K
Khách

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

27 tháng 11 2015

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

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

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

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

5 tháng 8 2023

\(A=\left(\dfrac{1}{\sqrt{x-1}}+\dfrac{1}{\sqrt{x-1}}\right)^2\cdot\dfrac{x^2-1}{2}-\sqrt{x^2-1}\) (ĐK: \(x>1\))

\(A=\left(\dfrac{2}{\sqrt{x-1}}\right)^2\cdot\dfrac{x^2-1}{2}-\sqrt{x^2-1}\)

\(A=\dfrac{4}{x-1}\cdot\dfrac{\left(x+1\right)\left(x-1\right)}{2}-\sqrt{x^2-1}\)

\(A=2\left(x+1\right)-\sqrt{\left(x+1\right)\left(x-1\right)}\)

\(A=\sqrt{x+1}\left(2\sqrt{x+1}-\sqrt{x-1}\right)\)

HQ
Hà Quang Minh
Giáo viên
5 tháng 8 2023

\(A=\left(\dfrac{1}{\sqrt{x-1}}+\dfrac{1}{\sqrt{x+1}}\right)^2\cdot\dfrac{x^2-1}{2}-\sqrt{x^2-1}\\ \Rightarrow A=\left(\dfrac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x^2-1}}\right)^2\cdot\dfrac{x^2-1}{2}-\sqrt{x^2-1}\\ \Rightarrow A=\dfrac{\left(\sqrt{x+1}+\sqrt{x-1}\right)^2}{2}-\sqrt{x^2-1}\\ \Rightarrow A=\dfrac{2x+2\sqrt{x^2-1}-2\sqrt{x^2-1}}{2}\\ \Rightarrow A=x\)

1 tháng 1 2021

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

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

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

 
25 tháng 12 2018

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

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

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

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

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

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

27 tháng 2 2020

\(\left(\frac{1}{x+1}-\frac{3}{x^3+1}+\frac{3}{x^2-x+1}\right):\frac{3x^2-3x+3}{\left(x+1\right)\left(x+2\right)}-\frac{2x-2}{x^2+2x}\left(x\ne-1;x\ne0;x\ne-2\right)\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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:

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