Question

Gấp

Answer 1

lên Gaumath mà tìm á 

Answer 2

a, Thay x = 3 vào A, ta có :

\(A=\dfrac{x-1}{x^2}=\dfrac{3-1}{3^2}=\dfrac{2}{9}\)

b, \(B=\dfrac{1}{x}-\dfrac{x}{2x+1}+\dfrac{2x^2-3x-1}{x\left(2x+1\right)}\left(MSC:x\left(2x+1\right)\right)\)

\(=\dfrac{2x+1}{x\left(2x+1\right)}-\dfrac{x^2}{x\left(2x+1\right)}+\dfrac{2x^2-3x-1}{x\left(2x+1\right)}\)

\(=\dfrac{2x+1-x^2+2x^2-3x-1}{x\left(2x+1\right)}\)

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

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

\(B=\dfrac{x-1}{2x+1}\)