Cho biểu thức sau: A=
a,Tìm điều kiện của x để A có nghĩa
b, Rút gọn A
c,Tìm x để A=
a) Ta có: $x^2+x-6=x^2+3x-2x-6=x(x+3)-2(x+3)=(x-2)(x+3)$
[tex]DK:\left\{\begin{matrix} x+3\neq 0 & \\ x-2\neq 0 & \end{matrix}\right.\\\Leftrightarrow \left\{\begin{matrix} x\neq -3 & \\ x\neq 2 & \end{matrix}\right.[/tex]
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b)[tex]\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}\\=\frac{x+2}{x+3}-\frac{5}{(x-2)(x+3)}-\frac{1}{x-2}\\=\frac{(x+2)(x-2)-5-x-3}{(x-2)(x+3)}\\=\frac{x^2-4-5-x-3}{(x-2)(x+3)}\\=\frac{x^2-x-12}{(x-2)(x+3)}\\=\frac{x^2+3x-4x-12}{(x-2)(x+3)}\\=\frac{x(x+3)-4(x+3)}{(x-2)(x+3)}\\=\frac{(x-4)(x+3)}{(x-2)(x+3)}\\=\frac{x-4}{x-2}[/tex]
c)[tex]A=\frac{-3}{4}\\\Rightarrow \frac{x-4}{x-2}=\frac{-3}{4}(x\neq -3;x\neq 2)\\\Leftrightarrow 4(x-4)=(-3)(x-2)\\\Leftrightarrow 4x-16=-3x+6\\\Leftrightarrow 4x+3x=6+16\\\Leftrightarrow 7x=22\\\Leftrightarrow x=\frac{22}{7}(TM)[/tex]
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