1. x2-2x-x+2=0
x(x-2)-(x-2)=0
(x-2)(x-1)=0
x-2=0
x-1=0
x=2 ;x=1
2. x2+4x+3x+12=0
(x2+4x)(3x+12)=0
x(x+4)+3(x+4)=0
x+3=0
x+4=0
x=-3; x=-4
Câu cuối mình ko hiểu x mũ mấy vậy bạn
1. x2 -3x+2=0
\(\Leftrightarrow\)(x-1)(x-2)=0
\(\Leftrightarrow\left[{}\begin{matrix}\\\end{matrix}\right.\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\\\end{matrix}\right.\begin{matrix}x=1\\x=2\end{matrix}\right.\)
vậy S=\(\left\{1;2\right\}\)
2. x2+7x+12=0
\(\Leftrightarrow\left(x+3\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-4\end{matrix}\right.\)
vậy S=\(\left\{-4;-3\right\}\)
3. x2 -3x-10=0
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
vậy S=\(\left\{-2;5\right\}\)
chúc bạn học tốt
Mình làm câu c) rõ hơn nè.
\(x^2-3x-10=0\)
\(\Leftrightarrow x^2-5x+2x-10=0\)
\(\Leftrightarrow\left(x^2-5x\right)+\left(2x-10\right)=0\)
\(\Leftrightarrow x.\left(x-5\right)+2.\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right).\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0+5\\x=0-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{5;-2\right\}.\)
Chúc bạn học tốt!