K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

29 tháng 4 2017

a)

\(\left\{{}\begin{matrix}x-1\ne0\\x+2\ne0\end{matrix}\right.\)

b)

x khác 1

c)

x khác 0; x khác 5

d) x khác 5 ; x khác -5

15 tháng 10 2020

1.

a, \(\left(x+3\right)\left(x-3\right)-\left(x-3\right)^2\)

\(=\left(x-3\right)\left(x+3-x+3\right)\)

\(=9\left(x-3\right)=9x-27\)

b, \(\left(2x+1\right)^2+2\left(2x+1\right)\left(x-1\right)+\left(x-1\right)^2\)

\(=\left(2x+1+x-1\right)^2=9x^2\)

c, \(x\left(x-3\right)\left(x+3\right)-\left(x^2+1\right)\left(x^2-1\right)\)

\(=x\left(x^2-9\right)-\left(x^4-1\right)\)

\(=x^3-9x-x^4+1=-x^4+x^3-9x+1\)

21 tháng 7 2020

\(B=\frac{5x}{x+2}-\frac{3x-23}{x-2}+\frac{40}{4-x^2}\)

a) ĐKXĐ : \(x\ne\pm2\)

\(B=\frac{5x}{x+2}-\frac{3x-23}{x-2}+\frac{40}{4-x^2}\)

\(B=\frac{5x}{x+2}-\frac{3x-23}{x-2}-\frac{40}{x^2-4}\)

\(B=\frac{5x}{x+2}-\frac{3x-23}{x-2}-\frac{40}{\left(x+2\right)\left(x-2\right)}\)

\(B=\frac{5x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{\left(3x-23\right)\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{40}{\left(x+2\right)\left(x-2\right)}\)

\(B=\frac{5x^2-10x}{\left(x+2\right)\left(x-2\right)}-\frac{\left(3x^2-17x-46\right)}{\left(x+2\right)\left(x-2\right)}-\frac{40}{\left(x+2\right)\left(x-2\right)}\)

\(B=\frac{5x^2-10x-\left(3x^2-17x-46\right)-40}{\left(x+2\right)\left(x-2\right)}\)

\(B=\frac{5x^2-10x-3x^2+17x+46-40}{\left(x+2\right)\left(x-2\right)}\)

\(B=\frac{2x^2+7x+6}{\left(x+2\right)\left(x-2\right)}=\frac{\left(x+2\right)\left(2x+3\right)}{\left(x+2\right)\left(x-2\right)}=\frac{2x+3}{x-2}\)

b) x2 - 1 = 0 <=> x2 = 1 <=> x = ±1

Với x = 1 

\(B=\frac{2\cdot1+3}{1-2}=-5\)

Với x = -1

\(B=\frac{2\cdot\left(-1\right)+3}{\left(-1\right)-2}=-\frac{1}{3}\)

2 tháng 2 2020

\(a,Đkxđ:x\ne\pm2\)

\(A=\frac{1}{x-2}+\frac{1}{x+2}+\frac{x^2+1}{x^2-4}\)

\(=\frac{x+2+x-2+x^2+1}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{x^2+2x+1}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{\left(x+1\right)^2}{x^2-4}\)

b, Ta có: \(\left(x-2\right)\left(x+2\right)< 0;\forall-2< 2< 2;x\ne-1\)

Mà: \(\left(x+1\right)^2>0\left(\forall x\ne-1\right)\)

\(\Rightarrow\frac{\left(x+1\right)^2}{\left(x+2\right)\left(x-2\right)}< 0;\forall-2< x< 2;x\ne-1\)

Vậy ............