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

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

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

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

=>\(x=3.15=45\)

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

\(\Leftrightarrow\frac{7,5-x}{3,5+x}=\frac{5}{6}\)

\(\Leftrightarrow6.\left(7,5-x\right)=5.\left(3,5+x\right)\)

\(\Leftrightarrow45-6x=17,5+5x\)

\(\Leftrightarrow45-17,5=5x+6x\)

\(\Leftrightarrow27,5=11x\)

\(\Leftrightarrow x=\frac{27,5}{11}=2,5\)

Vậy : \(x=2,5\)

b) Tương tự như câu a.

Chúc bạn học tốt nhé !!

a) (x + 15) : x = 4 : 3

=> x : x + 15 : x = \(\frac{4}{3}\)

=> 1 + 15 : x = \(\frac{4}{3}\)

=> 15 : x = \(\frac{4}{3}\)- 1 = \(\frac{1}{3}\)

=> x = 15 : \(\frac{1}{3}\)

=> x = 45

phượng quyên là otaku sa( hình đại diện so cute)

 =>x=45

\(\frac{x+15}{x}=\frac{4}{3}\)

\(\Rightarrow3\left(x+15\right)=4x\)

\(3x+45=4x\)

\(4x-3x=45\)

\(x=45\)

\(\frac{7,5-x}{3,5+x}=\frac{x+40}{x+70}\)

\(\Rightarrow\left(7,5-x\right)\left(x+70\right)=\left(3,5+x\right)\left(x+40\right)\)

\(7,5x+525-x^2-70x=3,5x+140+x^2+40x\)

\(4x+385-2x^2=110x\)

\(385-2x^2=106x\)

Đề sai òi bạn ~  nếu cái đề này thì x  = 412372022 ...... số vô tỉ =.= bậc lên cũng cũng không được số hữa tỉ nào auto ...

a: Ta có: \(\left|\dfrac{2}{5}-x\right|+\dfrac{1}{2}=3.5\)

\(\Leftrightarrow\left|x-\dfrac{2}{5}\right|=3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{2}{5}=3\\x-\dfrac{2}{5}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17}{5}\\x=-\dfrac{13}{5}\end{matrix}\right.\)

b: Ta có: \(\dfrac{21}{5}+3:\left|\dfrac{x}{4}-\dfrac{2}{3}\right|=6\)

\(\Leftrightarrow3:\left|\dfrac{1}{4}x-\dfrac{2}{3}\right|=6-\dfrac{21}{5}=\dfrac{9}{5}\)

\(\Leftrightarrow\left|\dfrac{1}{4}x-\dfrac{2}{3}\right|=\dfrac{5}{3}\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{4}x-\dfrac{2}{3}=\dfrac{5}{3}\\\dfrac{1}{4}x-\dfrac{2}{3}=-\dfrac{5}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{4}x=\dfrac{7}{3}\\\dfrac{1}{4}x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{28}{3}\\x=-4\end{matrix}\right.\)