Question

9
14. Show that (4/3m-3/4n)^2+2mn=16/9m^2+9/16n^2​

Answer 1

AnswEr :

☢ In the given question, we have to show that both sides of given equation are equal.

It simply means that we have to prove that L.H.S = R.H.S .

$\qquad$ ━━━━━━━━━━━━━━━━━━━━━

$\underline{\dag \: {\mathbf{ Taking \: L.H.S. \: of \: question}}}$

$\longrightarrow\displaystyle\sf \left( \frac{4}{3}m - \frac{3}{4}n \right) ^{2} + 2mn$

$\quad\qquad\qquad$$\footnotesize{\mathsf{\red{\bigstar \: {(Using \: identity \: : \: (a-b)^{2} = a^{2} + b^{2} - 2ab) }}}}$

$\longrightarrow\displaystyle\sf \left( \frac{16}{9}m^{2} - \frac{9}{16}n^{2} - 2 ( \frac{\cancel{3}}{\cancel{4}}m ) ( \frac{\cancel{4}}{\cancel{3}}n ) \right) + 2mn$

$\longrightarrow\displaystyle\sf \frac{16}{9}m^{2} - \frac{9}{16}n ^{2} - \cancel{2mn} + \cancel{2mn}$

$\longrightarrow\displaystyle\sf \frac{16}{9}m^{2} - \frac{9}{16}n ^{2}$

$\longrightarrow\bf\ R.H.S$

$\therefore\:\underline{\textsf{It \: implies \: that \textbf{L.H.S = R.H.S}}}$

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$\qquad$ $\dag\:\large{\underline{\mathfrak{Algebric \: Identities:}}}$

$\boxed{\begin{minipage}{7 cm}\boxed{\bigstar\:\:\textbf{\textsf{Algebric\:Identity}}\:\bigstar}\\\\1)\bf\:(A+B)^{2} = A^{2} + 2AB + B^{2}\\\\2)\sf\: (A-B)^{2} = A^{2} - 2AB + B^{2}\\\\3)\bf\: A^{2} - B^{2} = (A+B)(A-B)\\\\4)\sf\: (A+B)^{2} = (A-B)^{2} + 4AB\\\\5)\bf\: (A-B)^{2} = (A+B)^{2} - 4AB\\\\6)\sf\: (A+B)^{3} = A^{3} + 3AB(A+B) + B^{3}\\\\7)\bf\:(A-B)^{3} = A^{3} - 3AB(A-B) + B^{3}\\\\8)\sf\: A^{3} + B^{3} = (A+B)(A^{2} - AB + B^{2})\\\\\end{minipage}}$