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3 tháng 8 2023

(\(x+1\))2 - (\(x+1\)) = 0

(\(x+1\))(\(x+1-1\)) =0 

(\(x+1\))\(x\)   = 0

\(\left[{}\begin{matrix}x+1=0\\x=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=-1\\x=0\end{matrix}\right.\)

Vậy \(x\in\){ -1; 0}

Giải bằng cách phân tích đa thức thành nhân tử nhé mn

15 tháng 5 2021

minh biet

NM
5 tháng 3 2022

ta có : 

\(\left|x+1\right|+\left|x-1\right|=1+\left|\left(x-1\right)\left(x+1\right)\right|\)

\(\Leftrightarrow\left|x-1\right|\left|x+1\right|-\left|x-1\right|-\left|x+1\right|+1=0\)

\(\Leftrightarrow\left(\left|x-1\right|-1\right)\left(\left|x+1\right|-1\right)=0\Leftrightarrow\orbr{\begin{cases}\left|x-1\right|=1\\\left|x+1\right|=1\end{cases}}\)

\(\Leftrightarrow x\in\left\{-2,0,2\right\}\)

1) ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

Ta có: \(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{4}{x^2-1}\)

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

Suy ra: \(x^2+2x+1-\left(x^2-2x+1\right)=4\)

\(\Leftrightarrow x^2+2x+1-x^2+2x-1=4\)

\(\Leftrightarrow4x=4\)

hay x=1(loại)

Vậy: \(S=\varnothing\)

2) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

Ta có: \(\dfrac{x+2}{x-2}+\dfrac{x}{x+2}=2\)

\(\Leftrightarrow\dfrac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{2\left(x^2-4\right)}{\left(x-2\right)\left(x+2\right)}\)

Suy ra: \(x^2+4x+4+x^2-2x=2x^2-8\)

\(\Leftrightarrow2x^2+2x+4-2x^2-8=0\)

\(\Leftrightarrow2x-4=0\)

\(\Leftrightarrow2x=4\)

hay x=2(loại)

Vậy: \(S=\varnothing\)

9 tháng 12 2018

Câu (C) sai

24 tháng 1 2017

Tiếp

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

27 tháng 11 2021

lên google

17 tháng 10 2017

1. \(\dfrac{1}{x-1}-\dfrac{1}{x+1}\)

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

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

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

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

2. \(\dfrac{x}{x^2-1}-\dfrac{1}{x-1}\)

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

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

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

3. \(\dfrac{1}{x\left(x-y\right)}-\dfrac{1}{x\left(x-y\right)}\)

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

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

\(=\dfrac{x}{xy\left(x-y\right)}+\dfrac{-y}{xy\left(x-y\right)}\)

\(=\dfrac{x-y}{xy\left(x-y\right)}=\dfrac{1}{xy}\)

4. \(\dfrac{1}{x}-\dfrac{1}{x-1}\)

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

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

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

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

5. \(\dfrac{1}{x}-\dfrac{1}{x+1}\)

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

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

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

6. \(\dfrac{1}{2x^2-10x}-\dfrac{1}{x-5}\)

\(=\dfrac{1}{2x\left(x-5\right)}-\dfrac{1}{x-5}\)

\(=\dfrac{1}{2x\left(x-5\right)}-\dfrac{1.2x}{2x\left(x-5\right)}\)

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

\(=\dfrac{1-2x}{2x\left(x-5\right)}\)

7. \(\dfrac{x-1}{x^2-1}.\dfrac{x+1}{x+3}\)

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

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

8. \(\dfrac{2}{2x^2+10x}.\dfrac{x+5}{3x}\)

\(=\dfrac{2x\left(x+5\right)}{2x^2+10x.3x}\)

\(=\dfrac{2\left(x+5\right)}{2x\left(x+5\right)3x}\)

\(=\dfrac{2}{6x^2}=\dfrac{1}{3x^2}\)

AH
Akai Haruma
Giáo viên
7 tháng 7 2021

Lời giải:

a. ĐKXĐ: $x\neq 0;-1$

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

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

b. ĐKXĐ: $x\neq 0; 1;2$

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

\(=\frac{1}{x(x-1)}.\frac{(x-2)(x-1)}{3}=\frac{x-2}{3x}\)

c. ĐKXĐ: $x\neq 0; -1$
\(=\frac{x+1+x^2}{x(x+1)}.\frac{x(x+1)}{x}=\frac{x^2+x+1}{x}\)