a. 7/8 x 2/5 =7/20

b.9/4-5/6=17/12

c.4/7x2/5=8/35

d.3/5+2=13/5

e. 4-3/5=17/5

g.3x4/9=4/3

h.9/5:2=9/10

a) \(\dfrac{14}{40}=\dfrac{7}{20}\)

b) \(\dfrac{27}{12}-\dfrac{10}{12}=\dfrac{17}{12}\)

c) \(\dfrac{8}{35}\)

d) \(\dfrac{3}{5}+\dfrac{10}{5}=\dfrac{13}{5}\)

e) \(\dfrac{20}{5}-\dfrac{3}{5}=\dfrac{17}{5}\)

g) \(\dfrac{12}{9}=\dfrac{4}{3}\)

h) \(\dfrac{9}{5}x\dfrac{1}{2}=\dfrac{9}{10}\)

`@LQuyen`

\(\dfrac{7}{8}\times\dfrac{2}{5}=\dfrac{14}{40}\)

\(\dfrac{9}{4}-\dfrac{5}{6}=\dfrac{54}{24}-\dfrac{20}{24}=\dfrac{34}{24}\)

\(\dfrac{4}{7}\times\dfrac{2}{5}=\dfrac{8}{35}\)

\(\dfrac{3}{5}+\dfrac{2}{1}=\dfrac{3}{5}+\dfrac{10}{5}=\dfrac{13}{5}\)

\(\dfrac{4}{1}-\dfrac{3}{5}=\dfrac{20}{5}-\dfrac{3}{5}=\dfrac{17}{5}\)

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

\(\dfrac{9}{5}:2=\dfrac{9}{10}\)

\(a,5x\dfrac{7}{3}=\dfrac{5}{1}x\dfrac{7}{3}=\dfrac{35}{3};b,\dfrac{13}{4}:7=\dfrac{13}{4} :\dfrac{7}{1}=\dfrac{13}{4}x\dfrac{1}{7}=\dfrac{13}{28}\)

\(\dfrac{3}{7}+\dfrac{2}{5}+\dfrac{3}{4}=\dfrac{60}{140}+\dfrac{56}{140}+\dfrac{105}{140}=\dfrac{221}{140}\)

\(\dfrac{9}{7}-\dfrac{5}{11}x\dfrac{11}{7}=\dfrac{9}{7}-\dfrac{5}{7}=\dfrac{4}{7}\)

\(1,\dfrac{4x-4}{3}=\dfrac{7-x}{5}\\ \Leftrightarrow5\left(4x-4\right)=3\left(7-x\right)\\ \Leftrightarrow20x-20=21-3x\\ \Leftrightarrow17x=41\Leftrightarrow x=\dfrac{41}{17}\)

\(2,\dfrac{3x-9}{5}=\dfrac{3-x}{2}\\ \Leftrightarrow6x-18=15-5x\\ \Leftrightarrow11x=33\\ \Leftrightarrow x=3\)

\(3,\dfrac{2x-1}{5}-\dfrac{3-x}{3}=1\\ \Leftrightarrow\dfrac{6x-3-15+5x}{15}=1\\ \Leftrightarrow11x-18=1\\ \Leftrightarrow x=\dfrac{19}{11}\)

\(4,\dfrac{x-5}{3}+\dfrac{3x+4}{2}=\dfrac{5x+2}{6}\\ \Leftrightarrow2x-10+9x+12=5x+2\\ \Leftrightarrow6x=0\Leftrightarrow x=0\)

\(5,\dfrac{x-3}{2}+\dfrac{2x+3}{5}=\dfrac{2x+5}{10}\\ \Leftrightarrow5x-15+4x+6=2x+5\\ \Leftrightarrow7x=14\\ \Leftrightarrow x=2\)

Tick nha

2: Ta có: \(\dfrac{3x-9}{5}=\dfrac{3-x}{2}\)

\(\Leftrightarrow6x-18=15-5x\)

\(\Leftrightarrow11x=33\)

hay x=3

\(\dfrac{3}{7}+\dfrac{4}{1}=\dfrac{3}{7}+\dfrac{28}{7}=\dfrac{31}{7}\)

\(\dfrac{5}{9}\times3=\dfrac{5}{9}\times\dfrac{3}{1}=\dfrac{15}{9}=\dfrac{5}{3}\)

\(\dfrac{7}{8}-\dfrac{2}{9}=\dfrac{63}{72}-\dfrac{16}{72}=\dfrac{47}{72}\)

\(\dfrac{5}{1}-\dfrac{3}{4}=\dfrac{20}{4}-\dfrac{3}{4}=\dfrac{17}{4}\)

\(\dfrac{5}{7}:\dfrac{6}{1}=\dfrac{5}{7}\times\dfrac{1}{6}=\dfrac{5}{42}\)

\(\dfrac{5}{6}\times\dfrac{2}{7}=\dfrac{5\times2}{6\times7}=\dfrac{10}{42}=\dfrac{5}{21}\)

\(\dfrac{3}{1}+\dfrac{3}{4}=\dfrac{12}{4}+\dfrac{3}{4}=\dfrac{15}{4}\)

\(\dfrac{3}{5}\times\dfrac{4}{8}=\dfrac{3}{5}\times\dfrac{1}{2}=\dfrac{3\times1}{5\times2}=\dfrac{3}{10}\)

\(\dfrac{5}{1}:\dfrac{3}{8}=\dfrac{5}{1}\times\dfrac{8}{3}=\dfrac{40}{3}\)

a) \(\dfrac{3}{7}+4=\dfrac{3}{7}+\dfrac{4}{1}=\dfrac{3}{7}+\dfrac{28}{7}=\dfrac{31}{7}\)

b) \(\dfrac{5}{9}\times3=\dfrac{5\times3}{9}=\dfrac{15}{9}=\dfrac{5}{3}\)

c) \(\dfrac{7}{8}-\dfrac{2}{9}=\dfrac{63}{72}-\dfrac{16}{72}=\dfrac{47}{72}\)

d) \(5-\dfrac{3}{4}=\dfrac{5}{1}-\dfrac{3}{4}=\dfrac{20}{4}-\dfrac{3}{4}=\dfrac{17}{4}\)

e) \(\dfrac{5}{7}:6=\dfrac{5}{7}:\dfrac{6}{1}=\dfrac{5}{7}\times\dfrac{1}{6}=\dfrac{5}{42}\)

f) \(\dfrac{5}{6}\times\dfrac{2}{7}=\dfrac{5\times2}{6\times7}=\dfrac{10}{42}=\dfrac{5}{21}\)

g) \(3+\dfrac{3}{4}=\dfrac{3}{1}+\dfrac{3}{4}=\dfrac{12}{4}+\dfrac{3}{4}=\dfrac{15}{4}\)

h) \(\dfrac{3}{5}\times\dfrac{4}{8}=\dfrac{3\times4}{5\times8}=\dfrac{12}{40}=\dfrac{3}{10}\)

i) \(5:\dfrac{3}{8}=\dfrac{5}{1}:\dfrac{3}{8}=\dfrac{5}{1}\times\dfrac{8}{3}=\dfrac{40}{3}\)

\(\begin{array}{l} n) \Leftrightarrow \dfrac{{x + 1}}{7} + 1 + \dfrac{{x + 2}}{6} + 1 = \dfrac{{x + 3}}{5} + 1 + \dfrac{{x + 4}}{4} + 1\\ \Leftrightarrow \dfrac{{x + 8}}{7} + \dfrac{{x + 8}}{6} - \dfrac{{x + 8}}{5} - \dfrac{{x + 8}}{4} = 0\\ \Leftrightarrow \left( {x + 8} \right)\underbrace {\left( {\dfrac{1}{7} + \dfrac{1}{8} - \dfrac{1}{5} - \dfrac{1}{6}} \right)}_{ < 0} = 0\\ \Leftrightarrow x + 8 = 0\\ \Leftrightarrow x = - 8 \end{array}\)

k/

\(8-\dfrac{x-2}{3}=\dfrac{x}{4}\)

\(\Leftrightarrow\dfrac{96}{12}-\dfrac{4\left(x-2\right)}{12}=\dfrac{3x}{12}\)

\(\Leftrightarrow96-4x+8=3x\)

\(\Leftrightarrow96-4x+8-3x=0\)

\(\Leftrightarrow104-7x=0\)

\(\Leftrightarrow7x=104\)

\(\Leftrightarrow x=104:7\)

\(\Leftrightarrow x=\dfrac{104}{7}\)

Vậy tập nghiệm của phương trình là \(S=\left\{\dfrac{104}{7}\right\}\)

m/ 

\(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)

\(\Leftrightarrow\dfrac{3\left(3x+2\right)}{6}-\dfrac{3x+1}{6}=\dfrac{12x}{6}+\dfrac{10}{6}\)

\(\Leftrightarrow9x+6-3x-1-12x-10=0\)

\(\Leftrightarrow-6x-5=0\)

\(\Leftrightarrow-6x=5\)

\(\Leftrightarrow x=-\dfrac{5}{6}\)

Vậy tập nghiệm của phương trình là \(S=\left\{-\dfrac{5}{6}\right\}\)

k) Ta có: \(8-\dfrac{x-2}{2}=\dfrac{x}{4}\)

\(\Leftrightarrow\dfrac{32}{4}-\dfrac{2\left(x-2\right)}{4}=\dfrac{x}{4}\)

\(\Leftrightarrow32-2x+4-x=0\)

\(\Leftrightarrow28-x=0\)

hay x=28

Vậy: S={28}

m) Ta có: \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)

\(\Leftrightarrow\dfrac{3\left(3x+2\right)}{6}-\dfrac{3x+1}{6}=\dfrac{12x}{6}+\dfrac{10}{6}\)

\(\Leftrightarrow9x+6-3x-1=12x+10\)

\(\Leftrightarrow6x+5-12x-10=0\)

\(\Leftrightarrow-6x=5\)

hay \(x=-\dfrac{5}{6}\)

Vậy: \(S=\left\{-\dfrac{5}{6}\right\}\)

n) Ta có: \(\dfrac{x+1}{7}+\dfrac{x+2}{6}=\dfrac{x+3}{5}+\dfrac{x+4}{4}\)

\(\Leftrightarrow\dfrac{x+1}{7}+1+\dfrac{x+2}{6}+1=\dfrac{x+3}{5}+1+\dfrac{x+4}{4}+1\)

\(\Leftrightarrow\dfrac{x+8}{7}+\dfrac{x+8}{6}=\dfrac{x+8}{5}+\dfrac{x+8}{4}\)

\(\Leftrightarrow\dfrac{x+8}{7}+\dfrac{x+8}{6}-\dfrac{x+8}{5}-\dfrac{x+8}{4}=0\)

\(\Leftrightarrow\left(x+8\right)\left(\dfrac{1}{7}+\dfrac{1}{6}-\dfrac{1}{5}-\dfrac{1}{4}\right)=0\)

mà \(\dfrac{1}{7}+\dfrac{1}{6}-\dfrac{1}{5}-\dfrac{1}{4}\ne0\)

nên x+8=0

hay x=-8

Vậy: S={-8}

a/=4/9x10/3=40/27

b/=15/49

c/=4/9x2/3=8/27

a)\(=\dfrac{4}{9}\times\dfrac{10}{3}=\dfrac{40}{27}\)

b)\(\dfrac{3}{7}\times\dfrac{5}{7}=\dfrac{15}{49}\)

c)\(\dfrac{4}{9}:\dfrac{3}{2}=\dfrac{4}{9}\times\dfrac{2}{3}=\dfrac{8}{27}\)

a, =40/27

b, =15/49

c, =8/27

d, =3/9 +5/9

e, = 8/21 +2/6=5/7

g, =2/7+4/7=6/7

ét ô ét

a.25/27                                                                                                                 b.0                          c.0

\(b,\Rightarrow\dfrac{x}{2}-\dfrac{3x}{5}-\dfrac{13}{5}=-\dfrac{7}{5}-\dfrac{7x}{10}\\ \Rightarrow\dfrac{1}{2}x-\dfrac{3}{5}x+\dfrac{7}{10}x=\dfrac{6}{5}\\ \Rightarrow\dfrac{3}{5}x=\dfrac{6}{5}\Rightarrow x=2\\ c,\Rightarrow\dfrac{2x-3}{3}-\dfrac{5-3x}{6}=-\dfrac{1}{3}+\dfrac{3}{2}=\dfrac{7}{6}\\ \Rightarrow\dfrac{4x-6-5+3x}{6}=\dfrac{7}{6}\\ \Rightarrow7x-11=7\Rightarrow x=\dfrac{18}{7}\\ d,\Rightarrow\dfrac{2}{3x}+\dfrac{7}{x}=\dfrac{4}{5}+2+\dfrac{3}{12}=\dfrac{61}{20}\\ \Rightarrow\dfrac{23}{3x}=\dfrac{61}{20}\\ \Rightarrow183x=460\\ \Rightarrow x=\dfrac{460}{183}\\ e,\Rightarrow2\left(x-1\right)-\left(x-1\right)^2=0\\ \Rightarrow\left(x-1\right)\left(2-x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

e: Ta có: \(\left(x-1\right)^2=2\left(x-1\right)\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

1: Ta có: \(\dfrac{x+4}{4}+\dfrac{3x-7}{5}=\dfrac{7x+2}{20}\)

\(\Leftrightarrow5x+20+12x-28=7x+2\)

\(\Leftrightarrow17x-7x=2+8=10\)

hay x=1

2: Ta có: \(\dfrac{x}{6}+\dfrac{1-3x}{9}=\dfrac{-x+1}{12}\)

\(\Leftrightarrow\dfrac{6x}{36}+\dfrac{4\left(1-3x\right)}{36}=\dfrac{3\left(-x+1\right)}{36}\)

\(\Leftrightarrow6x+4-12x=-3x+3\)

\(\Leftrightarrow-6x+3x=3-4\)

hay \(x=\dfrac{1}{3}\)

3: Ta có: \(\dfrac{x-3}{3}-\dfrac{x+2}{12}=\dfrac{2x-1}{4}\)

\(\Leftrightarrow4x-12-x-2=6x-3\)

\(\Leftrightarrow3x-14-6x+3=0\)

\(\Leftrightarrow-3x=11\)

hay \(x=-\dfrac{11}{3}\)

4: Ta có: \(\dfrac{x-2}{4}-\dfrac{2x+3}{3}=\dfrac{x+6}{12}\)

\(\Leftrightarrow3x-6-8x-12=x+6\)

\(\Leftrightarrow-5x-x=6+18\)

hay x=-4

5: Ta có: \(\dfrac{2x-1}{12}-\dfrac{3-x}{18}=\dfrac{-1}{36}\)

\(\Leftrightarrow6x-3+2x-6=-1\)

\(\Leftrightarrow8x=8\)

hay x=1