Simplify- n2 - 1n2 - 4n - 1n - 6

Now, (n^2 + 1n^2) 4(n + 1n) + 6 =(n + 1n)^2 2.n.1n 4(n + 1n) + 6 [ Since a^2+b^2=(a+b)^2 2ab ] =(n + 1n)^2 4(n + 1n) + 4 =(n + 1n 2 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Factorisation by Common Factors

Simplify: n...

Question

Simplify: $(n^{2} + \frac{1}{n^{2}}) - 4 (n + \frac{1}{n}) + 6$

Open in App

Solution

Suggest Corrections

Similar questions

(n - \frac{1}{n}) (n + \frac{1}{n}) (n^{2} + \frac{1}{n^{2}})

lim n \to \infty \frac{(n^{2} + 5 n + 6)}{(n + 4) (n + 5)} i s e q u a l t o

ω

(2 - ω) (2 - ω^{2}) + 2 (3 - ω) (3 - ω^{2}) + . . . + (n - 1) (n - ω) (n - ω^{2})

\int \frac{5 x^{2} + 4 n - 1}{\sqrt{n^{2} - 4 n + 5}} d x

(l + 2 m + n)^{2} + (l - 2 m + n)^{2}

Join BYJU'S Learning Program