Rewrite the square source of the department \frac22516 as the department of square roots \frac\sqrt225\sqrt16. Take it the square source of both numerator and denominator.

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\displaystyle\frac54 Explanation:Remember\displaystyle\left(\sqrt\fracxy=\frac\sqrtx\sqrty\right.\displaystyle\therefore\sqrt\frac2516=\frac\sqrt25\sqrt16=\frac54

mark D. Jun 5, 2018\displaystyle\frac\sqrt25\sqrt128=\frac5\sqrt64\sqrt2=\frac58\sqrt2 You have to multiply top and also bottom by\displaystyle\sqrt2 ...

\displaystyle\frac5144 Explanation: \displaystyle\left(\sqrt25=5\right.\displaystyle\frac\sqrt25144=\frac5144

Tiger was unable to solve based upon your entry 10(sqrt(7/64)) her input is presently not sustained by the Square source Simplifier Primarily, a sqrt leveling drill must start with "sqrt" ...

\displaystyle\sqrt\frac25324=\frac518 Explanation: \displaystyle\sqrt\frac25324 =\displaystyle\sqrt\frac5\times52\times2\times3\times3\times3\times3 ...

Why is systems to inequality \sqrt1 - x - \sqrtx > \frac1\sqrt3 equal to expression <0, \frac3 - \sqrt56)?

https://math.stackexchange.com/questions/1904729/why-is-solution-to-inequality-sqrt1-x-sqrtx-frac1-sqrt3-equ

The number \sqrt1-x-\sqrtx can be an adverse and, in this case, the inequality doesn't hold. Squaring both sides have the right to be done just if both sides are nonnegative. Thus your inequality becomes \begincases 0\le x\le 1 \\<8px> \sqrt1-x-\sqrtx\ge0 \\<4px> 1-x-2\sqrtx(1-x)+x>\dfrac13 \endcases ...

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Rewrite the square root of the division \frac22516 as the department of square roots \frac\sqrt225\sqrt16. Take it the square source of both numerator and denominator.

\left< \beginarray l l 2 & 3 \\ 5 & 4 \endarray \right> \left< \beginarray l l together 2 & 0 & 3 \\ -1 & 1 & 5 \endarray \right>

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