\(\left(x-3\right)^2-x\left|x-3\right|=0\)
TH1 x\(\le\)3
=> (x-3)2-x(3-x)=0\(\Leftrightarrow\) x2-6x+9-3x+x2=0
\(\Leftrightarrow2x^2-9x+9=0\)
\(\Leftrightarrow2x^2-6x-3x+9=0\)
\(\Leftrightarrow2x\left(x-3\right)-3\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{3}{2}\end{matrix}\right.\) (thỏa mãn)
TH2 x>3
\(\left(x-3\right)^2-x\left(x-3\right)=0\) \(\Leftrightarrow x^2-6x+9-x^2+3x=0\)
\(\Leftrightarrow-3x+9=0\Leftrightarrow x=3\) (ktm)
Vậy x=3;x=\(\dfrac{3}{2}\)
x−3)2−x|x−3|=0(x−3)2−x|x−3|=0
TH1 x≤≤3
=> (x-3)2-x(3-x)=0⇔⇔ x2-6x+9-3x+x2=0
⇔2x2−9x+9=0⇔2x2−9x+9=0
⇔2x2−6x−3x+9=0⇔2x2−6x−3x+9=0
⇔2x(x−3)−3(x−3)=0⇔2x(x−3)−3(x−3)=0
⇔(x−3)(2x−3)=0⇔(x−3)(2x−3)=0
⇔[x−3=02x−3=0⇔⎡⎣x=3x=32⇔[x−3=02x−3=0⇔[x=3x=32 (thỏa mãn)
TH2 x>3
(x−3)2−x(x−3)=0(x−3)2−x(x−3)=0 ⇔x2−6x+9−x2+3x=0⇔x2−6x+9−x2+3x=0
⇔−3x+9=0⇔x=3⇔−3x+9=0⇔x=3 (ktm)
Vậy x=3;x=32
