Bài 1:

a: \(\frac{1}{2x^3y}=\frac{1\cdot6\cdot yz^3}{2x^3y\cdot6yz^3}=\frac{6yz^3}{12x^3y^2z^3}\)

\(\frac{2}{3xy^2z^3}=\frac{2\cdot4\cdot x^2}{3xy^2z^3\cdot4x^2}=\frac{8x^2}{12x^3y^2z^3}\)

\(\frac{5}{4yz}=\frac{5\cdot3\cdot x^3\cdot y\cdot z^2}{4yz\cdot3x^3yz^2}=\frac{15x^3yz^2}{12x^3y^2z^3}\)

b: \(\frac{x+1}{10x^3-40x}=\frac{x+1}{10x\left(x^2-4\right)}=\frac{x+1}{10x\left(x-2\right)\left(x+2\right)}\)

\(=\frac{\left(x+1\right)\cdot4\cdot x}{4x\cdot10x\cdot\left(x+2\right)\left(x-2\right)}=\frac{4x^2+4x}{40x^2\left(x+2\right)\left(x-2\right)}\)

\(\frac{5}{8x^3+16x^2}=\frac{5x}{8x^2\left(x+2\right)}\)

\(=\frac{5x\cdot5\cdot\left(x-2\right)}{8x^2\left(x+2\right)\cdot5\cdot\left(x-2\right)}=\frac{25x^2-50x}{40x^2\left(x+2\right)\left(x-2\right)}\)

Bài 2:

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

\(=\frac{\left(x-2\right)\cdot4x\cdot\left(x^2+x+1\right)}{3x\left(x-1\right)\cdot4x\cdot\left(x^2+x+1\right)}=\frac{\left(4x^2-8x\right)\left(x_{}^2+x+1\right)}{12x^2\left(x-1\right)\left(x^2+x+1\right)}\)

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

\(=\frac{\left(x^2-2\right)\cdot3}{4x^2\left(x-1\right)\left(x^2+x+1\right)\cdot3}=\frac{3x^2-6}{12x^2\left(x-1\right)\left(x^2+x+1\right)}\)

Bài 2:

a: \(\dfrac{1}{2x^3y}=\dfrac{6yz^3}{12x^3y^2z^3}\)

\(\dfrac{2}{3xy^2z^3}=\dfrac{2\cdot4x^2}{12x^3y^2z^3}=\dfrac{8x^2}{12x^3y^2z^3}\)

a) MTC: \(12x^3y^3\)

\(\dfrac{3}{4x^3y^2}=\dfrac{3\cdot3y}{4x^3y^2\cdot3y}=\dfrac{9y}{12x^3y^3}\)

\(\dfrac{2}{3xy^3}=\dfrac{2\cdot4x^2}{3xy^3\cdot4x^2}=\dfrac{8x^2}{12x^3y^3}\)

b) MTC: \(x\left(x-3\right)^2\)

\(\dfrac{5}{x^2-6x+9}=\dfrac{5}{\left(x-3\right)^2}=\dfrac{5x}{x\left(x-3\right)^2}\)

\(\dfrac{3}{x^2-3x}=\dfrac{3}{x\left(x-3\right)}=\dfrac{3\left(x-3\right)}{x\left(x-3\right)^2}=\dfrac{3x-9}{x\left(x-3\right)^2}\)

a)MTC:\(12x^5y^4\)

\(\dfrac{5}{x^5y^3}=\dfrac{5\cdot12y}{x^5y^3\cdot12y}=\dfrac{60y}{12x^5y^4}\)

\(\dfrac{7}{12x^3y^4}=\dfrac{7\cdot x^2}{12x^3y^4\cdot x^2}=\dfrac{7x^2}{12x^5y^4}\)

b)MTC:\(60x^4y^5\)

\(\dfrac{4}{15x^3y^5}=\dfrac{4\cdot4x}{15x^3y^5\cdot4x}=\dfrac{16x}{60x^4y^5}\)

\(\dfrac{11}{12x^4y^2}=\dfrac{11\cdot5y^3}{12x^4y^2\cdot5y^3}=\dfrac{55y^3}{60x^4y^5}\)

a: \(\frac{x+y}{x^2\left(y+z\right)}=\frac{\left(x+y\right)\cdot y^2z^2\left(x+z\right)\left(x+y\right)}{x^2y^2z^2\left(x+y\right)\left(y+z\right)\left(x+z\right)}=\frac{\left(x+z\right)\cdot y^2z^2\left(x+y\right)^2}{x^2y^2z^2\left(x+y\right)\left(y+z\right)\left(x+z\right)}\)

\(\frac{y+z}{y^2\left(x+z\right)}=\frac{\left(y+z\right)\cdot x^2z^2\left(x+y\right)\left(y+z\right)}{x^2y^2z^2\left(x+y\right)\left(y+z\right)\left(x+z\right)}=\frac{x^2z^2\left(y+z\right)^2\cdot\left(x+y\right)}{x^2y^2z^2\left(x+y\right)\left(y+z\right)\left(x+z\right)}\)

\(\frac{z+x}{z^2\left(x+y\right)}=\frac{\left(z+x\right)\cdot x^2y^2\cdot\left(x+z\right)\left(y+z\right)}{x^2y^2z^2\left(x+y\right)\left(x+z\right)\left(y+z\right)}=\frac{x^2y^2\left(x+z\right)^2\cdot\left(y+z\right)}{x^2y^2z^2\left(x+y\right)\left(y+z\right)\left(x+z\right)}\)

b: \(\frac{5x}{x^2+5x+6}=\frac{5x}{\left(x+2\right)\left(x+3\right)}=\frac{5x\left(x+5\right)}{\left(x+2\right)\left(x+3\right)\left(x+5\right)}\)

\(\frac{2x+3}{x^2+7x+10}=\frac{2x+3}{\left(x+2\right)\left(x+5\right)}=\frac{\left(2x+3\right)\left(x+3\right)}{\left(x+2\right)\left(x+3\right)\left(x+5\right)}\)

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

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

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

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

d: \(\frac{2}{x^2+5x}=\frac{2}{x\left(x+5\right)}=\frac{2\left(x+5\right)}{x\left(x+5\right)^2}=\frac{2x+10}{x\left(x+5\right)^2}\)

\(\frac{x+5}{x^2+10x+25}=\frac{x+5}{\left(x+5\right)^2}=\frac{x\left(x+5\right)}{x\left(x+5\right)^2}\)

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

Tại sao lại là x+2 và x-2

*Muốn quy đồng mẫu thức nhiều phân thức ta làm như sau:

-Phân tích các mẫu thức thành nhân tử rồi tìm ẫu tức chung.

-Tìm nhân tử phụ của mỗi mẫu thức.

-Nhân cả tử và mẫu của mỗi phân thức với nhân tử phụ tương ứng.

*Bài tập:

\(\dfrac{x}{x^2+2x+1}và\)\(\dfrac{3}{5x^2-5}\)

-Ta có:

x²+2x+1=(x+1)²=(x+1)(x+1)

5x²-5=5(x²-1)=5(x-1)(x+1)

\(\Rightarrow\)MTC:5(x-1)(x+1)(x+1)

-NTP:5(x-1)(x+1)(x+1):(x+1)(x+1)=5(x-1)

5(x-1)(x+1)(x+1):5(x-1)(x+1)=x+1

-Quy đồng mẫu thức:

\(\dfrac{x}{\left(x+1\right)\left(x+1\right)}\)=\(\dfrac{5x\left(x-1\right)}{5\left(x-1\right)\left(x+1\right)\left(x+1\right)}\)

\(\dfrac{3}{5\left(x-1\right)\left(x+1\right)}=\dfrac{3\left(x+1\right)}{5\left(x-1\right)\left(x+1\right)\left(x+1\right)}\)

a) \(MTC=a^2x^2b^2\)

\(NTP:a^2x^2b^2:a^2x=xb^2\)

\(a^2x^2b^2:x^2b=a^2b\)

\(a^2x^2b^2:b^2a=ax^2\)

Quy đồng :

\(\dfrac{a+x}{a^2x}=\dfrac{\left(a+x\right)\cdot xb^2}{a^2x.xb^2}=\dfrac{axb^2+x^2b^2}{a^2x^2b^2}\)

\(\dfrac{a+b}{x^2b}=\dfrac{\left(a+b\right)\cdot a^2b}{x^2b\cdot a^2b}=\dfrac{a^3b+a^2b^2}{a^2x^2b^2}\)

\(\dfrac{b+a}{b^2a}=\dfrac{\left(b+a\right)\cdot ax^2}{b^2a\cdot ax^2}=\dfrac{abx^2+a^2x^2}{a^2x^2b^2}\)