Câu a : \(\left(2x-3\right)^2=\left(x-2\right)^2\)
\(\Leftrightarrow\left(2x-3\right)^2-\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(2x-3-x+2\right)\left(2x-3+x-2\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\3x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{5}{3}\end{matrix}\right.\)
Vậy ........
b ) \(\left(3x+1\right)^2=\left(2x-1\right)^2\)
\(\Leftrightarrow\left(3x+1\right)^2-\left(2x-1\right)^2=0\)
\(\Leftrightarrow\left(3x+1-2x+1\right)\left(3x+1+2x-1\right)=0\)
\(\Leftrightarrow5x\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
Vậy.............
c ) \(x^3+2x^2+6x+12=0\)
\(\Leftrightarrow x^2\left(x+2\right)+6\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x^2+6=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x\left(loại\right)\end{matrix}\right.\) Do \(x^2+6>0\)
Vậy.........
a)\(\left(2x-3\right)^2=\left(x-2\right)^2\)
\(2x-3=x-2\)
\(2x-3-x+2=0\)
\(x-1=0\)
\(x=1\)
b)\(\left(3x+1\right)^2=\left(2x-1\right)^2\)
\(3x+1=2x-1\)
\(3x+1-2x+1=0\)
\(x+2=0\)
\(x=-2\)
c)\(x^3+2x^2+6x+12=0\)
\(\left(x^3+2x^2\right)+\left(6x+12\right)=0\)
\(x^2\left(x+2\right)+6\left(x+2\right)=0\)
\(\left(x^2+6\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+6=0\\x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=-6\\x=-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{6}\\x=-2\end{matrix}\right.\)
Vậy \(x=-\sqrt{6}\) hoặc \(x=-2\)
Mình sửa câu c, trường hợp \(x=-\sqrt{6}\) ,x chỉ = -2 thô ạ,loại trường hợp kia dùm
