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5 tháng 1 2021

\(A=\frac{x}{x-1}+\frac{3}{x+1}-\frac{5x}{x^2-1}\)

\(=\frac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{3\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{5x}{\left(x-1\right)\left(x+1\right)}\)

\(=\frac{x^2+x}{\left(x-1\right)\left(x+1\right)}+\frac{3x-3}{\left(x-1\right)\left(x+1\right)}-\frac{5x}{\left(x-1\right)\left(x+1\right)}\)

\(=\frac{x^2+x+3x-3-5x}{\left(x-1\right)\left(x+1\right)}=\frac{x^2-x-3}{\left(x-1\right)\left(x+1\right)}\)

1 tháng 4 2021

Với \(x\ne1\)ta có 

\(P=\left(\frac{4}{x-1}-\frac{7x+5}{x^3-1}\right):\left(1-\frac{x-4}{x^2+x+1}\right)\)

\(=\left[\frac{4x^2+4x+4-7x-5}{\left(x-1\right)\left(x^2+x+1\right)}\right]:\left(\frac{x^2+x+1-x-4}{x^2+x+1}\right)\)

\(=\frac{4x^2-3x-1}{\left(x-1\right)\left(x^2+x+1\right)}:\frac{x^2-3}{x^2+x+1}=\frac{4x+1}{x^2-3}\)

Bài 1: 

a: \(A=\dfrac{x+1+x}{x+1}:\dfrac{3x^2+x^2-1}{x^2-1}\)

\(=\dfrac{2x+1}{x+1}\cdot\dfrac{\left(x+1\right)\left(x-1\right)}{\left(2x+1\right)\left(2x-1\right)}=\dfrac{x-1}{2x-1}\)

b: Thay x=1/3 vào A, ta được:

\(A=\left(\dfrac{1}{3}-1\right):\left(\dfrac{2}{3}-1\right)=\dfrac{-2}{3}:\dfrac{-1}{3}=2\)

4 tháng 7 2017

a. A=\(1+\left(\frac{x+1}{x^3+1}-\frac{1}{x-x^2-1}-\frac{2}{x+1}\right):\frac{x^3-2x^2}{x^3-x^2+x}\)

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

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

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

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

b.\(\left|x-\frac{3}{4}\right|=\frac{5}{4}\Rightarrow\orbr{\begin{cases}x-\frac{3}{4}=\frac{5}{4}\\x-\frac{3}{4}=-\frac{5}{4}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=2\\x=-\frac{1}{2}\end{cases}}\)

Với \(x=2\Rightarrow A=\frac{2-1}{2+1}=\frac{1}{3}\)

Với \(x=-\frac{1}{2}\Rightarrow A=\frac{-\frac{1}{2}-1}{-\frac{1}{2}+1}=-3\)

29 tháng 11 2016

\(A=\frac{\left[x\left(x^2-x+1\right)\right]-\left[\left(x+1\right)\left(3-3x\right)\right]+\left[x+4\right]}{x^3+1}\)

\(A=\frac{\left(x^3-x^2+x\right)+3\left(x^2-1\right)+\left(x+4\right)}{x^3+1}=\frac{x^3+2x^2+2x+1}{x^3+1}\)

\(A=\frac{\left(x^3+1\right)+2x\left(x+1\right)}{x^3+1}=1+\frac{2x}{x^2-x+1}\)

29 tháng 11 2016

\(A=\frac{x}{x+1}-\frac{3-3x}{x^2-x+1}+\frac{x+4}{x^3+1}\)

\(A=\frac{x}{x+1}-\frac{3-3x}{x^2-x+1}+\frac{x+4}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(A=\frac{x\left(x^2-x+1\right)-\left(3+3x\right)\left(x+1\right)+\left(x+4\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(A=\frac{x^3-x^2+x-9x-3-3x^2+x+4}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(A=\frac{x^3-x^2-3x^2+x-9x+x+3+4}{x^3+1}\)

\(A=\frac{x^3+2x^2-4x+4}{\left(x+1\right)\left(x^2-x+1\right)}\)

15 tháng 10 2019

1. P = \(\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}\)                       ĐKXĐ: \(x\ne-3\),  \(x\ne2\)

       = \(\frac{x+2}{x+3}-\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{1}{x-2}\)

       = \(\frac{x^2-4}{\left(x+3\right)\left(x-2\right)}-\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{x+3}{x-2}\)

       = \(\frac{x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}\)

       = \(\frac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}\)

       = \(\frac{\left(x-4\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}\)

       = \(\frac{x-4}{x-2}\)

2. P=\(\frac{-3}{4}\)

<=> \(\frac{x-4}{x-2}=\frac{-3}{4}\)

<=> 4 ( x - 4 ) = -3  ( x - 2 )

<=> 4x - 16 = -3x + 6

<=> 7x = 2 

<=> x = \(\frac{22}{7}\)

3. \(x^2-9=0\)

<=> ( x -3 ) ( x + 3 ) = 0

<=> \(\orbr{\begin{cases}x=3\left(tm\right)\\x=-3\left(ktm\right)\end{cases}}\)

-> P = \(\frac{3-4}{3-2}\) = -1

11 tháng 12 2018

a)\(\frac{x^3-x}{3x+3}=\frac{x.\left(x^2-1\right)}{3.\left(x+1\right)}=\frac{x.\left(x-1\right).\left(x+1\right)}{3.\left(x+1\right)}=\frac{x.\left(x+1\right)}{3}=\frac{x^2+x}{3}\)

11 tháng 12 2018

Bạn có thể giúp mình 2 câu còn lại dc kh ạ 

11 tháng 6 2015

con này vừa hôm trc làm rồi mà bạn không nhận đc câu trả lời sao?? huhu :'((( gõ lâu muốn chết