a) ĐKXĐ: \(x\ne-5.\)
Thay \(x=7\) (TMĐK) vào biểu thức A ta có:
\(A=\dfrac{7-3}{7+5}=\dfrac{4}{12}=\dfrac{1}{3}.\)
Vậy \(x=7\) thì \(A=\dfrac{1}{3}.\)
b) \(B=\dfrac{x}{x-3}+\dfrac{2x}{x+3}+\dfrac{3x^2+9}{9-x^2}\left(ĐKXĐ:x\ne\pm3\right).\)
\(B=\dfrac{x}{x-3}+\dfrac{2x}{x+3}+\dfrac{-3x^2-9}{\left(x-3\right)\left(x+3\right)}.\)
\(B=\dfrac{x\left(x+3\right)+2x\left(x-3\right)-3x^2-9}{\left(x-3\right)\left(x+3\right)}.\)
\(B=\dfrac{x^2+3x+2x^2-6x-3x^2-9}{\left(x-3\right)\left(x+3\right)}.\)
\(B=\dfrac{-3x-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{-3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{-3}{x-3}.\)
c) Để \(B=\dfrac{2}{x}.\) \(\Rightarrow\dfrac{-3}{x-3}=\dfrac{2}{x}\left(ĐKXĐ:x\ne0;x\ne3\right).\)
\(\Leftrightarrow-3x=2x-6.\Leftrightarrow-5x=-6.\Leftrightarrow x=\dfrac{6}{5}\left(TMĐK\right).\)
d) Ta có: \(P=A.B.\)
\(\Rightarrow P=\dfrac{x-3}{x+5}.\dfrac{-3}{x-3}=\dfrac{-3}{x+5}\left(ĐKXĐ:x\ne3;x\ne-5\right).\)
Để biểu thức \(P\in Z.\) \(\Leftrightarrow\dfrac{-3}{x+5}\in Z.\Rightarrow x+5\inƯ\left(-3\right)=\left\{3;-3;1;-1\right\}.\)
Ta có bảng sau:
| x+5 | 3 | -3 | 1 | -1 |
| x | -2 (TMĐK). | -8 (TMĐK). | -4 (TMĐK). | -6 (TMĐK). |
Vậy \(x\in\left\{-2;-8;-4;-6\right\}.\)
