Ta có: a) \(2\left(x-3\right)=3\left(-2x+4\right)\)
\(\Leftrightarrow\) \(2x-6=12-6x\)
\(\Leftrightarrow\) \(2x+6x=12+6\)
\(\Leftrightarrow\) \(8x=18\)
\(\Leftrightarrow\) \(x=\frac{9}{4}\) (thỏa mãn)
Vậy phương trình có nghiệm duy nhất \(x=\frac{9}{4}\)
b) \(\left(x-2\right)^2-2x+4=0\)
\(\Leftrightarrow\) \(\left(x-2\right)^2-2\left(x-2\right)=0\)
\(\Leftrightarrow\) \(\left(x-2\right)\left(x-2-2\right)=0\) \(\Leftrightarrow\) \(\left(x-2\right)\left(x-4\right)=0\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x-2=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=4\end{matrix}\right.\) (thỏa mãn)
Vậy phương trình có tập nghiệm S\(=\left\{2;4\right\}\)
Nhớ tick đúng cho mình nha
\(a,2\left(x-3\right)=3\left(-2x+4\right)\\ \Leftrightarrow2x-6=-6x+12\\ \Leftrightarrow2x+6x=12+6\\ \Leftrightarrow8x=18\\ \Leftrightarrow x=\frac{9}{4}\\ b,\left(x-2\right)^2-2x+4=0\\ \Leftrightarrow x^2-4x+4-2x+4=0\\ \Leftrightarrow x^2-6x+8=0\\ \Leftrightarrow x^2-2x-4x+8=0\\ \Leftrightarrow x\left(x-2\right)-4\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)
