Question

Giải phương trình:

a) (2x-1)² = 3x(2x-1)

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

Answer 1

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

\(\left(2x-1\right)^2-3x\left(2x-1\right)=0\)

\(\left(2x-1\right)\left(2x-1-3x\right)=0\)

\(\left(2x-1\right)\left(-x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-1\end{matrix}\right.\)

Answer 2

\(a,\left(2x-1\right)^2=3x\left(2x-1\right)\\ \Leftrightarrow4x^2-4x+1=6x^2-3x\\ \Leftrightarrow6x^2-3x-4x^2+4x-1=0\\ \Leftrightarrow2x^2+x-1=0\\ \Leftrightarrow2x^2+2x-x-1=0\\ \Leftrightarrow2x\left(x+1\right)-\left(x+1\right)=0\\ \Leftrightarrow\left(x+1\right)\left(2x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{2}\end{matrix}\right.\)

\(b,ĐKXĐ:x\ne0,2\\ \dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x^2-2x}\\ \Leftrightarrow\dfrac{x\left(x+2\right)}{x\left(x-2\right)}-\dfrac{x-2}{x\left(x-2\right)}-\dfrac{2}{x\left(x-2\right)}=0\\ \Leftrightarrow\dfrac{x^2+2x-x+2-2}{x\left(x-2\right)}=0\\ \Rightarrow x^2+x=0\\ \Leftrightarrow x\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)

Answer 3

b) \(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x^2-2x}\left(đk:x\ne0,2\right)\)

\(x^2+2x-x+2=2\)

\(x^2+x=0\)

\(x\left(x+1\right)=0\)

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

Answer 4

\( (2x-1)^2 = 3x(2x-1)\)

\( (2x-1)^2 - 3x(2x-1)=0\)

\((2x-1)(2x-1-3x)=0\)

<=>x=1/2

      x=-1

Answer 5

\(a,\left(2x-1\right)^2=3x\left(2x-1\right)\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\-x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)

Vậy \(S=\left\{\dfrac{1}{2};1\right\}\)

\(b,\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x^2-2x}\left(ĐKXĐ:x\ne0;x\ne2\right)\)

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

\(\Leftrightarrow x\left(x+2\right)-x+2=2\)

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

\(\Leftrightarrow x^2+2x-x+2-2=0\)

\(\Leftrightarrow x^2-x=0\)

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

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

Vậy \(S=\left\{1\right\}\)