![\sqrt{2 {x}^{3} } + \sqrt[3]{2}{x}^{2} + 6x \div 2x](https://tex.z-dn.net/?f=+%5Csqrt%7B2+%7Bx%7D%5E%7B3%7D+%7D++%2B++%5Csqrt%5B3%5D%7B2%7D%7Bx%7D%5E%7B2%7D+%2B+6x+%5Cdiv+2x "\sqrt{2 {x}^{3} } + \sqrt[3]{2}{x}^{2} + 6x \div 2x")
solve this answer please
Answers
Answered by
1
Hello:
√2x³+∛2x²+6x÷2x;
(√2x³)²+(∛2x²)³+(6x÷2x);
2x³+2x²+3x;
x(2x²+2x+3)
bye :-)
√2x³+∛2x²+6x÷2x;
(√2x³)²+(∛2x²)³+(6x÷2x);
2x³+2x²+3x;
x(2x²+2x+3)
bye :-)
Answered by
1
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](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B2x%5E2%7D%2B%5Csqrt%5B3%5D%7B2x%5E2%7D%2B6x%7D%7B2x%7D%5C%5C%5C%5C%3D%5Cfrac%7B%282x%5E2%29%5E%5Cfrac%7B1%7D%7B2%7D%2B%282x%5E2%29%5E%5Cfrac%7B1%7D%7B3%7D%2B6x%7D%7B2x%7D%5C%5C%5C%5C%3D2%5E%7B%5Cfrac%7B1%7D%7B2%7D-1%7D+%5C%3A+x%5E%7B2%2A%5Cfrac%7B1%7D%7B2%7D-1%7D%2B2%5E%7B%5Cfrac%7B1%7D%7B3%7D-1%7D+%2A+x%5E%7B2%2A%5Cfrac%7B1%7D%7B3%7D-1%7D%2B3+%5C%5C%5C%5C%3D%5Csqrt%7B2%7D%2Ax%5E0%2B%5Cfrac%7B1%7D%7B%5Csqrt%5B3%5D%7B2%5E2+%5C%3A+x%7D%7D%2B3 "\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:
:-)
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