\(\dfrac{-3}{7}\)x=\(\dfrac{1}{6}\)

x=\(\dfrac{1}{6}\):\(\dfrac{-3}{7}\)

x=\(\dfrac{7}{-18}\)

\(a,\dfrac{3}{4}x-\dfrac{7}{12}=\dfrac{5}{6}-\dfrac{2}{3}\\ \Rightarrow\dfrac{3}{4}x-\dfrac{7}{12}=\dfrac{1}{6}\\ \Rightarrow\dfrac{3}{4}x=\dfrac{1}{6}+\dfrac{7}{12}\\ \Rightarrow\dfrac{3}{4}x=\dfrac{3}{4}\\ \Rightarrow x=\dfrac{3}{4}:\dfrac{3}{4}\\ \Rightarrow x=1\\ b,\dfrac{-5}{x}=\dfrac{20}{28}\\ \Rightarrow\dfrac{-5}{x}=\dfrac{5}{7}\\ \Rightarrow\dfrac{-5}{x}=\dfrac{-5}{-7}\\ \Rightarrow x=-7\\ c,2\dfrac{1}{3}:x=7\\ \Rightarrow\dfrac{7}{3}:x=7\\ \Rightarrow x=\dfrac{7}{3}:7\\ \Rightarrow x=\dfrac{1}{3}\)

\(d,\dfrac{-105}{12}< x< \dfrac{20}{7}\Rightarrow x\in\left\{-8;-7;...;2\right\}\)

a: \(\Leftrightarrow x\cdot\dfrac{3}{4}=\dfrac{3}{4}\)

hay x=1

b: \(\Leftrightarrow x=\dfrac{-28\cdot5}{20}=-7\)

c: \(\Leftrightarrow x=\dfrac{7}{3}:7=\dfrac{1}{3}\)

d: \(\Leftrightarrow-8< x< 3\)

hay \(x\in\left\{-7;-6;-5;-4;-3;-2;-1;0;1;2\right\}\)

\(a)\dfrac{3}{4}x-\dfrac{7}{12}=\dfrac{1}{6}\\ \dfrac{3}{4}x=\dfrac{1}{6}+\dfrac{7}{12}\\ \dfrac{3}{4}x=\dfrac{2}{3}\\ x=\dfrac{2}{3}:\dfrac{3}{4}\\ x=\dfrac{8}{9}\\ b)-\dfrac{5}{x}=\dfrac{20}{28}\\ -5\cdot28=x\cdot20=-140\\ x=-140:20\\ x=-7\\ c)\dfrac{7}{3}:x=7\\ x=\dfrac{7}{3}:7\\ x=\dfrac{1}{3}\)

a)\(x=\left(\dfrac{3}{56}\cdot\dfrac{28}{9}\right):\dfrac{-3}{7}=\dfrac{1}{6}:\dfrac{-3}{7}=-\dfrac{7}{18}\)

b)\(x=\left(\dfrac{7}{15}\cdot\dfrac{5}{3}\right)+\dfrac{3}{16}=\dfrac{7}{9}+\dfrac{3}{16}=\dfrac{139}{144}\)

c)\(x=\left(\dfrac{5}{6}-\dfrac{2}{5}\right).5=\dfrac{13}{6}\)

d)\(=>x\left(\dfrac{3}{4}-\dfrac{2}{5}\right)=\dfrac{1}{6}\cdot\left(\dfrac{3}{7}+\dfrac{5}{7}\right)\)

\(x\cdot\dfrac{7}{20}=\dfrac{4}{21}=>x=\dfrac{4}{21}\cdot\dfrac{20}{7}=\dfrac{80}{147}\)

=>x+2=-9

=>x=-11

\(=>\left(x+2\right).28=7.\left(-36\right)\\ =>\left(x+2\right).28=-252\\ =>x+2=-252:28\\ =>x+2=-9\\ =>x=-11\)

\(\Rightarrow28.\left(x+2\right)=-36.7\)

\(\Rightarrow28x+56+252=0\)

\(\Rightarrow28x=-308\)

\(\Rightarrow x=-11\)

\(a.\)

\(\dfrac{-28}{35}=\dfrac{16}{x}\)

\(\Rightarrow x=\dfrac{35\cdot16}{-28}=\dfrac{5\cdot7\cdot4\cdot4}{-7\cdot4}=-20\)

\(b.\)

\(\dfrac{x+7}{15}=\dfrac{-24}{36}\)

\(\Rightarrow x+7=\dfrac{15\cdot-24}{36}=\dfrac{5\cdot3\cdot-12\cdot2}{12\cdot3}=-10\)

\(\Leftrightarrow x=-17\)

a: =>4/x=y/-21=4/7

=>x=7; y=-12

b: =>xy=63

mà x>y

nên \(\left(x,y\right)\in\left\{\left(9;7\right);\left(21;3\right);\left(63;1\right);\left(-7;-9\right);\left(-3;-21\right);\left(-1;-63\right)\right\}\)

c: =>xy=45

mà x<y<0

nên \(\left(x,y\right)\in\left\{\left(-45;-1\right);\left(-15;-3\right);\left(-9;-5\right)\right\}\)

Lời giải:

a. $\frac{x}{7}=\frac{6}{21}$

$x=\frac{6}{21}.7$

$x=2$

b.

$\frac{-5}{y}=\frac{20}{28}$

$y=-5:\frac{20}{28}$

$y=-7$

c.

$\frac{-4}{8}=\frac{-7}{y}$

$y=-7:\frac{-4}{8}$

$y=14$

a, \(\dfrac{x}{7}=\dfrac{6}{21}\Leftrightarrow\dfrac{3x}{21}=\dfrac{6}{21}\Rightarrow x=2\)

b, \(\dfrac{-5}{y}=\dfrac{20}{28}\Leftrightarrow\dfrac{20}{-4y}=\dfrac{20}{28}\Leftrightarrow y=-7\)

c, \(\dfrac{-4}{8}=-\dfrac{7}{y}\Rightarrow-4y=-56\Leftrightarrow y=14\)

a) Ta có: \(\dfrac{x}{7}=\dfrac{6}{21}\)

nên \(x=\dfrac{6\cdot7}{21}=\dfrac{42}{21}=2\)

b) Ta có: \(\dfrac{-5}{y}=\dfrac{20}{28}\)

nên \(y=\dfrac{-5\cdot28}{20}=\dfrac{-140}{20}=-7\)

c) Ta có: \(\dfrac{-4}{8}=\dfrac{-7}{y}\)

nên \(y=\dfrac{-7\cdot8}{-4}=\dfrac{-56}{-4}=14\)

Bài 2: 

\(=\dfrac{28}{25}\cdot\dfrac{15}{7}\cdot5=\dfrac{75}{25}\cdot4=12\)

Bài 1: 

a: \(x+\dfrac{7}{8}=\dfrac{13}{2}:4=\dfrac{13}{8}\)

nên x=13/8-7/8=6/8=3/4

b: \(x:\dfrac{5}{3}=\dfrac{6}{5}-\dfrac{2}{3}=\dfrac{18-10}{15}=\dfrac{8}{15}\)

nên \(x=\dfrac{8}{15}\cdot\dfrac{5}{3}=\dfrac{8}{9}\)

a: =>\(\dfrac{x}{-5}=\dfrac{y}{-7}=\dfrac{z}{2}=\dfrac{x-y+z}{-5+7+2}=\dfrac{-28}{4}=-7\)

=>x=35; y=49; z=-14

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

`x/-5=y/-7=z/2=(x-y+z)/((-5)-(-7)+2)=-28/4=-7`

`-> x/-5=y/-7=z/2=-7`

`-> x=-7*-5=35, y=-7*-7=49, z=-7*2=-14`