Bài 4:
a: \(P=\dfrac{x^3-3}{x^2-2x-3}-\dfrac{2\left(x-3\right)}{x+1}+\dfrac{x+3}{3-x}\)
\(=\dfrac{x^3-3}{\left(x-3\right)\left(x+1\right)}-\dfrac{2\left(x-3\right)^2}{\left(x+1\right)\left(x-3\right)}-\dfrac{\left(x+3\right)\left(x+1\right)}{\left(x-3\right)\left(x+1\right)}\)
\(=\dfrac{x^3-3-2x^2+12x-18-x^2-4x-3}{\left(x-3\right)\left(x+1\right)}=\dfrac{x^3-3x^2+8x-24}{\left(x-3\right)\left(x+1\right)}=\dfrac{x^2+8}{x+1}\)
b: \(P-3=\dfrac{x^2+8-3x-3}{x+1}=\dfrac{x^2-3x+5}{x+1}=\dfrac{\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{4}}{x+1}\)
x>0 nên x+1>1>0
mà \(\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{4}>=\dfrac{11}{4}>0\)
nên P-3>0
=>P>3
c: Để P nguyên thì \(x^2+8⋮x+1\)
=>\(x^2-1+9⋮x+1\)
=>\(9⋮x+1\)
=>\(x+1\in\left\{1;-1;3;-3;9;-9\right\}\)
=>\(x\in\left\{0;-2;2;-4;8;-10\right\}\)
Bài 5:
a: \(P=\dfrac{2x}{x+3}\cdot\left(1+\dfrac{1}{x+2}\right)+\dfrac{9x+1}{x^2+3x+2}\)
\(=\dfrac{2x}{x+3}\cdot\dfrac{x+3}{x+2}+\dfrac{9x+1}{\left(x+1\right)\left(x+2\right)}\)
\(=\dfrac{2x}{x+2}+\dfrac{9x+1}{\left(x+1\right)\left(x+2\right)}=\dfrac{2x\left(x+1\right)+9x+1}{\left(x+2\right)\left(x+1\right)}\)
\(=\dfrac{2x^2+11x+1}{\left(x+1\right)\left(x+2\right)}\)
b: Thay x=2 vào P, ta được:
\(P=\dfrac{2\cdot2^2+11\cdot2+1}{\left(2+1\right)\left(2+2\right)}=\dfrac{8+22+1}{3\cdot4}=\dfrac{31}{12}\)
c: P=4
=>\(4\left(x+1\right)\left(x+2\right)=2x^2+11x+1\)
=>\(4x^2+12x+8-2x^2-11x-1=0\)
=>\(2x^2+x+7=0\)(1)
\(\Delta=1^2-4\cdot2\cdot7=1-56=-55< 0\)
=>Phương trình (1) vô nghiệm
