cho biểu thức P =[tex](\frac{2x}{x+3}+\frac{x}{x-3}+\frac{3x^{2}}{9-x^{2}}) : (\frac{2x-2}{x-3} - 1)[/tex]
a) Rút gọn P
b) Tìm x để P < -1/2
a)P =[tex](\frac{2x}{x+3}+\frac{x}{x-3}+\frac{3x^{2}}{9-x^{2}}) : (\frac{2x-2}{x-3} - 1)[/tex]
=[tex](\frac{2x}{x+3}+\frac{x}{x-3}-\frac{3x^{2}}{(x+3)(x-3)}) : (\frac{2x-2}{x-3} - 1)[/tex]
[tex]=\frac{2x(x-3)+x\left ( x+3 \right )-3x^{2}}{\left ( x+3 \right )\left ( x-3 \right )}:\frac{2x-2-x+3}{x-3}[/tex]
[tex]=\frac{2x^{2}-6x+x^{2}+3x-3x^{2}}{\left ( x+3 \right )\left ( x-3 \right )}.\frac{x-3}{x+1}[/tex]
[tex]=\frac{-3x}{\left ( x-3 \right )\left ( x+3 \right )}.\frac{x-3}{x+1}[/tex]
[tex]=\frac{-3x}{\left ( x+3 \right )\left ( x+1 \right )}[/tex]
b) P[tex]<\frac{1}{2}[/tex]
[tex]\Leftrightarrow \frac{-3x}{\left ( x+3 \right )(x-1)}+\frac{1}{2}<0[/tex]
[tex]\Leftrightarrow \frac{-3x}{2\left ( x+3 \right )\left ( x+1 \right )}+\frac{\left ( x+3 \right )\left ( x+1\right )}{2\left ( x+3 \right )\left ( x+1 \right )}<0[/tex]
[tex]\Leftrightarrow \frac{-6x+x^{2}+4x+3}{2\left ( x+3 \right )\left ( x+1 \right )}<0[/tex]
[tex]\Leftrightarrow \frac{x^{2}-2x+3}{2\left ( x+3 \right )\left ( x+1 \right )}<0[/tex]
Vì [tex]x^{2}-2x+3=\left ( x-1 \right )^{2}+2>0[/tex]
Nên [tex]2\left ( x+3 \right )\left ( x+1 \right )<0[/tex]
Vì x+3>x+1 nên chỉ có 1 TH
[tex]\left\{\begin{matrix} x+3>0\\ x+1<0 \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x>-3\\ x<-1 \end{matrix}\right.\Leftrightarrow -3<x<-1[/tex]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
