c, x3-2x2+x=0
=> x(x-1)2=0
=>\(\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
b,4x2-3x-7=(x+1)(4x-7)=0
=>\(\orbr{\begin{cases}x+1=0\\4x-7=0\end{cases}}\)=>\(\orbr{\begin{cases}x=-1\\x=\frac{7}{4}\end{cases}}\)
a) Ta có: 8y4 \(\ge\)0
7y2 \(\ge\)0
5 > 0
=> 8y4 + 7y2 + 5 > 0
=> 8y4 + 7y2 + 5 ko có nghiệm