\(a.\) \(\frac{x}{3.15}\) \(=\) \(\frac{0.15}{7.2}\) 

<=> \(x=\) \(\frac{0.15\cdot3.15}{7.2}\) 

<=>\(x=0.065625\)

\(b.\) \(\frac{-2.6}{x-11}=\frac{-12}{42}\) 

<=>\(x-11=\frac{-2.6\cdot-12}{42}\)

<=>\(x-11=0.742857142\)

<=>\(x=11.74285714\)

\(3\frac{4}{5}:2x=0,25:2\frac{2}{3}\)

\(\Leftrightarrow\frac{19}{5}:2x=0,25:\frac{8}{3}\)

\(\Leftrightarrow\frac{19}{5}:2x=0,25\cdot\frac{3}{8}\)

\(\Leftrightarrow\frac{19}{5}:2x=\frac{3}{32}\)

\(\Leftrightarrow2x=\frac{19}{5}:\frac{3}{32}=\frac{19}{5}\cdot\frac{32}{3}=\frac{608}{15}\)

\(\Leftrightarrow x=\frac{608}{15}:2=\frac{608}{30}=\frac{304}{15}\)

a: \(\Leftrightarrow4x+\dfrac{3}{4}=2\cdot\dfrac{2}{5}+0.01\cdot10=\dfrac{9}{10}\)

=>4x=3/20

hay x=3/80

b: \(\Leftrightarrow\left|x\right|=4+\dfrac{1}{8}-9=-\dfrac{39}{8}\)(vô lý)

c: 2x(x-2/3)=0

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-\dfrac{2}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)

d: \(\dfrac{37-x}{x+13}=\dfrac{3}{7}\)

=>259-7x=3x+39

=>-10x=-220

hay x=22

a) 0,4:x=x:0,9

0,4.0,9=x²

0,36 =x²

\(\sqrt{0,36}\)=x

0,6 =x

a) 0,4 : x = x : 0,9 x^2=0,9 . 0,4 x^2=(0,6)^2 =>x=+_6

a,  0,4 : x = x : 0,9

<=> x² = 0,4 . 0,9

<=> x² = 0,36

<=> x = 0,6 hoặc -0,6

b, \(13\frac{1}{3}\div1\frac{1}{3}=26\div\left(2x-1\right)\)

\(\Leftrightarrow\frac{40}{3}\div\frac{4}{3}=26\div\left(2x-1\right)\)

\(\Leftrightarrow10=26\div\left(2x-1\right)\)

\(\Leftrightarrow2x-1=\frac{13}{5}\)

\(\Leftrightarrow2x=\frac{18}{5}\)

\(\Leftrightarrow x=\frac{9}{5}\)

c, \(0,2\div1\frac{1}{5}=\frac{2}{3}\div\left(6x+7\right)\)

\(\Leftrightarrow\frac{1}{5}\div\frac{6}{5}=\frac{2}{3}\div\left(6x+7\right)\)

\(\Leftrightarrow\frac{1}{6}=\frac{2}{3}\div\left(6x+7\right)\)

\(\Leftrightarrow6x+7=4\)

\(\Leftrightarrow6x=-3\)

\(\Leftrightarrow x=\frac{-1}{2}\)

d, \(\frac{37-x}{x+13}=\frac{3}{7}\) 

\(\Leftrightarrow7\left(37-x\right)=3\left(x+13\right)\)

\(\Leftrightarrow259-7x=3x+39\)

\(\Leftrightarrow-10x=-220\)

\(\Leftrightarrow x=22\)

a) \(\left(\frac{5}{7}x-\frac{1}{4}\right)\left(\frac{-3}{4}x+\frac{1}{2}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}\frac{5}{7}x-\frac{1}{4}=0\\\frac{-3}{4}x+\frac{1}{2}=0\end{cases}}\Rightarrow\orbr{\begin{cases}\frac{5}{7}x=\frac{1}{4}\\\frac{-3}{4}x=\frac{-1}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{7}{20}\\x=\frac{2}{3}\end{cases}}\)

Vậy \(x=\frac{7}{20}\) hoặc x=\(\frac{2}{3}\)

b) \(\left(\frac{4}{5}+x\right)\left(x-\frac{8}{13}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}\frac{4}{5}+x=0\\x-\frac{8}{13}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-4}{5}\\x=\frac{8}{13}\end{cases}}\)

Vậy x=-4/5 hoặc x=8/13

c) \(\left(2x-\frac{1}{2}\right)\left(x-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-\frac{1}{2}=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=3\end{cases}}\)

Vậy x=1/4 hoặc x=3

\(x+\frac{7}{2}x+x=\frac{1}{2}\)

\(2x+\frac{7}{2}x=\frac{1}{2}\)

\(\left(2+\frac{7}{2}\right)x=\frac{1}{2}\)

\(\frac{11}{2}x=\frac{1}{2}\)

\(x=\frac{1}{2}:\frac{11}{2}\)

\(x=\frac{1}{11}\)

1a) \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)

=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{cases}}\)

=> \(\orbr{\begin{cases}-\frac{5}{2}x=-\frac{3}{2}\\\frac{11}{2}x=\frac{1}{2}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{1}{11}\end{cases}}\)

b) \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)

=>\(\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)

=> \(\orbr{\begin{cases}\frac{5}{4}x-\frac{7}{2}=\frac{5}{8}x+\frac{3}{5}\\\frac{5}{4}x-\frac{7}{2}=-\frac{5}{8}x-\frac{3}{5}\end{cases}}\)

=> \(\orbr{\begin{cases}\frac{5}{8}x=\frac{41}{10}\\\frac{15}{8}x=\frac{29}{10}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)

c) TT

a, \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)

=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\-\frac{3}{2}x-\frac{1}{2}=4x-1\end{cases}}\)

=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}-4x=-1\\-\frac{3}{2}x-\frac{1}{2}-4x=-1\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)

\(b,\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)

=> \(\left|\frac{5}{4}x-\frac{7}{2}\right|-0=\left|\frac{5}{8}x+\frac{3}{5}\right|\)

=> \(\frac{\left|5x-14\right|}{4}=\frac{\left|25x+24\right|}{40}\)

=> \(\frac{10(\left|5x-14\right|)}{40}=\frac{\left|25x+24\right|}{40}\)

=> \(\left|50x-140\right|=\left|25x+24\right|\)

=> \(\orbr{\begin{cases}50x-140=25x+24\\-50x+140=25x+24\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)

c, \(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)

=> \(\orbr{\begin{cases}\frac{7}{5}x+\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\\-\frac{7}{5}x-\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\end{cases}}\)

=> \(\orbr{\begin{cases}x=-\frac{55}{4}\\x=-\frac{25}{164}\end{cases}}\)

Bài 2 : a. |2x - 5| = x + 1

 TH1 : 2x - 5 = x + 1

    => 2x - 5 - x = 1

    => 2x - x - 5 = 1

    => 2x - x = 6

    => x = 6

TH2 : -2x + 5 = x + 1

   => -2x + 5 - x = 1

   => -2x - x + 5 = 1

   => -3x = -4

   => x = 4/3

Ba bài còn lại tương tự

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

⇒ \(\frac{1}{3}x=\frac{1}{3}+\frac{2}{5}\)

⇒ \(\frac{1}{3}x=\frac{11}{15}\)

⇒ \(x=\frac{11}{15}:\frac{1}{3}\)

⇒ \(x=\frac{11}{5}\)

Vậy \(x=\frac{11}{5}.\)

2) \(2,5:7,5=x:\frac{3}{5}\)

⇒ \(\frac{5}{2}:\frac{15}{2}=x:\frac{3}{5}\)

⇒ \(\frac{1}{3}=x:\frac{3}{5}\)

⇒ \(x=\frac{1}{3}.\frac{3}{5}\)

⇒ \(x=\frac{1}{5}\)

Vậy \(x=\frac{1}{5}.\)

4) \(\left|x\right|+\left|x+2\right|=0\)

Có: \(\left\{{}\begin{matrix}\left|x\right|\ge0\\\left|x+2\right|\ge0\end{matrix}\right.\forall x.\)

⇒ \(\left|x\right|+\left|x+2\right|=0\)

⇒ \(\left\{{}\begin{matrix}x=0\\x+2=0\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=0\\x=0-2\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

Vô lí vì \(x\) không thể nhận cùng lúc 2 giá trị khác nhau.

⇒ \(x\in\varnothing\)

Vậy không tồn tại giá trị nào của \(x\) thỏa mãn yêu cầu đề bài.

10) \(5-\left|1-2x\right|=3\)

⇒ \(\left|1-2x\right|=5-3\)

⇒ \(\left|1-2x\right|=2\)

⇒ \(\left[{}\begin{matrix}1-2x=2\\1-2x=-2\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}2x=1-2=-1\\2x=1+2=3\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=\left(-1\right):2\\x=3:2\end{matrix}\right.\)

⇒ \(\left[{}\begin{matrix}x=-\frac{1}{2}\\x=\frac{3}{2}\end{matrix}\right.\)

Vậy \(x\in\left\{-\frac{1}{2};\frac{3}{2}\right\}.\)

Chúc bạn học tốt!

9, \(13\frac{1}{3}:1\frac{1}{3}=26:\left(2x-1\right)\)

\(\frac{40}{3}:\frac{4}{3}=26:\left(2x-1\right)\)

\(10=26:\left(2x-1\right)\)

\(2x-1=26:10\)

\(2x-1=2,6\)

\(2x=2,6+1\)

\(2x=3,6\)

\(x=3,6:2\)

\(x=1,8\)