Question
Show that: $\big(\frac{4\text{m}}{3}−\frac{3\text{n}}{4}\big)^2+2\text{mn}=\frac{16\text{m}^2}{9}+\frac{9\text{n}^2}{16}$
✓
Answer
$\big(\frac{4\text{m}}{3}−\frac{3\text{n}}{4}\big)^2+2\text{mn}=\frac{16\text{m}^2}{9}+\frac{9\text{n}^2}{16}$
$\text{LHS}=\big(\frac{4\text{m}}{3}−\frac{3\text{n}}{4}\big)^2+2\text{mn}$
$=\big(\frac{4\text{m}}{3}−\frac{3\text{n}}{4}\big)^2+2\times \frac{4\text{m}}{3}\times \frac{3\text{n}}{4}$
$=\big(\frac{4\text{m}}{3}\big)^2+\big(\frac{3\text{n}}{4}\big)^2 $
$\big[\because (\text{a}−\text{b}\big)^2+2\text{ab}=\text{a}^2+\text{b}^2\big]$
$=\frac{16\text{m}^2}{9}+\frac{9\text{n}^2}{16}$
$=\text{RHS}$ Because $LHS$ is equal to $RHS,$ the given equation is verified.
$\text{LHS}=\big(\frac{4\text{m}}{3}−\frac{3\text{n}}{4}\big)^2+2\text{mn}$
$=\big(\frac{4\text{m}}{3}−\frac{3\text{n}}{4}\big)^2+2\times \frac{4\text{m}}{3}\times \frac{3\text{n}}{4}$
$=\big(\frac{4\text{m}}{3}\big)^2+\big(\frac{3\text{n}}{4}\big)^2 $
$\big[\because (\text{a}−\text{b}\big)^2+2\text{ab}=\text{a}^2+\text{b}^2\big]$
$=\frac{16\text{m}^2}{9}+\frac{9\text{n}^2}{16}$
$=\text{RHS}$ Because $LHS$ is equal to $RHS,$ the given equation is verified.
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