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

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

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

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

\(\Leftrightarrow20x=1000\)

\(\Rightarrow x=50\)

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

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

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

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

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

\(\Leftrightarrow27,5=11x\)

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

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)

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

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

\(3x+45=4x\)

\(4x-3x=45\)

\(x=45\)

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

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

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

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

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

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: \(\dfrac{1}{4}:x=3\dfrac{4}{5}:40\dfrac{8}{15}\)

\(\Leftrightarrow x=\dfrac{1}{4}\cdot\dfrac{\dfrac{608}{15}}{3+\dfrac{4}{5}}\)

\(\Leftrightarrow x=\dfrac{152}{15}:\dfrac{19}{5}=\dfrac{8}{3}\)

b: Ta có: \(\left(x+1\right):\dfrac{5}{6}=\dfrac{20}{3}\)

\(\Leftrightarrow x+1=\dfrac{50}{9}\)

hay \(x=\dfrac{41}{9}\)

c: Ta có: \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)

\(\Leftrightarrow x^2-1=63\)

\(\Leftrightarrow x^2=64\)

hay \(x\in\left\{8;-8\right\}\)

c. \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\) 

    \(7.9=\left(x-1\right).\left(x+1\right)\) 

    \(63=x^2-1\) 

             \(x^2=63+1\) 

             \(x^2=64\) 

             \(x^2=8^2\)

             \(x=8\)          

\(a,\Leftrightarrow\left|x+\dfrac{2}{5}\right|=\dfrac{7}{4}\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{2}{5}=\dfrac{7}{4}\left(x\ge-\dfrac{2}{5}\right)\\x+\dfrac{2}{5}=-\dfrac{7}{4}\left(x< -\dfrac{2}{5}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{27}{20}\left(tm\right)\\x=-\dfrac{43}{20}\left(tm\right)\end{matrix}\right.\)

\(b,\Leftrightarrow\left|x-\dfrac{13}{10}\right|=\dfrac{13}{10}\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{13}{10}=\dfrac{13}{10}\left(x\ge\dfrac{13}{10}\right)\\x-\dfrac{13}{10}=-\dfrac{13}{10}\left(x< \dfrac{13}{10}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13}{5}\left(tm\right)\\x=0\left(tm\right)\end{matrix}\right.\)

\(c,\Leftrightarrow\left|\dfrac{3}{4}-\dfrac{1}{2}x\right|=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{4}-\dfrac{1}{2}x=\dfrac{1}{2}\left(x\le\dfrac{3}{2}\right)\\\dfrac{1}{2}x-\dfrac{3}{4}=\dfrac{1}{2}\left(x>\dfrac{3}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\x=\dfrac{5}{2}\left(tm\right)\end{matrix}\right.\)

\(d,\Leftrightarrow\left|5-2x\right|=4\Leftrightarrow\left[{}\begin{matrix}5-2x=4\left(x\le\dfrac{5}{2}\right)\\2x-5=4\left(x>\dfrac{5}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\x=\dfrac{9}{2}\left(tm\right)\end{matrix}\right.\)

\(đ,\Leftrightarrow\left\{{}\begin{matrix}x-3,5=0\\x-1,3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3,5\\x=1,3\end{matrix}\right.\left(vô.lí\right)\Leftrightarrow x\in\varnothing\)

\(e,\Leftrightarrow\left\{{}\begin{matrix}x-2021=0\\x-2022=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2021\\x=2022\end{matrix}\right.\left(vô.lí\right)\Leftrightarrow x\in\varnothing\)

\(f,\Leftrightarrow\left|x\right|=\dfrac{1}{3}-x\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}-x\left(x\ge0\right)\\x=x-\dfrac{1}{3}\left(x< 0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\left(tm\right)\\0x=-\dfrac{1}{3}\left(vô.lí\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{1}{6}\)

\(g,\Leftrightarrow\left[{}\begin{matrix}x-2=x\left(x\ge2\right)\\2-x=x\left(x< 2\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}0x=2\left(vô.lí\right)\\x=1\left(tm\right)\end{matrix}\right.\Leftrightarrow x=1\)