Explanation
We can write
$$ {9.7}^{2} = {(10 - 0.3)}^{2} $$It is the form of $$ {(a - b)}^{2} $$, where $$ a = 10, b = 0.3 $$Applying the formula $$ { (a - b) }^{ 2 } = {a}^{2} + { b }^{ 2 } - 2ab$$$$ {9.7}^{2} = {(10 - 0.3)}^{2} = {10}^{2} + { 0.3 }^{ 2 } -2\times 10 \times 0.3 = 100 + 0.09 - 6 = 94.09 $$
Given, $$\left(\dfrac{7}{8}{x}+\dfrac{4}{5}{y}\right )^{2}$$.
We know, $$(x+y)^2=x^2+2xy+y^2$$.
Then,
$$\left(\dfrac{7}{8}{x}+\dfrac{4}{5}{y}\right )^{2}$$
$$=(\dfrac{7}{8}{x})^2+2(\dfrac{7}{8}{x})(\dfrac{4}{5}{y})+(\dfrac{4}{5}{y})^2$$
$$=\dfrac{49}{64}{x^{2}} +\dfrac{7}{5}{xy}+\dfrac{16}{25}{y^{2}} $$.
Therefore, option $$D$$ is correct.
Find the missing term in the following problem:
$$\left(\displaystyle \frac{3x}{4}\, -\, \displaystyle \frac{4y}{3} \right )^2\, =\, \displaystyle \frac{9x^2}{16}\, +\, ..........\, +\, \displaystyle \frac{16y^2}{9}$$.
Given, $$\left(\displaystyle \frac{3x}{4}\, -\, \displaystyle \frac{4y}{3} \right )^2\, $$.
We know, $$(a-b)^2=a^2-2ab+b^2$$.
$$\left(\displaystyle \frac{3x}{4}\, -\, \displaystyle \frac{4y}{3} \right )^2\, $$
$$=\left( \displaystyle \frac{3x}{4} \right )^2\, -\, 2\, \left(\displaystyle \frac{3x}{4} \right )\left(\displaystyle \frac{4y}{3} \right )\, +\, \left(\displaystyle \frac{4y}{3}\right)^2$$
$$=\, \displaystyle \frac{9x^2}{16}\, -\, 2xy\, +\, \displaystyle \frac{16y^2}{9}$$
$$=\, \displaystyle \frac{9x^2}{16}\, +\, (-\, 2xy)\, +\, \displaystyle \frac{16y^2}{9}$$.
Hence, the missing term is $$-2xy$$.
Therefore, option $$B$$ is correct.
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