Question

pls answer it's urgent​

Answer 1

Question:-

$\dfrac{ (\sqrt{x})^{ \frac{3}{5} } \times ( \sqrt{x})^ \frac{2}{5} + 2 \sqrt{x} }{2( \sqrt{x})^ \frac{2}{3} \times {( \sqrt{x} )}^{ \frac{1}{3} } + \sqrt{x} }$

Let us solve numerator first then the denominator,

Numerator:-

$( \sqrt{x})^{ \frac{3}{5} } \times {( \sqrt{x}) }^{ \frac{2}{5} } + 2 \sqrt{x}$

Applying BODMAS rule - Brackets of Division Multiplication Addition Subtraction.

This is the order in which operations are performed

${a}^{m} \times {a}^{n} = {a}^{m + n}$

${( \sqrt{x} )}^{ \frac{3}{5} + \frac{2}{5} } + 2 \sqrt{x}$

${( \sqrt{x}) }^{ \frac{5}{5} } + 2 \sqrt{x}$

$\sqrt{x} + 2 \sqrt{x} = 3 \sqrt{x}$

Denominator:-

$2{( \sqrt{x}) }^{ \frac{2}{3} } \times {( \sqrt{x}) }^{ \frac{1}{3} } + \sqrt{x}$

$2 \times {( \sqrt{x}) }^{ \frac{2}{3} } \times {( \sqrt{x}) }^{ \frac{1}{3} } + \sqrt{x}$

Applying the exponent law again,

$2 \times {( \sqrt{x}) }^{ \frac{2}{3} + \frac{1}{3} } + \sqrt{x}$

$2 \times ( \sqrt{x})^{ \frac{3}{3} } + \sqrt{x}$

$2 \sqrt{x} + \sqrt{x} = 3 \sqrt{x}$

Hence now substitute these in the numerator and denominator,

$\dfrac{3 \sqrt{x} }{3 \sqrt{x} } = 1$

Option (2) 1 is correct

Hope it helps!

Happy day!

Answer 2

$\sf \Large Solution!!$

Before solving this question, you should know the product rule and the quotient rule of exponents.

Product rule of exponents → $\sf m^{a}\times m^{b}=m^{a+b}$

Quotient rule of exponents → $\sf \dfrac{m^{a}}{m^{b}}=m^{a-b}$

$\sf \to \dfrac{(\sqrt{x})^{\frac{3}{5}}\times (\sqrt{x})^{\frac{2}{5}}+2\sqrt{x}}{2(\sqrt{x})^{\frac{2}{3}}\times (\sqrt{x})^{\frac{1}{3}}+\sqrt{x}}$

$\sf \to \dfrac{(\sqrt{x})^{\frac{3}{5}+\frac{2}{5}}+2\sqrt{x}}{2(\sqrt{x})^{\frac{2}{3}+\frac{1}{3}}+\sqrt{x}}$

$\sf \to \dfrac{(\sqrt{x})^{\frac{3+2}{5}}+2\sqrt{x}}{2(\sqrt{x})^{\frac{2+1}{3}}+\sqrt{x}}$

$\sf \to \dfrac{(\sqrt{x})^{\frac{\cancel{5}}{\cancel{5}}}+2\sqrt{x}}{2(\sqrt{x})^{\frac{\cancel{3}}{\cancel{3}}}+\sqrt{x}}$

$\sf \to \dfrac{\sqrt{x}+2\sqrt{x}}{2\sqrt{x}+\sqrt{x}}$

$\sf \to 1$

We have to select 2 alternatives. We have the answer 1. We know that if a base has a power 0, the answer is 1. Paying attention to the first option, we observe that the whole power is 0.

(√x)⁰ = 1

$\sf \large Correct\:options:-$

$\blue{\sf \to (1)\:(\sqrt{x})^{0}}$

$\blue{\sf \to (2)\:1}$