Ta có
\(x\left(-2x+1\right)=-1\)
\(\Rightarrow x=1,-1;\left(-2x+1\right)=1;-1\\
\)
Mà \(\left(-2x+1\right)\le0\\
\left(-2x+1\right)=-1\Leftrightarrow x=1\)
\(\Leftrightarrow-2x^2+x+1=0\)
\(\Leftrightarrow-2x+2x.\frac{1}{2}+\frac{1}{4}-\frac{1}{4}+1=0\)
\(\Leftrightarrow\left(-2x-\frac{1}{2}\right)^2+\frac{3}{4}=0\)
Ta có : \(\left(-2x-\frac{1}{2}\right)^2>0\)
\(\Rightarrow\left(-2x-\frac{1}{2}\right)^2+\frac{3}{4}>0_{ }\)
Kết luận vậy pt VÔ NGHIỆM
-2x^2+x=-1
-2x^2+x+1=0
-1/2(2x^2+x+1)=0
x^2-1/2x-1/2=0
x^2-1/2x+1/16-1/16-1/2=0
(x-1/4)^2-9/16=0
(x-1/4-3/4)(x-1/4+3/4)=0
(x-1)(x+1/2)=0
=> x-1=0 hoặc x+1/2=0
x=1 hoặc x=-1/2