a: THay x=4 vào B, ta được:
\(B=1-\frac{1}{4+3}=1-\frac17=\frac67\)
b: \(A=\frac{2x^2+4x+13}{x^2-9}+\frac{3}{3-x}-\frac{x}{x+3}\)
\(=\frac{2x_{}^2+4x+13}{\left(x-3\right)\left(x+3\right)}-\frac{3}{x-3}-\frac{x}{x+3}\)
\(=\frac{2x^2+4x+13-3\left(x+3\right)-x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{2x^2+4x+13-3x-9-x^2+3x}{\left(x-3\right)\left(x+3\right)}=\frac{x^2+4x+4}{\left(x-3\right)\left(x+3\right)}=\frac{\left(x+2\right)^2}{\left(x-3\right)\left(x+3\right)}\)
P=A:B
\(=\frac{\left(x+2\right)^2}{\left(x-3\right)\left(x+3\right)}:\frac{x+2}{x+3}\)
\(=\frac{\left(x+2\right)^2}{\left(x-3\right)\left(x+3\right)}\cdot\frac{x+3}{x+2}=\frac{x+2}{x-3}\)
d: \(P\left(-2x^2+9x-9\right)\)
\(=\frac{x+2}{x-3}\cdot\left(-1\right)\cdot\left(2x^2-9x+9\right)\)
\(=\frac{-\left(x+2\right)}{x-3}\cdot\left(x-3\right)\left(2x-3\right)=-\left(2x-3\right)\left(x+2\right)\)
\(=-\left(2x^2+4x-3x-6\right)=-\left(2x^2+x-6\right)\)
\(=-2\left(x^2+\frac12x-3\right)=-2\left(x^2+2\cdot x\cdot\frac14+\frac{1}{16}-\frac{49}{16}\right)\)
\(=-2\left(x+\frac14\right)^2+\frac{49}{16}\le\frac{49}{16}\forall x\)
Dấu '=' xảy ra khi \(x+\frac14=0\)
=>\(x=-\frac14\)
