ii: Ta có: 15:4=x:3.5

nên \(x=\dfrac{3.5\cdot15}{4}=\dfrac{105}{8}\)

\(a,\frac{x+15}{x}=\frac{4}{3}\Rightarrow4x=3x+45\Leftrightarrow x=45\)

\(b,\frac{7,5-x}{3,5+x}=\frac{5}{6}\Rightarrow17,5+5x=45-6x\Leftrightarrow11x=27,5\Rightarrow x=2,5\)

\(c,\frac{x-20}{x-10}=\frac{x+40}{x+70}\Rightarrow\left(x-20\right)\left(x+70\right)=\left(x-10\right)\left(x+40\right)\)

\(\Leftrightarrow x^2+50x-1400=x^2+30x-400\)

\(\Leftrightarrow20x=1000\)

\(\Rightarrow x=50\)

a. \(\frac{\left(x+15\right)}{x}=\frac{4}{3}\Leftrightarrow4x=3\left(x+15\right)\Leftrightarrow4x=3x+45\Leftrightarrow x=45\)

Vậy x=45

b. \(\frac{7,5-x}{3,5+x}=\frac{5}{6}\Leftrightarrow5\left(3,5+x\right)=6\left(7,5-x\right)\Leftrightarrow17,5+5x=45-6x\Leftrightarrow11x=27,5\Leftrightarrow x=2,5\)

Vậy x=2,5

c. \(\frac{x+20}{x-10}=\frac{x+40}{x+70}\Leftrightarrow\left(x+40\right)\left(x-10\right)=\left(x+20\right)\left(x+70\right)\)

\(\Leftrightarrow x^2+30x-400=x^2+90x+1400\Leftrightarrow-60x=-30\Leftrightarrow x=-30\)

Vậy x=-30

a).\(\dfrac{x}{15}\)=\(\dfrac{4}{3}\)\(\Leftrightarrow\)x.3=15.4\(\Leftrightarrow\)3x=60\(\Leftrightarrow\)x=20

b)x:\(\dfrac{4}{3}\)=\(\dfrac{5}{6}\)\(\Leftrightarrow\)\(\dfrac{3x}{4}=\dfrac{5}{6}\)\(\Leftrightarrow\)3x.6=4.5\(\Leftrightarrow\)18x=20\(\Leftrightarrow\)\(\dfrac{10}{9}\)

c)\(\dfrac{7}{2}\)-x=\(\dfrac{5}{6}\)\(\Leftrightarrow\)\(\dfrac{7}{2}-\dfrac{2x}{2}\)=\(\dfrac{5}{6}\)\(\Leftrightarrow\)\(\dfrac{7-2x}{2}\)=\(\dfrac{5}{6}\)\(\Leftrightarrow\)(7-2x).6=2.5\(\Leftrightarrow\)42-12x=10

\(\Leftrightarrow\)-12x=-32\(\Leftrightarrow\)x=\(\dfrac{8}{3}\)

a) \(\dfrac{x}{3}=\dfrac{-10}{6}\)

\(x\times6=-10\times3\)

\(x\times6=-30\)

\(x=-5\)

b) \(\dfrac{-8}{x}=\dfrac{-9}{15}\)

\(x\times-9=15\times-8\)

\(x\times-9=-120\)

\(x=\dfrac{40}{3}\)

c) \(\dfrac{2,7}{0,9}=\dfrac{-8}{x}\)

\(x\times2,7=-8\times0,9\)

\(x\times2,7=-7,2\)

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

d) \(\dfrac{4}{9}=\dfrac{x}{12}\)

\(x\times9=12\times4\)

\(x\times9=48\)

\(x=\dfrac{48}{9}\)

\(x=\dfrac{16}{3}\)

\(a,\left(x+15\right):x=4:3\)

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

=>\(\dfrac{15}{x}=\dfrac{1}{3}\)

=>\(x=3.15=45\)

Vậy x=45

b)\(\dfrac{7,5-x}{3,5+x}=\dfrac{5}{6}\)

\(6\left(7,5-x\right)=5\left(3,5+x\right)\)

=>\(45-6x=17,5+5x\)

=>\(-11x=-27,5\)

=>\(x=2,5\)

Vậy...

c)\(\dfrac{x-20}{x-10}=\dfrac{x+40}{x+70}\)

=>\(\left(x-20\right)\left(x+70\right)=\left(x-10\right)\left(x+40\right)\)

=>\(x^2+50x-140=x^2+30x-40\)

=>\(20x=100\)

=>\(x=50\)

a)x*4/5:2=6/7               b)6/5:x:5/4=10/15

x*4/5=6/7 x 2                6/5:x=10/15 x 5/4

x*4/5=12/7                     6/5:x=5/6

x=12/7:4/5                       x=6/5:5/6

x=15/7                            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

    Vậy x = 15/7

Bài 2: 

a: 

1: \(\dfrac{a}{b}=\dfrac{c}{d}\)

\(\Leftrightarrow\dfrac{a+b}{a}=\dfrac{c+d}{c}\)

hay \(\dfrac{a}{a+b}=\dfrac{c}{c+d}\)