b/ Ta có \(G\left(x\right)=3x\left(x-1\right)-x+1\)
=> \(G\left(x\right)=3x\left(x-1\right)-\left(x-1\right)\)
=> \(G\left(x\right)=\left(x-1\right)\left(3x-1\right)\)
Khi G (x) = 0
=> \(\left(x-1\right)\left(3x-1\right)=0\)
=> \(\orbr{\begin{cases}x-1=0\\3x-1=0\end{cases}}\)=> \(\orbr{\begin{cases}x=1\\3x=1\end{cases}}\)=> \(\orbr{\begin{cases}x=1\\x=\frac{1}{3}\end{cases}}\)
Vậy G (x) có 2 nghiệm là \(\hept{\begin{cases}x=1\\x=\frac{1}{3}\end{cases}}\).
c/ Ta có \(H\left(x\right)=x^2-4x+3\)
=> \(H\left(x\right)=x^2-x-3x+3\)
=> \(H\left(x\right)=\left(x^2-x\right)-\left(3x-3\right)\)
=> \(H\left(x\right)=x\left(x-1\right)-3\left(x-1\right)\)
=> \(H\left(x\right)=\left(x-1\right)\left(x-3\right)\)
Khi H (x) = 0
=> \(\left(x-1\right)\left(x-3\right)=0\)
=> \(\orbr{\begin{cases}x-1=0\\x-3=0\end{cases}}\)=> \(\orbr{\begin{cases}x=1\\x=3\end{cases}}\)
Vậy H (x) có 2 nghiệm: \(\hept{\begin{cases}x=1\\x=3\end{cases}}\)
