bài 4
a)
\(\dfrac{4}{x+2}+\dfrac{3}{x-2}-\dfrac{5x+2}{x^2-4}\\ =\dfrac{4}{x+2}+\dfrac{3}{x-2}-\dfrac{5x+2}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{4\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5x+2}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{4x-8+3x+6-5x-2}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{2x-4}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{2}{x+2}\)
b)
với x=-3 (tmdk) ta có
\(\dfrac{2}{-3+2}=\dfrac{2}{-1}=-2\)
bài 5
a) \(\dfrac{x+1}{x+3}\left(ĐK:x\ne-3\right)\)
với x=5 (tmđk) ta có
\(\dfrac{5+1}{5+3}=\dfrac{6}{8}=\dfrac{3}{4}\)
b)
\(\dfrac{3}{x-3}-\dfrac{6x}{9-x^2}+\dfrac{x}{x+3}\left(ĐK:x\ne3;x\ne-3\right)\)
\(=\dfrac{3}{x-3}+\dfrac{6x}{\left(x-3\right)\left(x+3\right)}+\dfrac{x}{x+3}\\ =\dfrac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{6x}{\left(x-3\right)\left(x+3\right)}+\dfrac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{3x+9+6x+x^2-3x}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{x^2+6x+9}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{x+3}{x-3}\)
c)
\(P=A\cdot B\\ =>P=\dfrac{x+1}{x+3}\cdot\dfrac{x+3}{x-3}\\ =\dfrac{x+1}{x-3}\)