7/3:0,2X=7/6:0,8

35/3X=35/24

X=35/24:35/3

X=1/8

7/3 : 0,2x = 7/6 : 0,8

7/3 : 0,2x = 35/24

        0,2x = 7/3 : 35/24

        0,2x = 8/5

             x = 8/5 : 0,2

             x = 8

7/3 : 0.2 x = 35/24

0.2 x = 35/24 : 7/3

0.2 x = 5/8

x=5/8 : 0.2

x= 25/8

Vậy x=25/8

=>\(\dfrac{7}{3}:0,2x=\dfrac{7}{6}.\dfrac{5}{4}\)

=>\(\dfrac{7}{3}:0,2x=\dfrac{35}{24}\)

=>0,2\(.x=\dfrac{7}{3}:\dfrac{35}{24}\)

=>0,2\(.x=\dfrac{7}{3}.\dfrac{24}{35}\)

=>0,2.\(x=\dfrac{8}{5}\)

=>\(x=\dfrac{8}{5}:0,2\)

=>\(x=\dfrac{8}{5}.5\)

=>\(x=8\)

Vậy x=8

7) vì \(\dfrac{x}{5}\)=\(\dfrac{y}{6}\)=\(\dfrac{z}{7}\)và x-y+z=36

Nên theo tính chất của dãy tỉ số bằng nhau ta có:

 \(\dfrac{x}{5}\)=\(\dfrac{y}{6}\)=\(\dfrac{z}{7}\)=\(\dfrac{x-y+z}{5-6+7}\)=\(\dfrac{36}{6}\)=6

 \(\Rightarrow\)x=6.5=30

     y=6.6=36

     z=6.7=42

vậy x=30,y=36,z=42

a: x=4/27-2/3=4/27-18/27=-14/27

b: =>3/4x-1/4x=1/6+7/3

=>1/2x=1/6+14/6=5/2

hay x=5

c: =>13/10x=7/2+5/2=6

=>x=13/10:6=13/60

d: (3x+2)(-2/5x-7)=0

=>3x+2=0 hoặc 2/5x+7=0

=>x=-2/3 hoặc x=-35/2

a) x = 4/27 - 2/3

    x = -14/27

a: x=4/27-2/3=4/27-18/27=-14/27

b: =>3/4x-1/4x=1/6+7/3

=>1/2x=1/6+14/6=5/2

hay x=5

c: =>13/10x=7/2+5/2=6

=>x=13/10:6=13/60

d: (3x+2)(-2/5x-7)=0

=>3x+2=0 hoặc 2/5x+7=0

=>x=-2/3 hoặc x=-35/2

\(\Rightarrow x-1=\dfrac{6}{7}+\dfrac{1}{7}\times\left(\dfrac{2}{7}+\dfrac{5}{7}\right)=\dfrac{6}{7}+\dfrac{1}{7}\times1\\ \Rightarrow x-1=\dfrac{6}{7}+\dfrac{1}{7}=1\\ \Rightarrow x=1+1=2\)

= 2

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

\(\Leftrightarrow x-\dfrac{3}{5}=\dfrac{19}{42}\)

\(\Leftrightarrow x=\dfrac{19}{42}+\dfrac{3}{5}\)

\(\Leftrightarrow x=\dfrac{221}{210}\)

=>x-3/5=12/42+7/42=19/42

=>x=19/42+3/5=95/210+126/210=221/210

x-3/5=2/7+1/6

x-3/5=12/42+7/42

x-3/5=19/42

x=19/42+3/5

x=95/210 + 126/210

x=221/210

Bài 1: 

a) Ta có: \(\dfrac{7^4\cdot3-7^3}{7^4\cdot6-7^3\cdot2}\)

\(=\dfrac{7^3\cdot\left(7\cdot3-1\right)}{7^3\cdot2\left(7\cdot3-1\right)}\)

\(=\dfrac{1}{2}\)

c) Ta có: \(E=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\)

\(\Leftrightarrow\dfrac{1}{3}\cdot E=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{101}}\)

\(\Leftrightarrow E-\dfrac{1}{3}\cdot E=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{101}}\right)\)

\(\Leftrightarrow E\cdot\dfrac{2}{3}=1-\dfrac{1}{3^{101}}\)

\(\Leftrightarrow E=\dfrac{3-\dfrac{3}{3^{101}}}{2}=\dfrac{1-\dfrac{1}{3^{100}}}{2}\)

\(a)\dfrac{x^2}{6}=\dfrac{36}{x}\)

\(=>x^3=36.6\)

\(=>x^3=6^3\)

\(=>x=6\)

(câu b thiếu dữ kiện)

                  áp dụng dãy tỉ số bằng nhau ta có 

                      x/3=y/7=z/2=x+y+z/3+7+2=-16/12=-4/3

                   =>x/3=-4/3=>x=-4/3X3=-4

                  =>y/7=-4/3=>y=-4/3X7=-9,(3)

                 =>z/2=-4/3=>z=-4/3X2=-2(6)

b, \(\dfrac{x}{3}\)= \(\dfrac{y}{7}\)= \(\dfrac{z}{2}\)  Mà x+ y+ z= -16
             
    _xl mk bị thiếu mất vế sau ^^_