(a) Find the value of the expression \[\left( 81{{x}^{2}}+16{{y}^{2}}-72xy \right),\] when \[x=\text{ }\frac{2}{3}\]and \[y=\text{ }\frac{3}{4}\] |
(b) If a = 2 and b = 5, then verify \[{{(a+b)}^{2}}=\text{ }{{a}^{2}}+{{b}^{2\text{ }}}+2ab.\] |
Answer:
(a)\[81\text{ }{{x}^{2}}+16{{y}^{2}}-72xy={{\left( 9x \right)}^{2}}+{{\left( 4y \right)}^{2}}-2\times 9x\times 4y\] \[={{(9x-4y)}^{2}}\] \[[\because \,{{a}^{2}}+{{b}^{2}}-2ab={{(a-b)}^{2}}]\] Now, putting \[x=\frac{2}{3}\]and \[y=\frac{3}{4},\]then \[={{\left( 9\times \frac{2}{3}-4\times \frac{3}{4} \right)}^{2}}\] \[={{(6-3)}^{2}}={{3}^{2}}=9\] (b) Putting a = 2 and b = 5, then L.H.S \[={{\left( a+b \right)}^{2}}\] \[={{\left( 2+5 \right)}^{2}}={{7}^{2}}=49~\] and R.H.S =\[={{a}^{2}}+{{b}^{2}}+2ab\] \[={{2}^{2}}+{{5}^{2}}+2\times 2\times 5\] = 4 + 25 + 20 = 49 Hence, L.H.S = R.H.S =49
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