Ravi Kant Kumar

If x, y, a, b are positive real numbers, prove that : 1/(1+x^{a-b}) + 1/(1+x^{b-a}) = 1

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} […]

If x, y, a, b are positive real numbers, prove that : 1/(1+x^{a-b}) + 1/(1+x^{b-a}) = 1 Read More »

Prove that : (3^-3 × 6^2 × √98)/(5^2 × 3√1/25 × (15)^-4/3 × 3^1/3) = 28√2

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

Prove that : (3^-3 × 6^2 × √98)/(5^2 × 3√1/25 × (15)^-4/3 × 3^1/3) = 28√2 Read More »

Prove that : (64/125)^-2/3 + 1/(256/625)^1/4 + (√25/3√64)^0 = 61/16

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

Prove that : (64/125)^-2/3 + 1/(256/625)^1/4 + (√25/3√64)^0 = 61/16 Read More »

Prove that : (2^1/2 × 3^1/2 × 4^1/4/10^-1/5 × 5^3/5)/(3^4/3 × 5^-7/5/4^-3/5 × 6) = 10

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

Prove that : (2^1/2 × 3^1/2 × 4^1/4/10^-1/5 × 5^3/5)/(3^4/3 × 5^-7/5/4^-3/5 × 6) = 10 Read More »

Prove that : (1/4)^-2 – 3 × 8^2/3 × 4^0 + (9/16)^-1/2 = 16/3

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

Prove that : (1/4)^-2 – 3 × 8^2/3 × 4^0 + (9/16)^-1/2 = 16/3 Read More »

Prove that : (√(3×5^-3)/(cube root √3^(-1)√5))× 6th root √(3×5^6) = 3/5

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

Prove that : (√(3×5^-3)/(cube root √3^(-1)√5))× 6th root √(3×5^6) = 3/5 Read More »