đkxđ \(x\ne1\)

\(\Leftrightarrow\left(m-1\right)\left(x-1\right)=2m-2\)

\(\Leftrightarrow mx-m-x+1=2m-2\)

\(\Leftrightarrow mx-x=3m-3\)

\(\Leftrightarrow x\left(m-1\right)=3\left(m-1\right)\)(*)

Biện luận

+ Nếu m = 1 pt (*) 0x = 0 (vsn)

+ Nếu m khác 1 pt (*) -2x = -6 (cn)

Kết luận m khác 1 thì pt có nghiệm

               m=1 thì pt vsn

\(\frac{x+3}{x-3}-\frac{17}{x^2-9}=\frac{x-3}{x+3}\left(x\ne\pm3\right)\)

\(\Leftrightarrow\frac{x+3}{x-3}-\frac{17}{\left(x-3\right)\left(x+3\right)}-\frac{x-3}{x+3}=0\)

\(\Leftrightarrow\frac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}-\frac{17}{\left(x-3\right)\left(x+3\right)}-\frac{\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}=0\)

\(\Leftrightarrow\frac{x^2+6x+9}{\left(x-3\right)\left(x+3\right)}-\frac{17}{\left(x-3\right)\left(x+3\right)}-\frac{x^2-6x+9}{\left(x-3\right)\left(x+3\right)}=0\)

\(\Leftrightarrow\frac{x^2+6x+9-17-x^2+6x-9}{\left(x-3\right)\left(x+3\right)}=0\)

\(\Leftrightarrow\frac{12x-17}{\left(x-3\right)\left(x+3\right)}=0\)

\(\Rightarrow12x-17=0\)

\(\Leftrightarrow12x=17\)

\(\Leftrightarrow x=\frac{17}{12}\left(tmđk\right)\)

\(\left(3x-5\right)\left(-2x-7\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x-5=0\\-2x-7=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=5\\-2x=7\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{-7}{2}\end{cases}}}\)

\(9x^2-1=\left(1+3x\right)\left(2x-3\right)\)

\(\Leftrightarrow9x^2-1=2x-3+6x^2-9x\)

\(\Leftrightarrow9x^2-1=-7x-3+6x^2\)

\(\Leftrightarrow9x^2-1+7x+3-6x^2=0\)

\(\Leftrightarrow3x^2+2+7x=0\)

\(\Leftrightarrow3x^2+6x+x+2=0\)

\(\Leftrightarrow3x\left(x+2\right)+\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(3x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\3x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-\frac{1}{3}\end{cases}}\)

\(\left(8x-4x^2-1\right)\left(x^2+2x+1\right)=4\left(x^2+x+1\right)\)

\(\Leftrightarrow8x^3+16x^2+8x-4x^4-8x^3-4x^2-x^2-2x-1=4x^2+4x+4\)

\(\Leftrightarrow11x^2+6x-4x^4-1=4x^2+4x+4\)

\(\Leftrightarrow11x^2+6x-4x^2-1-4x^2-4x-4=0\)

\(\Leftrightarrow7x^2+2x-4x^4-4=0\)

\(\Leftrightarrow\left(-4x^3-4x^2+3x+5\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left(-4x^2-8x-5\right)\left(x-1\right)\left(x-1\right)=0\)

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\)

https://h7.net/hoi-dap/toan-8/giai-phuong-trinh-x-2-x-x-1-2-5-4-faq445177.html

haiz nghỉ tránh dịch mà cô cho bài tới tấp hà

Ai giúp mình với! Cần gấp!

Ta co:

\(\left|x-2016\right|+\left|x-2018\right|=\left|x-2016\right|+\left|2018-x\right|\ge\left|x-2016+2018-x\right|=2\)

\(\left|x-2017\right|\ge0\)

\(\Rightarrow\left|x-2016\right|+\left|x-2017\right|+\left|x-2018\right|\ge2\)

Dau "=" xay ra tai \(\hept{\begin{cases}2016\le x\le2018\\x=2017\end{cases}}\)

Vay x=2017

Bình thường A xđ \(\Leftrightarrow\left(x^2+1\right)\left(x^2+4x+5\right)\ne0\)

Ta có \(x^2+4x+5=\left(x+2\right)^2+1\)

Mà \(\left(x+2\right)^2\ge0\forall x\)

\(\Rightarrow x^2+4x+5>1\)(1)

Lại có \(x^2\ge0\forall x\)

\(\Rightarrow x^2+1>0\)(2)

(1)(2) \(\Rightarrow\left(x^2+1\right)\left(x^2+4x+5\right)>0\)hay \(\left(x^2+1\right)\left(x^2+4x+5\right)\ne0\)

\(4x^2+8x-5=0\)

\(\Leftrightarrow\left(2x+5\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x+5=0\\2x-1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}2x=-5\\2x=1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{5}{2}\\x=\frac{1}{2}\end{cases}}\)

cảm ơn nhi nhé!