Question Number 198511 by Hridiana last updated on 21/Oct/23
$$\mathrm{3}+{xy}\sqrt{\frac{{z}}{{xy}}} \\ $$$${xz}+{zx} \\ $$$${how}\:{to}\:{solve} \\ $$$$\mathrm{3}+\pi\left\{\frac{{x}}{\mathrm{2}}>\mathrm{0}\right\}=\left\{\frac{{zy}}{{x}+{z}}>{x}\right\} \\ $$$${operations} \\ $$$$\left\{{x}^{\mathrm{2}+\frac{{x}}{\mathrm{2}}} >\mathrm{0}\right\}\:{is}\:{the}\:{comparator} \\ $$
Commented by HomeAlone last updated on 21/Oct/23
$$\mathrm{5}{x}\:{comparator}\:{based}\:{by}\:{a}\:{friction} \\ $$
Answered by HomeAlone last updated on 21/Oct/23
$$\mathrm{4}+{x} \\ $$$${x}=\mathrm{1} \\ $$$$\sqrt{\mathrm{148}×\frac{\mathrm{10}}{\mathrm{300}+\sqrt{\mathrm{9}}}} \\ $$$$\mathrm{2}.\mathrm{210088}+\mathrm{0}.\mathrm{0081991} \\ $$$$\mathrm{2}.\mathrm{218287} \\ $$