a) Ta có: \(x^2+4x+4=x^2-6x+9\)
\(\Leftrightarrow4x+4=-6x+9\)
\(\Leftrightarrow4x+6x=9-4\)
\(\Leftrightarrow10x=5\)
hay \(x=\dfrac{1}{2}\)
b) Ta có: \(B=-x^2+2x-2\)
\(=-\left(x^2-2x+2\right)\)
\(=-\left(x^2-2x+1\right)-1\)
\(=-\left(x-1\right)^2-1< 0\forall x\)
Bài 1:
\(pt\Leftrightarrow10x=5\Leftrightarrow x=\dfrac{1}{2}\)
Vậy \(S=\left\{\dfrac{1}{2}\right\}\)
Bài 2:
\(B=x^2+2x-2\)
Lấy \(x=1\Rightarrow B=1>0\)
Vậy \(B=x^2+2x-2< 0\forall x\in R\) ( vô lí)
a) Ta có: x2+4x+4=x2−6x+9
⇔4x+4=−6x+9
⇔4x+6x=9−4
⇔10x=5
hay
