Question

[KAl(SO4)2·12H2O]

Bài 1: rút gọn phân thức

a) \(\frac{14xy^2\left(2x-3y\right)}{21x^2y\left(2x-3y\right)^2}\)

b) \(\frac{8xy\left(3x-1\right)^2}{12x^3\left(1-3x\right)}\)

c) \(\frac{20x^2-45}{\left(2x+3\right)^2}\)

d) \(\frac{5x^2-10xy}{2\left(2y-x\right)^3}\)

e) \(\frac{80x^3-125x}{3\left(x-3\right)-\left(x-3\right)\left(8-4x\right)}\)

f) \(\frac{9-\left(x+5\right)^2}{x^2+4x+4}\)

g) \(\frac{32x-8x^2+2x^3}{x^3+64}\)

h) \(\frac{5x^3+5x}{x^4-1}\)

Bài 2: Quy đồng mẫu thức của các phân thức sau

a) \(\frac{7x-1}{2x^2+6x};\frac{5-3x}{x^2-9}\)

b) \(\frac{x+1}{x-x^2};\frac{x+2}{2-4x+2x^2}\)

c) \(\frac{4x^2-3x+5}{x^3-1};\frac{2x}{x^2+x+1};\frac{6}{x-1}\)

d) \(\frac{7}{5x};\frac{4}{x-2y};\frac{x-y}{8y^2-2x^2}\)

Answer 1

[Agatsuma Zenitsu]

Bài 2: \(a,\frac{7x-1}{2x^2+6x}=\frac{7x-1}{2x\left(x+3\right)}=\frac{\left(7x-1\right)\left(x-3\right)}{2x\left(x+3\right)\left(x-3\right)}\) 

 \(\frac{5-3x}{x^2-9}=\frac{5-3x}{\left(x-3\right)\left(x+3\right)}=\frac{\left(5-3x\right)2x}{2x\left(x-3\right)\left(x+3\right)}\)

\(b,\frac{x+1}{x-x^2}=\frac{x+1}{x\left(1-x\right)}=-\frac{x+1}{x\left(x+1\right)}=-\frac{2\left(x-1\right)\left(x+1\right)}{2x\left(x-1\right)^2}\) 

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

\(c,\frac{4x^2-3x+5}{x^3-1}=\frac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}\) 

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

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

\(d,\frac{7}{5x}=\frac{7.2\left(2y-x\right)\left(2y+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)

\(\frac{4}{x-2y}=-\frac{4}{2y-x}=-\frac{4.2.5x\left(2x+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)

\(\frac{x-y}{8y^2-2x^2}=\frac{x-y}{2\left(4y^2-x^2\right)}=\frac{x-y}{2\left(2y-x\right)\left(2y+x\right)}=\frac{5x\left(x-y\right)}{2.5x.\left(2y-x\right)\left(2y+x\right)}\)