Math, asked by raghuveer3, 1 year ago


 \sqrt{2 {x}^{2} } +  \sqrt[3]{ {2x}^{2} }  + 6x \div 2x

Answers

Answered by kvnmurty
3
The question is not very clear. Perhaps 2x divides the entire expression, all terms.  So we need to simplify the terms.

\frac{\sqrt{2x^2}+\sqrt[3]{2x^2}+6x}{2x}\\\\=\frac{(2x^2)^\frac{1}{2}+(2x^2)^\frac{1}{3}+6x}{2x}\\\\=2^{\frac{1}{2}-1} \: x^{2*\frac{1}{2}-1}+2^{\frac{1}{3}-1} * x^{2*\frac{1}{3}-1}+3 \\\\=\sqrt{2}*x^0+\frac{1}{\sqrt[3]{2^2 \: x}}+3



kvnmurty: :-)
Answered by patel25
0
Answer is in the image
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