Question

Answer it. Pls proper explanation required

Answer 1

Answer:

${x}^{1 \div 2} \times {x}^{1 \div 4} \times {x}^{1 \div 4} = k \\ {x}^{1 \div 2 + 1 \div 4 + 1 \div 4 = k } \\ {x}^{1} = k \\ k \div x = 1$

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Answer 2

Solution :-

Some properties before solving :-

xⁿ = (x) × (x) × (x) ........ {n - times}

$x^a \times x^b = x^{(a + b)}$

$x^a \div x^b = x^{(a - b)}$

$(x^m)^n = x^{mn}$

$x^{\frac{1}{n}} = \sqrt[n]{x}$

As given :-

$\sqrt{x \times \sqrt{x} \sqrt{x}} = k$

By simplifying further :-

$\rightarrow \sqrt{x \times \sqrt{x}^2} = k$

$\rightarrow \sqrt{x \times x } = k$

$\rightarrow \sqrt{x^2 } = k$

$\rightarrow x = k$

So

$\dfrac{k}{x} = \dfrac{k}{k} \\\\ = 1$