Question

Giải Phương Trình sau :

Answer 1

1 \(\Leftrightarrow\left|x-1\right|=-x^2+3x-2\) 

\(\Leftrightarrow\left[{}\begin{matrix}x-1=-x^2+3x-2\left(1\right)\left(Khix\ge1\right)\\x-1=x^2-3x+2\left(2\right)\left(Khix< 1\right)\end{matrix}\right.\) 

Từ (1) \(\Leftrightarrow x^2-3x+2+x-1=0\Leftrightarrow x^2-2x+1=0\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\left(TM\right)\)

Từ (2) \(\Leftrightarrow x^2-3x+2-x+1=0\Leftrightarrow x^2-4x+3=0\Leftrightarrow x^2-x-3x+3=0\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\left(L\right)\\x=3\left(L\right)\end{matrix}\right.\) Vậy ...

2 Đặt \(a=2x^2+x+3,b=2x^2-x+3\Rightarrow x=\dfrac{a-b}{2}\) 

PT trở thành \(\dfrac{9\left(a-b\right)}{2a}-\dfrac{a-b}{2b}=8\Rightarrow9b\left(a-b\right)-a\left(a-b\right)=16ab\Leftrightarrow9ab-9b^2-a^2+ab=16ab\)

\(\Leftrightarrow a^2+6ab+9b^2=0\Leftrightarrow\left(a+3b\right)^2=0\) \(\Leftrightarrow a+3b=0\Leftrightarrow a=-3b\)

\(\Rightarrow2x^2+x+3=-3\left(2x^2-x+3\right)\Leftrightarrow2x^2+x+3=-6x^2+3x-9\Leftrightarrow8x^2-2x+12=0\)

\(\Leftrightarrow4x^2-x+6=0\Leftrightarrow4x^2-x+\dfrac{1}{16}+\dfrac{95}{16}=0\Leftrightarrow\left(2x-\dfrac{1}{4}\right)^2=-\dfrac{95}{16}\) Vô lí \(\Rightarrow\) Ko có x 

Vậy...