x2-7x-8=x2-8x+x-8=(x2+x)-(8x+8)=x(x+1)-8(x+1)=(x+1)(x-8)
a) \(x^2+3x-4=0\)
\(\Rightarrow\left(x^2-x\right)+\left(4x-4\right)=0\\ \Rightarrow x\left(x-1\right)+4\left(x-1\right)=\left(x-1\right)\left(x+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-1=0\\x+4=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)
b) \(x^2-7x-8=0\)
\(\Rightarrow\left(x^2+x\right)-\left(8x+8\right)=0\\ \Rightarrow x\left(x+1\right)-8\left(x+1\right)=0\\ \Rightarrow\left(x-8\right)\left(x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-8=0\\x+1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=8\\x=-1\end{matrix}\right.\)
c) \(x^2+4x-5=0\)
\(\Rightarrow\left(x^2-x\right)+\left(5x-5\right)=0\\ \Rightarrow x\left(x-1\right)+5\left(x-1\right)=0\\ \Rightarrow\left(x-1\right)\left(x+5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-1=0\\x+5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)
d) \(3x^2-5x+2=0\)
\(\Rightarrow\left(3x^2-3x\right)-\left(2x-2\right)=0\\ \Rightarrow3x\left(x-1\right)-2\left(x-1\right)=0\\ \Rightarrow\left(x-1\right)\left(3x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-1=0\\3x-2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{2}{3}\end{matrix}\right.\)
e) \(4x^2-7x+3=0\)
\(\Rightarrow\left(4x^2-4x\right)-\left(3x-3\right)=0\\ \Rightarrow4x\left(x-1\right)-3\left(x-1\right)=0\\ \Rightarrow\left(x-1\right)\left(4x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-1=0\\4x-3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{3}{4}\end{matrix}\right.\)
1: Ta có: \(x^2+3x-4=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=1\end{matrix}\right.\)
2: Ta có: \(x^2-7x-8=0\)
\(\Leftrightarrow\left(x-8\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-1\end{matrix}\right.\)
3: Ta có: \(x^2+4x-5=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)
4: Ta có: \(3x^2-5x+2=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{2}{3}\end{matrix}\right.\)
5: Ta có: \(4x^2-7x+3=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{3}{4}\end{matrix}\right.\)
