Bài 1: Phân thức đại số.

a: \(=\dfrac{x+3-x-4}{x-2}=\dfrac{-1}{x-2}\)

b: \(=\dfrac{x^2+x+4x+6}{2x\left(x+3\right)}=\dfrac{\left(x+3\right)\left(x+2\right)}{2x\left(x+3\right)}=\dfrac{x+2}{2x}\)

c: \(=\dfrac{3x-x+6}{x\left(2x+6\right)}=\dfrac{2x+6}{x\left(2x+6\right)}=\dfrac{1}{x}\)

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

h: \(=\dfrac{x^2-x^2-2x-2x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{-4x+4}{\left(x-2\right)\left(x+2\right)}\)

Bài 3:

a: ĐKXĐ: x<>0; x<>4

b: \(A=\dfrac{2\left(x-4\right)}{x\left(x-4\right)}=\dfrac{2}{x}\)

Khi x=2 thì A=2/2=1

Đề bài là gì bạn?

Phan tich da thuc thanh nhan tu

`4x^2-6x`

`=2x.2x-2x.3`

`=2x(2x-3)`

\(=\dfrac{x-y}{y\left(x+y\right)}+\dfrac{3x+y}{x\left(x-y\right)}\cdot\dfrac{x-y}{x+y}\)

\(=\dfrac{x-y}{y\left(x+y\right)}+\dfrac{3x+y}{x\left(x+y\right)}\)

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

a: ĐKXĐ: x<>0; x<>2

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

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

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

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

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

c: Khi x=1/2 thì \(C=\left(\dfrac{1}{2}+1\right):\left(2\cdot\dfrac{1}{2}\right)=\dfrac{3}{2}\)

a: ĐKXĐ: x<>0; x<>2

b: \(A=\dfrac{\left(x-2\right)\left(x+2\right)}{x\left(x^2-4x+4\right)}=\dfrac{\left(x-2\right)\left(x+2\right)}{x\left(x-2\right)^2}=\dfrac{x+2}{x\left(x-2\right)}\)

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

\(=\dfrac{3\left(2x-1\right)}{2x\left(2x-1\right)}+\dfrac{2x\left(3x-3\right)}{2x\left(2x-1\right)}-\dfrac{2x^2+1}{2x\left(2x-1\right)}\)
\(=\dfrac{6x-3+6x^2-6x-2x^2-1}{2x\left(2x-1\right)}\)

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

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

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

\(=\dfrac{6x-3+6x^2-6x-2x^2-1}{2x\left(2x-1\right)}=\dfrac{4x^2-4}{2x\left(2x-1\right)}=\dfrac{4\left(x^2-1\right)}{2x\left(2x-1\right)}=\dfrac{2\left(x^2-1\right)}{x\left(2x-1\right)}\)

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

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

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

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

\(=\left[\dfrac{x^2+y^2}{x^2y^2}+\dfrac{2}{x-y}\cdot\dfrac{y-x}{xy}\right]:\dfrac{x^2-y^2}{x^2y^2}\)

\(=\dfrac{x^2+y^2-2xy}{x^2y^2}\cdot\dfrac{x^2y^2}{\left(x-y\right)\left(x+y\right)}=\dfrac{x-y}{x+y}\)

Chọn A

Đáp án: A
Vì: \(\dfrac{\left(-1\right)^2}{-1.2+3}=\dfrac{1}{1}=1\)
       \(\dfrac{3^2}{2.3+3}=\dfrac{9}{9}=1\)

A