Question

find value of ( 967 + 289 ) ^ 2 + ( 967 - 289 ) ^ 2 / ( 967 × 967 + 289 × 289 )​

Answer 1

ANSWER:

To Find:

• Value of (967 + 289)^2 + (967 - 289)^2 / (967 × 967 + 289 × 289)

Solution:

We need to find, value of,

$\implies \dfrac{(967 + 289)^2 + (967 - 289)^2}{(967\times967 + 289\times289)}$

Let us assume 967 as p and 289 as q.

So,

$\implies \dfrac{(967 + 289)^2 + (967 - 289)^2}{(967\times967 + 289\times289)}$

$\implies \dfrac{(p + q)^2 + (p - q)^2}{(p\times p + q\times q)}$

We know that,

$\hookrightarrow a\times a = a^2$

So,

$\implies \dfrac{(p + q)^2 + (p - q)^2}{(p\times p + q\times q)}$

$\implies \dfrac{(p + q)^2 + (p - q)^2}{(p^2 + q^2)}$

We know that,

$\hookrightarrow (a\pm b)=a^2+b^2\pm2ab.$

So,

$\implies \dfrac{(p + q)^2 + (p - q)^2}{(p^2 + q^2)}$

$\implies \dfrac{p^2+q^2+2pq+p^2+q^2-2pq}{(p^2 + q^2)}$

Cancelling 2pq,

$\implies \dfrac{2p^2+2q^2}{(p^2 + q^2)}$

Taking 2 common,

$\implies \dfrac{2(p^2+q^2)}{(p^2 + q^2)}$

Cancelling p^2 + q^2,

$\implies 2$

Hence,

$\implies\bf\dfrac{(967 + 289)^2 + (967 - 289)^2}{(967\times967 + 289\times289)}=2$

Answer 2

Answer:

2

Step-by-step explanation:

Value of ( 967 + 289 ) ^ 2 + ( 967 - 289 ) ^ 2 / ( 967 × 967 + 289 × 289 ) = 2

Hope it helps!!

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