Question

1) x² + x/ 4x + 4

2) x² - 6x + 9/ ( x - 3)³

3) x² + 2x + 1/ 2x² + 2x

4) x²y²z²t³/ 3xy³zt²

5) 2xy ( x+ 3)²/ 10 x³y² ( x+3)⁵

mink đag cần gấp(đây là dạng bài rút gọn phân thức)

Answer 1

Lời giải:

1. \(\frac{x^2+x}{4x+4}=\frac{x(x+1)}{4(x+1)}=\frac{x}{4}\)

2. \(\frac{x^2-6x+9}{(x-3)^3}=\frac{x^2-2.x.3+3^2}{(x-3)^3}=\frac{(x-3)^2}{(x-3)^3}=\frac{1}{x-3}\)

3. \(\frac{x^2+2x+1}{2x^2+2x}=\frac{(x+1)^2}{2x(x+1)}=\frac{x+1}{2x}\)

4. \(\frac{x^2y^2z^2t^3}{3xy^3zt^2}=\frac{xzt}{3y}\)

5. \(\frac{2xy(x+3)^2}{10x^3y^2(x+3)^5}=\frac{2xy(x+3)^2}{2xy.(x+3)^2.5x^2y(x+3)^3}=\frac{1}{5x^2y(x+3)^3}\)

Answer 2

1

Ta có: \(\frac{x^2+x}{4x+4}=\frac{x\left(x+1\right)}{4\left(x+1\right)}=\frac{x\left(x+1\right):\left(x+1\right)}{4\left(x+1\right):\left(x+1\right)}=\frac{x}{4}\)

2

Ta có: \(\frac{x^2-6x+9}{\left(x-3\right)^3}=\frac{\left(x-3\right)^2}{\left(x-3\right)^3}=\frac{\left(x-3\right)^2:\left(x-3\right)^2}{\left(x-3\right)^3:\left(x-3\right)^2}=\frac{1}{x-3}\)

3

Ta có: \(\frac{x^2+2x+1}{2x^2+2x}=\frac{\left(x+1\right)^2}{2x\left(x+1\right)}=\frac{\left(x+1\right)^2:\left(x+1\right)}{2x\left(x+1\right):\left(x+1\right)}=\frac{x+1}{2x}\)