Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: ĐKXĐ: x<>0; x<>1
\(P=\dfrac{x^2-1+x+2-x^2}{x\left(x-1\right)}\cdot\dfrac{x-1}{x}\)
\(=\dfrac{x+1}{x\cdot x}=\dfrac{x+1}{x^2}\)
b: |2x-1|=3
=>2x-1=3 hoặc 2x-1=-3
=>x=-1(nhận) hoặc x=2(nhận)
Khi x=-1 thì \(P=\dfrac{-1+1}{\left(-1\right)^2}=0\)
Khi x=2 thì \(P=\dfrac{2+1}{2^2}=\dfrac{3}{4}\)
a: ĐKXĐ: x<>1; x<>-1
\(A=\dfrac{x^2+2x+1}{\left(x-1\right)\left(x+1\right)}-\dfrac{1}{x-1}\)
\(=\dfrac{x+1}{x-1}-\dfrac{1}{x-1}=\dfrac{x}{x-1}\)
b: x^2+3x+2=0
=>x=-1(loại) hoặc x=-2(nhận)
Khi x=-2 thì A=-2/(-3)=2/3
a) A = \(\dfrac{1}{x-1}-\dfrac{4}{x+1}+\dfrac{8x}{\left(x-1\right)\left(x+1\right)}\)
= \(\dfrac{x+1-4x+4+8x}{\left(x-1\right)\left(x+1\right)}=\dfrac{5x+5}{\left(x-1\right)\left(x+1\right)}=\dfrac{5}{x-1}\) => đpcm
b) \(\left|x-2\right|=3=>\left[{}\begin{matrix}x-2=3< =>x=5\left(C\right)\\x-2=-3< =>x=-1\left(L\right)\end{matrix}\right.\)
Thay x = 5 vào A, ta có:
A = \(\dfrac{5}{5-1}=\dfrac{5}{4}\)
c) Để A nguyên <=> \(5⋮x-1\)
| x-1 | -5 | -1 | 1 | 5 |
| x | -4(C) | 0(C) | 2(C) | 6(C) |
a: ĐKXĐ: x<>2; x<>-2; x<>0; x<>3
b: \(P=\left(\dfrac{-\left(x+2\right)}{x-2}+\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{x+2}\right)\cdot\dfrac{x^2\left(2-x\right)}{x\left(x-3\right)}\)
\(=\dfrac{-x^2-4x-4+4x^2+x^2-4x+4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{-x\left(x-2\right)}{x-3}\)
\(=\dfrac{4x^2-8x}{\left(x+2\right)}\cdot\dfrac{-x}{\left(x-3\right)}=\dfrac{-4x^2\left(x-2\right)}{\left(x+2\right)\left(x-3\right)}\)
c: 2(x-1)=6
=>x-1=3
=>x=4
Thay x=4 vào P, ta đc:
\(P=\dfrac{-4\cdot4^2\cdot\left(4-2\right)}{\left(4+2\right)\left(4-3\right)}=\dfrac{-64\cdot2}{6}=\dfrac{-128}{6}=-\dfrac{64}{3}\)
