a: \(2x^2+8=0\)
mà \(2x^2+8>=8>0\forall x\)
nên \(x\in\varnothing\)
b: \(\left(x-4\right)\left(5x-2\right)-3\left(x-4\right)=0\)
=>\(\left(x-4\right)\left(5x-2-3\right)=0\)
=>(x-4)(5x-5)=0
=>5(x-1)(x-4)=0
=>(x-1)(x-4)=0
=>\(\left[{}\begin{matrix}x-1=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)
c: Sửa đề: \(\left(x+2\right)\left(x^2-2x+4\right)+x\left(5-x\right)\left(5+x\right)=-17\)
=>\(x^3+8+x\left(25-x^2\right)=-17\)
=>\(x^3+8+25x-x^3=-17\)
=>25x+8=-17
=>25x=-17-8=-25
=>\(x=-\dfrac{25}{25}=-1\)
d: \(\left(x^2+x+1\right)\left(x^2+x+2\right)-12=0\)
=>\(\left(x^2+x\right)^2+3\left(x^2+x\right)+2-12=0\)
=>\(\left(x^2+x\right)^2+3\left(x^2+x\right)-10=0\)
=>\(\left(x^2+x+5\right)\left(x^2+x-2\right)=0\)
mà \(x^2+x+5=x^2+x+\dfrac{1}{4}+\dfrac{19}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{19}{4}>=\dfrac{19}{4}>0\forall x\)
nên \(x^2+x-2=0\)
=>(x+2)(x-1)=0
=>\(\left[{}\begin{matrix}x+2=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
