K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
27 tháng 2 2021
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: \(S=\varnothing\)
17 tháng 6 2022
a: \(\dfrac{x-1}{x^2-x+1}-\dfrac{x+1}{x^2+x+1}=\dfrac{10}{x\left(x^4+x^2+1\right)}\)
\(\Leftrightarrow x\left(x-1\right)\left(x^2+x+1\right)-x\left(x+1\right)\left(x^2-x+1\right)=10\)
\(\Leftrightarrow x\left(x^3-1\right)-x\left(x^3+1\right)=10\)
=>-2x=10
hay x=-5
d: \(\Leftrightarrow\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+...+\dfrac{1}{\left(x+7\right)\left(x+8\right)}=\dfrac{1}{14}\)
\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+8}=\dfrac{1}{14}\)
\(\Leftrightarrow\left(x+1\right)\left(x+8\right)=14\left(x+8\right)-14\left(x+1\right)\)
\(\Leftrightarrow x^2+9x+8=14x+112-14x-14=98\)
\(\Leftrightarrow x^2+9x-90=0\)
\(\Leftrightarrow x\in\left\{6;-15\right\}\)
LT
0
21 tháng 4 2019
b,\(\frac{2}{x-1}=\frac{6}{x+1}\)
\(2x+2=6x-6\)
\(4x=8\)
\(x=2\)
9 tháng 9 2021
a: Ta có: \(5\left(4x-1\right)+2\left(1-3x\right)-6\left(x+5\right)=10\)
\(\Leftrightarrow20x-5+2-6x-6x-30=10\)
\(\Leftrightarrow8x=43\)
hay \(x=\dfrac{43}{8}\)
b: ta có: \(2x\left(x+1\right)+3\left(x-1\right)\left(x+1\right)-5x\left(x+1\right)+6x^2=0\)
\(\Leftrightarrow2x^2+2x+3x^2-3-5x^2-5x+6x^2=0\)
\(\Leftrightarrow6x^2-3x-3=0\)
\(\Leftrightarrow2x^2-x-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{2}\end{matrix}\right.\)
`x.(x-1)-x.(x+1)=5`
`x^2-x-x^2-x=5`
`(x^2-x^2)+(-x-x)=5`
`-2x=5`
`x=5:(-2)`
`x=-5/2`
Vậy `s=-5/2`
\(x\left(x-1\right)-x\left(x+1\right)=5\)
\(\Leftrightarrow x^2-x-x^2-x=5\)
\(\Leftrightarrow-2x=5\)
\(\Leftrightarrow x=\dfrac{-5}{2}\)