A/ Đặt \(f\left(x\right)=x^2+7x-8\)
Khi f (x) = 0
=> \(x^2+7x-8=0\)
=> \(x^2+8x-x-8=0\)
=> \(\left(x^2-x\right)+\left(8x-8\right)=0\)
=> \(x\left(x-1\right)+8\left(x-1\right)=0\)
=> \(\left(x-1\right)\left(x+8\right)=0\)
=> \(\orbr{\begin{cases}x-1=0\\x+8=0\end{cases}}\)=> \(\orbr{\begin{cases}x=1\\x=-8\end{cases}}\)
Vậy f (x) có 2 nghiệm \(\hept{\begin{cases}x=1\\x=-8\end{cases}}\)
B/ Đặt \(g\left(x\right)=5x^2+9x+4\)
Khi g (x) = 0
=> \(5x^2+9x+4=0\)
=>\(5x^2+5x+4x+4=0\)
=> \(\left(5x^2+5x\right)+\left(4x+4\right)=0\)
=> \(5x\left(x+1\right)+4\left(x+1\right)=0\)
=> \(\left(x+1\right)\left(5x+4\right)=0\)
=> \(\orbr{\begin{cases}x+1=0\\5x+4=0\end{cases}}\)=> \(\orbr{\begin{cases}x=-1\\5x=-4\end{cases}}\)=> \(\orbr{\begin{cases}x=-1\\x=\frac{-4}{5}\end{cases}}\)
Vậy g (x) có 2 nghiệm: \(\hept{\begin{cases}x=1\\x=\frac{-4}{5}\end{cases}}\)
