \[ \frac{1}{1+x^{a-b}} + \frac{1}{1+x^{b-a}} \] Solution \[ = \frac{1}{1+\frac{x^a}{x^b}} + \frac{1}{1+\frac{x^b}{x^a}} \] \[ = \frac{1}{\frac{x^b + x^a}{x^b}} + \frac{1}{\frac{x^a + x^b}{x^a}} \] \[ = \frac{x^b}{x^a + x^b} + \frac{x^a}{x^a + x^b} \] \[ = \frac{x^b + x^a}{x^a + x^b} \] \[ = 1 \] Answer \[ \boxed{1}

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