a) ADTCDTSBN

có: \(\frac{x}{2}=\frac{z}{4}=\frac{x+z}{2+4}=\frac{18}{6}=3.\)

=> x/2 = 3 => x = 6

y/3 = 3 => y = 9

z/4 = 3 => z = 12

KL:...

b,c làm tương tự nha

d) ta có: \(\frac{x}{5}=\frac{y}{-6}=\frac{z}{7}=\frac{2x}{10}\)

ADTCDTSBN

có: \(\frac{2x}{10}=\frac{y}{-6}=\frac{z}{7}=\frac{2x+y-z}{10+\left(-6\right)-7}=\frac{49}{-3}\)

=>...

e) ADTCDTSBN

có: \(\frac{x+1}{2}=\frac{y+2}{3}=\frac{z+3}{4}=\frac{x+1+y+2+z+3}{2+3+4}=\frac{\left(x+y+z\right)+\left(1+2+3\right)}{9}\)

\(=\frac{21+6}{9}=\frac{27}{9}=3\)

=>...

g) ta có: \(\frac{x}{4}=\frac{y}{3}=k\Rightarrow\hept{\begin{cases}x=4k\\y=3k\end{cases}}\)

mà xy = 12 => 4k.3k = 12

                          12.k² = 12

                              k² = 1

                        => k = 1 hoặc k = -1

=> x = 4.1 = 4

y = 3.1 = 3

x=4.(-1) = -4 

y=3.(-1) = -3

KL:...

h) ta có: \(\frac{x}{5}=\frac{y}{3}\Rightarrow\frac{x^2}{25}=\frac{y^2}{9}\)

ADTCDTSBN

có: \(\frac{x^2}{25}=\frac{y^2}{9}=\frac{x^2-y^2}{25-9}=\frac{16}{16}=1\)

=>...

mn. giúp em nha 

\(a.\dfrac{6}{5}=\dfrac{18}{x}\Rightarrow x=\dfrac{18\cdot5}{6}=15\\ \text{Vậy}\text{ }x=15.\)

\(b.\dfrac{3}{4}=\dfrac{-21}{x}\Rightarrow x=\dfrac{-21\cdot4}{3}=28\\ \text{ }\text{ }\text{ }\text{ }\text{Vậy }x=28.\)

\(c.\dfrac{x}{4}=\dfrac{21}{28}\Rightarrow x=\dfrac{21\cdot4}{28}=3\\ \text{Vậy }x=3.\)

\(d.\dfrac{-8}{2x}=\dfrac{3}{-9}\Rightarrow x=\dfrac{-8\cdot\left(-9\right)}{3}:2=12\\ \text{Vậy }x=12.\)

\(e.\dfrac{-4}{11}=\dfrac{x}{22}=\dfrac{40}{z}\\ \Rightarrow x=\dfrac{-4\cdot22}{11}=-8\\ \Rightarrow z=\dfrac{22\cdot40}{-8}=-110\\ \text{Vậy }x=-8;z=-110.\)

\(f.\dfrac{-3}{4}=\dfrac{x}{20}=\dfrac{21}{y}\\ \Rightarrow x=\dfrac{-3\cdot20}{4}=-15\\ \Rightarrow y=\dfrac{21\cdot20}{-15}=-28\\ \text{Vậy }x=-15;y=-28.\)

\(g.\dfrac{-4}{8}=\dfrac{x}{-10}=\dfrac{-7}{y}=\dfrac{z}{-24}\\ \Rightarrow x=\dfrac{-4\cdot\left(-10\right)}{8}=5\\ \Rightarrow y=\dfrac{-7\cdot\left(-10\right)}{5}=14\\ \Rightarrow z=\dfrac{-7\cdot\left(-24\right)}{14}=12\\ \text{Vậy }x=5;y=14;z=12.\)

\(h.\dfrac{x}{4}=\dfrac{9}{x}\\ \Rightarrow x\cdot x=9\cdot4\\ \Rightarrow x\cdot x=36\\ \Rightarrow x\cdot x=6\cdot6\\ \text{Vậy }\text{cả hai }x=6.\)

Bài 2 : a, x = -36/9 = -4

b, đề sai 

c, <=> -2 =< x =< -3 => x = -1 

Bài 1: 

a: 2/8=9/36; 2/9=8/36; 8/2=36/9; 9/2=36/8

b: -2/4=9/-18; -2/9=4/-18; 4/-2=-18/9; 9/-2=-18/4

Bài 2: 

a: =>x/3=-4/3

hay x=-4

Câu b đề sai rồi bạn

a ) \(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\)và \(x+z=18\)

Áp dụng t/c dãy tỏ số bằng nhau ta có :

\(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=\frac{x+z}{2+4}=\frac{18}{6}=3\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{2}=3\\\frac{y}{3}=3\\\frac{z}{4}=3\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=6\\y=9\\z=12\end{cases}}\)

b ) \(\frac{x}{5}=\frac{y}{-6}=\frac{z}{7}\) và \(y-x=39\)

Áp dụng t/c dãy tỉ số bằng nhau ta có :

\(\frac{x}{5}=\frac{y}{-6}=\frac{z}{7}=\frac{y-x}{-6-5}=\frac{39}{-11}\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{5}=\frac{39}{-11}\\\frac{y}{-6}=\frac{39}{-11}\\\frac{z}{7}=\frac{39}{-11}\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=\frac{195}{11}\\y=-\frac{234}{11}\\z=\frac{273}{11}\end{cases}}\)

Bài 2:

\(\dfrac{a+b}{a-b}=\dfrac{c+a}{c-a}\)

\(\Rightarrow\dfrac{a+b}{c+a}=\dfrac{a-b}{c-a}=\dfrac{a+b+a-b}{c+a+c-a}=\dfrac{a}{c}\) (T/c dãy tỷ số = nhau)

\(\Rightarrow\dfrac{a+b}{c+a}=\dfrac{a}{c}\Rightarrow c\left(a+b\right)=a\left(c+a\right)\)

\(\Rightarrow ac+bc=ac+a^2\Rightarrow a^2=bc\)

a) 4x - 15 = -75 -x

   4x+x = -75 + 15

   5x = 60

     x= 60: 5

  => x= 12

b) 3| x-7| = 21

      |x-7|= 21:3

      |x-7|=7

  => x-7 =7 hoặc x-7=-7

 +) x-7=7

     x=7+7=14

  +) x-7=-7

      x= -7+7=0

=> x=14 hoặc x=0

c) Áp dụng t/c phân số bằng nhau 

=> x= \(\frac{-3.\left(-2\right)}{6}\)=\(\frac{6}{6}\)=1

Thay x=1 => y= \(\frac{\left(-2\right).\left(-18\right)}{1}\)=\(\frac{36}{1}\)=36

Thay y =36 => z=\(\frac{\left(-18\right).24}{36}\)=\(\frac{-432}{36}\)=-12

vậy (x,y,z)= (1;36;-12)

(câu d dài quá vs lại cx dễ nên bn tự lm nha mk chỉ giúp đến đây thôi)

a\(\left(x-3\right)^2-\left(x+2\right)^2-5\left(\frac{1}{5}-7\right)=-30\)

=>(x-3-x-2)(x-3+x+2)-x+35=-30

=>-5(2x-1)-x+35=-30

=>-10x+5-x+35=-30

=>-11x+40=-30

=>-11x=-70 =>x=70/11

d)\(\left(x+3\right)^2-\left(x+5\right)\left(x-5\right)=2\)

\(=>\left(x+3\right)^2-x^2+25=2\)

\(=>\left(z+3-z\right)\left(z+3+z\right)+25=2\)

\(=>3\left(2z+3\right)+25-2=0\)

\(=>6z+9+23=0\)

\(=>6x+32=0=>6x=-32=>x=-\frac{16}{3}\)

e)\(3\left(x+2\right)^2+\left(2x-1\right)^2-7\left(x+3\right)\left(x-3\right)=36\)

\(=>3\left(x^2+4x+4\right)+\left(4x^2-4x+1\right)-7\left(x^2-9\right)=36\)

\(=>3x^2+12x+12+4x^2-4x+1-7x^2+63\)

\(=>8x+76=36=>8x=36-76=>x=-40\div8=-5\)

g)\(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=5\)

\(=>x^3-1-x\left(x^2-4\right)=5=>x^3-1-x^3+4x=5\)

\(=>4x-1=5=>4x=6=>x=\frac{3}{2}\)