Question

please solve this sum.​

Answer 1

Step-by-step explanation:

(2)

$= > {(a + b)}^{2} - {(a - b)}^{2} \\ = > {a}^{2} + {b}^{2} + 2ab - {a}^{2} - {b}^{2} + 2ab \\ = > 4ab$

Answer 2

2 answer:-

$\sf\;(a+b)^2-(a-b)^2\;\;\;\;\;using\;identity\;a^2-b^2=(a+b)(a-b)\\\\==>(a+b+a-b)(a+b-a+b)\\\\==>(2a)(2b)\\==>{4ab}$

3 answer:-

$\sf\;x^2+\frac{1}{x^2}=27,find \;value\;of\;x-\frac{1}{x}\\\\\\(x-\frac{1}{x})^2=x^2+\frac{1}{x^2}-2\times\;x\times\frac{1}{x}\\\\(x-\frac{1}{x})^2=x^2+\frac{1}{x^2}-2\\\\\\x^2+\frac{1}{x^2}=27[GIVEN]\\\\(x-\frac{1}{x})^2=27-2\\\\(x-\frac{1}{x})^2=25\\\\(x-\frac{1}{x})=\sqrt{25} \\\\(x-\frac{1}{x})=5$

4 ANSWER:-

$\sf\;a^2-3a+1=0\\\\a^2+1=3a\\\\a+\frac{1}{a}=3\\\\(a+\frac{1}{1})^2=3^2\\\\\\(a+\frac{1}{1})^2=6$

5 ANSWER:-

$\sf\;(10+0.3)^2=(10)^2+(0.3)^2+2(10)(0.3)\\\\(10+0.3)^2=100+0.09+6\\\\(10+0.3)^2=106.09$