\(A=\dfrac{2x}{x+3}-\dfrac{x+1}{3-x}-\dfrac{3-11x}{x^2-9}\)
\(=\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}-\dfrac{3-11x}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{2x\left(x-3\right)+\left(x+1\right)\left(x+3\right)-3+11x}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{2x^2-6x+x^2+4x+3-3+11x}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{3x^2+9x}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x}{x-3}\)
b) Ta đã rút gọn được \(A=\frac{3x}{x-3}\)
TH1: \(x-3>0\rightarrow x> 3\). Khi đó:
\( \frac{3x}{x-3}<2\)
\(\Leftrightarrow 3x< 2(x-3)\Leftrightarrow x< -6\) (vô lý)
TH2: \(x-3> 0\rightarrow x< 3\). Khi đó:
\(\frac{3x}{x-2}<2 \Leftrightarrow 3x> 2(x-3)\) (nhân với một số âm thì phải đổi dấu)
\(\Leftrightarrow x> -6\)
Vậy \(3> x> -6\) thì \(A< 2\)
a)
\(\dfrac{2x}{x+3}-\dfrac{x+1}{3-x}-\dfrac{3-11x}{x^2-9}=\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}-\dfrac{3-11x}{x^2-9}\\ =\dfrac{2x\left(x-3\right)+\left(x+1\right)\left(x+3\right)-3+11x}{x^2-9}\\ =\dfrac{2x^2-6x+x^2+4x+3-3+11x}{x^2-9}\\ =\dfrac{3x^2+9x}{x^2-9}=\dfrac{3x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x}{x-3}\)
b)
\(A< 2\\ \Leftrightarrow\dfrac{3x}{x-3}< 2\\ \Leftrightarrow3x< 2x-6\\ \Leftrightarrow x< -6\)
vậy khi x<-6 thì A<2
c)
\(A=\dfrac{3x}{x-3}=\dfrac{3x-9+9}{x-3}\\ A=3+\dfrac{9}{x-3}\)
A nguyên khi \(\dfrac{9}{x-3}\) nguyên
\(\Rightarrow x-3=\left\{-9;-3;-1;1;3;9\right\}\)
| x-3 | -9 | -3 | -1 | 1 | 3 | 9 |
| x | -6 | 0 | 2 | 4 | 6 | 12 |
vậy khi x={-6;0;2;4;6;12} thì A có GT nguyên
a,
\(A=\dfrac{2x}{x+3}-\dfrac{x+1}{3-x}-\dfrac{3-11x}{x^2-9}\)
\(A=\dfrac{2x^2-6x}{\left(x-3\right)\left(x+3\right)}+\dfrac{x^2+4x+3}{\left(x-3\right)\left(x+3\right)}-\dfrac{3-11x}{\left(x-3\right)\left(x+3\right)}\)
\(A=\dfrac{3x^2+9x+1}{\left(x-3\right)\left(x+3\right)}\)
