a: Sai

b: Đúng

c: Sai

a: \(\dfrac{5}{7}=\dfrac{5\cdot11}{7\cdot11}=\dfrac{55}{77}\)

\(\dfrac{9}{11}=\dfrac{9\cdot7}{11\cdot7}=\dfrac{63}{77}\)

b: \(\dfrac{36}{42}=\dfrac{6}{7}=\dfrac{6\cdot9}{7\cdot9}=\dfrac{54}{63}\)

\(-\dfrac{12}{54}=\dfrac{-2}{9}=\dfrac{-2\cdot7}{9\cdot7}=-\dfrac{14}{63}\)

c: \(\dfrac{-11}{30}=\dfrac{-11\cdot4}{30\cdot4}=\dfrac{-44}{120}\)

\(\dfrac{-17}{-40}=\dfrac{17}{40}=\dfrac{17\cdot3}{40\cdot3}=\dfrac{51}{120}\)

d: \(\dfrac{36}{42}=\dfrac{6}{7}=\dfrac{6\cdot3}{7\cdot3}=\dfrac{18}{21}\)

\(\dfrac{-12}{36}=\dfrac{-1}{3}=\dfrac{-1\cdot7}{3\cdot7}=\dfrac{-7}{21}\)

Bài 1:

a) 

\(\dfrac{1}{2}=\dfrac{1\times6}{2\times6}=\dfrac{6}{12}\)

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

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

b)

\(\dfrac{1}{3}=\dfrac{1\times15}{3\times15}=\dfrac{15}{45}\)

\(\dfrac{2}{15}=\dfrac{2\times3}{15\times3}=\dfrac{6}{45}\)

\(\dfrac{4}{45}\) (giữ nguyên)

c)

\(\dfrac{1}{8}=\dfrac{1\times3}{8\times3}=\dfrac{3}{24}\)

\(\dfrac{2}{3}=\dfrac{2\times8}{3\times8}=\dfrac{16}{24}\)

\(\dfrac{5}{2}=\dfrac{5\times12}{2\times12}=\dfrac{60}{24}\)

d)

\(\dfrac{2}{7}=\dfrac{2\times4}{7\times4}=\dfrac{8}{28}\)

\(\dfrac{9}{4}=\dfrac{9\times7}{4\times7}=\dfrac{63}{28}\)

\(\dfrac{5}{28}\) (giữ nguyên)

Bài 2:

a)

\(4=\dfrac{4}{1}=\dfrac{4\times12}{1\times12}=\dfrac{48}{12}\)

\(\dfrac{9}{4}=\dfrac{9\times3}{4\times3}=\dfrac{27}{12}\)

b)

\(\dfrac{5}{8}=\dfrac{5\times30}{8\times30}=\dfrac{150}{240}\)

\(\dfrac{25}{30}=\dfrac{5}{6}=\dfrac{5\times40}{6\times40}=\dfrac{200}{240}\)

\(2=\dfrac{2}{1}=\dfrac{2\times240}{1\times240}=\dfrac{480}{240}\).

giúp mình với ạ!

\(1,\\ a,=xy^2-\dfrac{3}{2}y^3+\dfrac{5}{4}x^2\\ b,=\left(x-7\right)\left(x+7\right):\left(x-7\right)=x+7\\ 2,\dfrac{1}{a^2}-ab=\dfrac{1-a^3b}{a^2};\dfrac{1}{a^2}\text{ giữ nguyên}\\ 3,=\dfrac{-7}{t}\\ 4,=\dfrac{1-x+1-y}{x-y}=\dfrac{2-x-y}{x-y}\)

Bài 1:

\(a,\left(16x^3y^2-24x^2y^3+20x^4\right):16x^2=16x^2\left(xy^2-\dfrac{3}{2}y^3+\dfrac{5}{4}x^2\right):16x^2=xy^2-\dfrac{3}{2}y^3+\dfrac{5}{4}x^2\)

\(b,\left(x^2-49\right):\left(x-7\right)=\left[\left(x-7\right)\left(x+7\right)\right]:\left(x-7\right)=x+7\)

Bài 2:

\(\dfrac{1}{a^2}-ab=\dfrac{1-a^2b}{a^2}\)

\(\dfrac{1}{a^2}\)

Bài 3:

\(\dfrac{7\left(t-z\right)}{t\left(z-t\right)}=\dfrac{-7\left(z-t\right)}{t\left(z-t\right)}=\dfrac{-7}{t}\)

Bài 4:

\(\dfrac{x-1}{y-x}+\dfrac{1-y}{x-y}=\dfrac{x-1}{y-x}-\dfrac{1-y}{y-x}=\dfrac{x-1-1+y}{y-x}=\dfrac{x+y-2}{y-x}\)

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

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

2/5 và 5/7 

\(\dfrac{2}{5}=\dfrac{2\times7}{5\times7}=\dfrac{14}{35}\)

\(\dfrac{5}{7}=\dfrac{5\times5}{7\times5}=\dfrac{25}{35}\)

x 1 à bn

Ta có: \(\dfrac{5}{7} = \dfrac{{5.4}}{{7.4}} = \dfrac{{20}}{{28}}\) và \(\dfrac{{ - 3}}{4} = \dfrac{{ - 3.7}}{{4.7}} = \dfrac{{ - 21}}{{28}}\)

Như vậy, \(\dfrac{{20}}{{28}} + \dfrac{{ - 21}}{{28}} = \dfrac{{20 + \left( { - 21} \right)}}{{28}} =  \dfrac{-1}{{28}}\)

\(\dfrac{5}{7}=\dfrac{75}{105}\)

\(\dfrac{-3}{21}=\dfrac{-15}{105}\)

\(\dfrac{-8}{15}=\dfrac{-56}{105}\)

5/7 =75/105

-3/21= -15/105

-8/15=-56/105