Question

làm hộ mk với mai mk phải nộp rồi,cảm ơn truovcs nhá

Answer 1

Bài 10:

a: Thay x=3 vào A, ta được:

\(A=\left(\dfrac{1}{3+2}+\dfrac{1}{3^2-4}\right)=\dfrac{1}{5}+\dfrac{1}{5}=\dfrac{2}{5}\)

b: Ta có: P=AB

\(=\left(\dfrac{1}{x+2}+\dfrac{1}{x^2-4}\right)\cdot\dfrac{x^2+2x}{x-1}\)

\(=\dfrac{x-2+1}{\left(x+2\right)\left(x-2\right)}\cdot\dfrac{x\left(x+2\right)}{x-1}\)

\(=\dfrac{x-1}{x-2}\cdot\dfrac{x}{x-1}\)

\(=\dfrac{x}{x-2}\)

c: Để \(P=\dfrac{2}{3}\) thì \(\dfrac{x}{x-2}=\dfrac{2}{3}\)

\(\Leftrightarrow3x=2x-4\)

hay x=-4(nhận)