Explanation
$$ {391}^{2} = {(400 - 9)}^{2} $$It is the form of $$ {(a - b)}^{2} $$, where $$ a = 400, b = 9 $$Applying the formula $$ { (a - b) }^{ 2 } = {a}^{2} + { b }^{ 2 } - 2ab$$$$ {391}^{2} = {(400 - 9)}^{2} = {400}^{2} + { 9 }^{ 2 } -2\times 400 \times 9 = 1,60,000 + 81 - 7200 = 152881 $$
$$607$$ can be written as $$600+7$$
$$\therefore {607}^{2} = {(600 + 7)}^{2} $$It is the form of $$ {(a+b)}^{2} $$, where $$ a = 600, b = 7 $$Applying the formula $$ { (a+b) }^{ 2 } = {a}^{2} + { b }^{ 2 } + 2ab $$$$ {607}^{2} = {(600 + 7)}^{2} = {600}^{2} + { 7 }^{ 2 } +2\times 600 \times 7 = 360000 + 49 + 8400 = 368449 $$
Given, $$(4a+3b)^{2} -(4a-3b)^{2}+48 ab$$.
We know, $$(a+b)^2=a^2+2ab+b^2$$
and $$(a-b)^2=a^2-2ab+b^2$$.
Given, $$\left(\dfrac{2x}{7}-\dfrac{7y}{4}\right)^{2}$$.
We know, $$(a-b)^2=a^2+2ab+b^2$$.
Then,
$$\left(\dfrac{2x}{7}-\dfrac{7y}{4}\right)^{2}$$
$$=(\dfrac{2x}{7})^2-2(\dfrac{2x}{7})(\dfrac{7y}{4})+(\dfrac{7y}{4})^2$$
$$=\dfrac{4x^{2}}{49}- xy+ \dfrac{49y^{2}}{16} $$.
Therefore, ption $$C$$ is correct.
Given, $$\left(\dfrac{7}{8}{x}+\dfrac{4}{5}{y}\right )^{2}$$.
We know, $$(x+y)^2=x^2+2xy+y^2$$.
$$\left(\dfrac{7}{8}{x}+\dfrac{4}{5}{y}\right )^{2}$$
$$=(\dfrac{7}{8}{x})^2+2(\dfrac{7}{8}{x})(\dfrac{4}{5}{y})+(\dfrac{4}{5}{y})^2$$
$$=\dfrac{49}{64}{x^{2}} +\dfrac{7}{5}{xy}+\dfrac{16}{25}{y^{2}} $$.
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.
Please disable the adBlock and continue. Thank you.