\(\Leftrightarrow-\dfrac{16}{279}< \dfrac{x}{9}< =\dfrac{2}{3}\)

\(\Leftrightarrow\dfrac{x}{9}=0\)

hay x=0

\(x+7\dfrac{4}{5}=9\dfrac{7}{15}\)

\(=>x+\dfrac{39}{5}=\dfrac{142}{15}\)

\(=>x=\dfrac{142}{15}-\dfrac{39}{5}=\dfrac{142}{15}-\dfrac{117}{15}\)

\(=>x=\dfrac{25}{15}=\dfrac{5}{3}\)

___________

\(x-4\dfrac{9}{11}=\dfrac{2121}{2222}\)

\(=>x-\dfrac{53}{11}=\dfrac{21}{22}\)

\(=>x=\dfrac{21}{22}+\dfrac{53}{11}=\dfrac{21}{22}+\dfrac{106}{22}\)

\(=>x=\dfrac{127}{22}\)

\(x+7\dfrac{4}{5}=9\dfrac{7}{15}\)

\(x=9\dfrac{7}{15}-7\dfrac{4}{5}\)

\(x=\left(9-7\right)+\left(\dfrac{7}{15}-\dfrac{4}{5}\right)\)

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

\(x=\dfrac{2121:101}{2222:101}+4\dfrac{9}{11}\)

\(x=\dfrac{21}{22}+4\dfrac{18}{22}\)

\(x=\dfrac{21+4\cdot22+18}{22}=\dfrac{127}{22}\)

\(\dfrac{15-x}{14}=\dfrac{5}{7}\\ \left(15-x\right)\times7=5\times14\\ \left(15-x\right)\times7=70\\ 15-x=70:7\\ 15-x=10\\ x=15-10\\ x=5\)

__

\(\dfrac{x}{9}=\dfrac{17}{3}\\ x\times3=17\times9\\ x\times3=153\\ x=153:3\\ x=51\)

__

\(15-\dfrac{x}{2}=\dfrac{3}{7}\\ \dfrac{x}{2}=15-\dfrac{3}{7}\\ \dfrac{x}{2}=\dfrac{102}{7}\\ x\times7=102\times2\\ x\times7=204\\ x=204:7\\ x=\dfrac{204}{7}\)

???

`x xx4/5 :2=6/7`

`x xx 4/5xx1/2=6/7`

`x xx2/5=6/7`

`x=6/7:2/5`

`x=6/7xx5/2`

`x=15/7`

b)

`6/5:x:5/4=10/15`

`6/5:x xx4/5=2/3`

`6/5:x=2/3:4/5`

`6/5:x=2/3xx5/4`

`6/5:x=5/6`

`x=6/5:5/6`

`x=6/5xx6/5`

`x=36/25`

a) x × 4/5 : 2 = 6/7

x × 4/5 = 6/7 × 2

x × 4/5 = 12/7

x = 12/7 : 4/5

x = 15/7

b) 6/5 : x : 5/4 = 10/15

6/5 : x : 5/4 = 2/3

6/5 : x = 2/3 × 5/4

6/5 : x = 5/6

x = 6/5 : 5/6

x = 36/25

a: \(A=\dfrac{-7}{28}\cdot\dfrac{15}{25}=\dfrac{-1}{4}\cdot\dfrac{3}{5}=\dfrac{-3}{20}\)

b: \(B=\dfrac{-5\cdot7}{14\cdot\left(-3\right)}=\dfrac{35}{42}=\dfrac{5}{6}\)

c: \(C=\dfrac{-1}{5}-\dfrac{1}{5}\cdot\dfrac{3}{5}=\dfrac{-1}{5}-\dfrac{3}{25}=\dfrac{-8}{25}\)

d: \(D=\dfrac{-3}{4}-\dfrac{1}{4}=-1\)

e: \(E=\dfrac{-4}{5}\left(1-\dfrac{15}{16}\right)=\dfrac{-4}{5}\cdot\dfrac{1}{16}=\dfrac{-1}{20}\)

f: \(F=\dfrac{6-7}{4}\cdot\dfrac{4+12}{22}=\dfrac{-1}{4}\cdot\dfrac{8}{11}=\dfrac{-2}{11}\)

a: =>1/x=4

hay x=1/4

b: =>x+7=-10

=>x=-17

c: =>x²=36

=>x=6 hoặc x=-6

d: =>(x-3)²=16

=>x-3=4 hoặc x-3=-4

=>x=7 hoặc x=-1

\(a,\dfrac{1}{-x}=-4\\ \Rightarrow\left(-x\right)\left(-4\right)=1\\ \Rightarrow4x=1\\ \Rightarrow x=\dfrac{1}{4}\\ b,\dfrac{x+7}{15}=\dfrac{-24}{36}\\ \Rightarrow\dfrac{x+7}{15}=\dfrac{-2}{3}\\ \Rightarrow3\left(x+7\right)=-2.15\\ \Rightarrow3x+21=-30\\ \Rightarrow3x=-51\\ \Rightarrow x=-17\)

\(c,\dfrac{x}{-3}=\dfrac{-12}{x}\\ \Rightarrow x.x=\left(-3\right)\left(-12\right)\\ \Rightarrow x^2=36\\ \Rightarrow x=\pm6\)

\(d,\dfrac{-2}{x-3}=\dfrac{x-3}{-8}\\ \Rightarrow\left(x-3\right)\left(x-3\right)=\left(-2\right)\left(-8\right)\\ \Rightarrow\left(x-3\right)^2=16\\ \Rightarrow\left[{}\begin{matrix}x-3=4\\x-3=-4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=7\\x=-1\end{matrix}\right.\)

a.\(\Leftrightarrow4x=1\Leftrightarrow x=\dfrac{1}{4}\)

b.\(\Leftrightarrow\left(x+7\right).36=-360\Leftrightarrow x+7=-10\Rightarrow x=-17\)

c.\(x^2=36\Leftrightarrow x=6;-6\)

d.\(-2.-8=\left(x-3\right)\left(x-3\right)\Leftrightarrow16=x^2-6x+9\Leftrightarrow x^2-6x-7=0\Rightarrow x=7;-1\)

x=\(\dfrac{4}{15}\) : \(\dfrac{-2}{3}\)

x=\(\dfrac{-2}{5}\)

a: Ta có: \(x\cdot\dfrac{-2}{3}=\dfrac{4}{15}\)

\(\Leftrightarrow x=\dfrac{4}{15}:\dfrac{-2}{3}=\dfrac{4}{15}\cdot\dfrac{-3}{2}=\dfrac{-2}{5}\)

b: Ta có: \(x\cdot\dfrac{-7}{19}=\dfrac{-13}{24}\)

\(\Leftrightarrow x=\dfrac{13}{24}:\dfrac{7}{19}=\dfrac{247}{168}\)

Mn ơi giúp mk với , please !!!

1. Ta có: \(\dfrac{x}{-7}=\dfrac{y}{4}\Rightarrow\dfrac{2x}{-14}=\dfrac{3y}{12}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta được:

\(\dfrac{2x-3y}{-14-12}=\dfrac{-78}{-26}=3\)

=> \(\left\{{}\begin{matrix}x=-21\\y=12\end{matrix}\right.\)

2. Ta có:

- \(\dfrac{x}{y}=\dfrac{9}{7}\Rightarrow\dfrac{x}{9}=\dfrac{y}{7}\)

- \(\dfrac{y}{z}=\dfrac{7}{3}\Rightarrow\dfrac{y}{7}=\dfrac{z}{3}\)

=> \(\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta được:

\(\dfrac{x-y+z}{9-7+3}=\dfrac{-15}{5}=-3\)

=> \(\left\{{}\begin{matrix}x=-27\\y=-21\\z=-9\end{matrix}\right.\)

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