Question

answer this maths question ​

Answer 1

AnsWer :

Option 2.

QuestioN :

Square of binomial ( 7x / 5 - 5y / 7 ) is

SolutioN :

We have, Expression.

$\tt \bullet \: \: \: Square \: of \: binomial \Big( \frac{7x}{5} - \frac{5y}{7} \Big) \: is.$

Compare With General Expression.

$\tt \bullet \: \: \: \sum\limits^{n}_{r= 0} {}^{n}C_r\,x^{n-r}\,y^r$

Now, Putting given value.

$\tt:\implies\Big( \frac{7x}{5} - \frac{5y}{7} \Big)^{2}$

Where as,

• n = 2.

And, Let

• x = 7x / 5.
• y = 5y / 7.

$\rule{90}3$

★ According to Question,

When,

• n = 2.
• r = 0.

$\tt:\implies {}^{n}C_r\,x^{n-r}\,y^r$

$\tt:\implies {}^{2}C_0\,x^{2-0}\,y^0$

$\tt:\implies 1 \times x^{2} \times 1$

$\tt:\implies {x}^{2}$

$\rule{90}2$

When,

• r = 1.

$\tt:\implies {}^{2}C_1\,x^{2-1}\,y^1$

$\tt:\implies {}^{2}C_1\,x^{1}\,y^1$

$\tt:\implies {}^{2}C_1\,xy$

We know,

• nC1 = n.

$\tt:\implies 2xy$

$\rule{90}2$

When,

• r = 2.

$\tt:\implies {}^{2}C_2\,x^{2-2}\,y^2$

$\tt:\implies {}^{2}C_2\,x^{0}\,y^2$

$\tt:\implies {}^{2}C_2\,y^2$

$\tt:\implies y^2$

So, we can also write as,

$\tt \longrightarrow ( x - y )^n = [ x + ( - y ) ]^n$

From above observation, we can write as.

$\tt \longrightarrow ( x - y )^2 = {x}^{2} - 2xy + {y}^{2}$

★ Now,

$\tt:\implies\Bigg\lgroup \dfrac{7x}{5} - \dfrac{5y}{7} \Bigg\rgroup^{2} = { \Big(\dfrac{7x}{5} \Big)}^{2} -2.\Big( \dfrac{7x}{5} . \dfrac{5y}{7}\Big) -\Big( \dfrac{5y}{7} \Big)^2$

$\tt:\implies\Bigg\lgroup \dfrac{7x}{5} - \dfrac{5y}{7} \Bigg\rgroup^{2} = \dfrac{49x^2}{25} -\dfrac{ \cancel{70xy}}{ \cancel{35}} +\dfrac{25y^2}{49}$

$\tt:\implies\Bigg\lgroup \dfrac{7x}{5} - \dfrac{5y}{7} \Bigg\rgroup^{2} = \dfrac{49x^2}{25} -2xy +\dfrac{25y^2}{49}$

Therefore, the value of square of binomial ( 7x / 5 - 5y/ 7 ) is 49x² / 25 - 2xy + 25y² / 49.

Answer 2

Answer:

option 2 is the answer

hope it helps u