a: \(A=\frac{x}{x+5}-\frac{7x-15}{25-x^2}+\frac{3}{x-5}\)
\(=\frac{x}{x+5}+\frac{7x-15}{\left(x+5\right)\left(x-5\right)}+\frac{3}{x-5}\)
\(=\frac{x\left(x-5\right)+7x-15+3\left(x+5\right)}{\left(x+5\right)\left(x-5\right)}=\frac{x^2-5x+7x-15+3x+15}{\left(x-5\right)\left(x+5\right)}\)
\(=\frac{x^2+5x}{\left(x-5\right)\left(x+5\right)}=\frac{x\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}=\frac{x}{x-5}\)
b:|2x-1|=9
=>2x-1=9 hoặc 2x-1=-9
=>2x=10 hoặc 2x=-8
=>x=5(loại) hoặc x=-4(nhận)
Thay x=-4 vào A, ta được:
\(A=\frac{-4}{-4-5}=\frac{-4}{-9}=\frac49\)
c: \(P=A\cdot B\)
\(=\frac{x}{x-5}\cdot\left(x-5\right)\left(x+3\right)=x\left(x+3\right)\)
\(=x^2+3x+\frac94-\frac94=\left(x+\frac32\right)^2-\frac94\ge-\frac94\forall x\)
Dấu '=' xảy ra khi x+3/2=0
=>x=-3/2(nhận)
