Question

tìm nghiệm của đa thức h(x)= x(x-1)+1

Answer 1

\(x^2-x+1=0\)

\(\Leftrightarrow x^2-2x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=0\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=-\dfrac{3}{4}\)  (Vô nghiệm)

Answer 2

\(h\left(x\right)=0\)

\(\Leftrightarrow x\cdot\left(x-1\right)+1=0\)

\(\Leftrightarrow x^2-x+1=0\)

\(\Leftrightarrow x^2-2\cdot\dfrac{1}{2}x+\left(\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\)

PTVN