`MSC:12`

`2/3=(2xx4)/(3xx4)=8/12` và `5/12`

`2/3=`\(\dfrac{2\times4}{3\times4}\)`=8/12` và `5/12`

\(\dfrac{1}{2}+\dfrac{2}{4}\\ \dfrac{2}{4}=\dfrac{1\cdot2}{2\cdot2}=\dfrac{2}{4}\\ \dfrac{3}{-12}+\dfrac{4}{12}\\ \dfrac{3}{-12}=\dfrac{-3}{12}\\ \dfrac{4}{6}=\dfrac{4\cdot2}{6\cdot2}=\dfrac{8}{12}\\ \dfrac{-3}{12}+\dfrac{8}{13}=\dfrac{5}{12}\)

\(\dfrac{8}{13}?\)

\(\dfrac{-3}{4}=\dfrac{-3.9}{4.9}=\dfrac{-27}{36}\)

\(\dfrac{5}{9}=\dfrac{5.4}{9.4}=\dfrac{20}{36}\)

\(\dfrac{2}{3}=\dfrac{2.12}{3.12}=\dfrac{24}{36}\)

a) \(\dfrac{2}{36}=\dfrac{1}{18}\)

\(\dfrac{8}{12}=\dfrac{2}{3}\)

b) \(\dfrac{10}{25}=\dfrac{2}{5}\)

\(\dfrac{14}{40}=\dfrac{7}{20}\)

1) 

\(\dfrac{3}{4}=\dfrac{3\times3}{4\times3}=\dfrac{9}{12}\) 

\(\dfrac{2}{12}\) (giữ nguyên)

2)

\(\dfrac{1}{4}=\dfrac{1\times3}{4\times3}=\dfrac{3}{12}\)

\(\dfrac{2}{3}=\dfrac{2\times4}{3\times4}=\dfrac{8}{12}\)

a) \(\dfrac{3}{4}\) và \(\dfrac{2}{12}\)

\(\dfrac{3}{4}=\dfrac{3\times3}{4\times3}=\dfrac{9}{12}\) ; giữ nguyên \(\dfrac{2}{12}\)

Quy đồng mẫu số hai phân số \(\dfrac{3}{4}\) và \(\dfrac{2}{12}\) được hai phân số \(\dfrac{9}{12}\) và \(\dfrac{2}{12}\).

b) \(\dfrac{1}{4}\) và \(\dfrac{2}{3}\)

\(\dfrac{1}{4}=\dfrac{1\times3}{4\times3}=\dfrac{3}{12}\) ; \(\dfrac{2}{3}=\dfrac{2\times4}{3\times4}=\dfrac{8}{12}\)

Quy đồng mẫu số hai phân số \(\dfrac{1}{4}\) và \(\dfrac{2}{3}\) được hai phân số \(\dfrac{3}{12}\) và \(\dfrac{8}{12}\).

\(\dfrac{2}{6}=\dfrac{2\times5}{6\times5}=\dfrac{10}{30}\\ \dfrac{1}{5}=\dfrac{1\times6}{5\times6}=\dfrac{6}{30}\\ \dfrac{1}{2}=\dfrac{1\times15}{2\times15}=\dfrac{15}{30}\)

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

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

\(BCNN\left( {5,2} \right) = 10\)

\(\begin{array}{l}\dfrac{{ - 3}}{5} = \dfrac{{ - 3.2}}{{5.2}} = \dfrac{{ - 6}}{{10}}\\\dfrac{{ - 1}}{2} = \dfrac{{ - 1.5}}{{2.5}} = \dfrac{{ - 5}}{{10}}\end{array}\)