\(A=2x\left(x-1\right)+8-x=0\)
\(A=2x^2-2x+8-x=0\)
\(A=2x^2-3x+8=0\)
Dùng hàm MODE 5 3 (Giải pt bậc 2) của máy tính CASIO.
Ra hai nghiệm :
\(x_1=\dfrac{3}{4}+1,854049622i\) và \(x_2=\dfrac{3}{4}-1,854049622i\)
Vậy pt trên vô nghiệm.
\(2x\left(x-1\right)+8-x\)
\(\Leftrightarrow2x^2-2x+8-x\)
\(\Leftrightarrow2x^2-3x+8\)
\(\Leftrightarrow2x^2-3x+8\)
\(\Leftrightarrow x^2+x^2-2x-x+1+7\)
\(\Leftrightarrow\left(x^2-2x+1\right)+\left(x^2-x+7\right)\)
\(\Leftrightarrow\left(x-1\right)^2+\left(x^2-x+\dfrac{1}{4}+\dfrac{27}{4}\right)\)
\(\Leftrightarrow\left(x-1\right)^2+\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{27}{4}\)
\(\Leftrightarrow\left(x-1\right)^2+\left(x-\dfrac{1}{2}\right)^2+\dfrac{27}{4}\)
Ta có:
\(\left(x-1\right)^2\ge0\)
\(\left(x-\dfrac{1}{2}\right)^2\ge0\)
\(\Rightarrow\left(x-1\right)^2+\left(x-\dfrac{1}{2}\right)^2\ge0\)
\(\Rightarrow\left(x-1\right)^2+\left(x-\dfrac{1}{2}\right)^2+\dfrac{27}{4}\ge\dfrac{27}{4}>0\)
\(\Rightarrow\)Pt vô nghiệm
Vậy \(S=\varnothing\)