\(2\frac{3}{4}x=3\frac{1}{7}:0,01\)

\(\Leftrightarrow\frac{11}{4}x=\frac{22}{7}:\frac{1}{100}\)

\(\Leftrightarrow\frac{11}{4}x=\frac{2200}{7}\)

\(\Leftrightarrow x=\frac{2200}{7}:\frac{11}{4}\)

\(\Leftrightarrow x=\frac{800}{7}\)

Vậy ..................

\(2\frac{3}{4}x=3\frac{1}{7}:0,01\)

\(\Rightarrow\frac{11}{4}x=\frac{22}{7}:\frac{1}{100}\)

\(\Rightarrow\frac{11}{4}x=\frac{2200}{7}\)

\(\Rightarrow x=\frac{2200}{7}:\frac{11}{4}\)

\(\Rightarrow x=\frac{800}{7}\)

Vậy \(x=\frac{800}{7}.\)

Chúc bạn học tốt!

\(a)\frac{3}{4}+\frac{2}{5}x=\frac{29}{60}\)

  \(\)TỰ LÀM NHA HIHI

 MI SUỐT NGÀY NGỒI MÁY TÍNH LƯỚT FACE, LÚC NÀO ĐI QUA CŨNG THẤY

có chỗ x^7 hả

\(2\frac{3}{4}x=3\frac{1}{7}:0,01\)

=> \(\frac{11}{4}x=\frac{22}{7}:\frac{1}{100}\)

=> \(x=\frac{\frac{22}{7}\cdot100}{\frac{11}{4}}=\frac{\frac{2200}{7}}{\frac{11}{4}}=\frac{2200}{7}\cdot\frac{4}{11}=\frac{800}{7}\)

\(2\frac{1}{3}:\frac{1}{3}=\frac{7}{9}:x\)

=> \(\frac{7}{3}:\frac{1}{3}=\frac{7}{9}:x\)

=> \(\frac{7}{3}\cdot\frac{3}{1}=\frac{7}{9}:x\)

=> \(\frac{7}{9}:x=7\)

=> \(x=\frac{7}{9}:7=\frac{7}{9}\cdot\frac{1}{7}=\frac{1}{9}\)

\(3,2x+\left(-1,2\right)x+2,7=-4,9\)

=> \(\left[3,2+\left(-1,2\right)\right]x=-7,6\)

=> 2.x = -7,6

=> x = -3,8

\(x:\frac{9}{14}=\frac{7}{3}:x\)

=> \(x\cdot\frac{14}{9}=\frac{7}{3}\cdot\frac{1}{x}\)

=> \(\frac{14x}{9}=\frac{7}{3x}\)

nên sửa lại đề này

\(\frac{37-x}{x+13}=\frac{3}{7}\)

=> 7(37 - x) = 3(x + 13)

=> 259 - 7x = 3x + 39

=> 259 - 7x - 3x - 39 = 0

=> 220 - 10x = 0

=> 220  = 10x

=> x = 22

\(\frac{x}{15}=\frac{-60}{x}\)=> x² = -900 => x không thỏa mãn

a) \(2\frac{3}{4}.x=3\frac{1}{7}:0,01\)

\(=>2\frac{3}{4}.x=2200=>\frac{11}{4}x=2200=>x=800\)

b) \(2\frac{1}{3}:\frac{1}{3}=\frac{7}{9}:x\)

\(=>7=\frac{7}{9}:x\)

hay \(\frac{7}{9}:x=7=>x=\frac{1}{9}\)

c) \(3,2x+\left(-1,2\right)x+2,7=-4,9\)

\(=>2x+2,7=-4,9\)

\(=>2x=-7,6=>x=-3,8\)

d) \(x:\frac{9}{14}=\frac{7}{3}:x\)

\(=>x.x=\frac{7}{3}.\frac{9}{14}\)

\(=>x^2=\frac{3}{2}\)

....................( ko hiểu)

e) \(\frac{37-x}{x+13}=\frac{3}{7}\)

\(=>7.\left(37-x\right)=3.\left(x+13\right)\)

\(=>259-7x=3x+39\)

\(-7x-3x=39-269\)

\(=>-10x=-220=>x=22\)

f) \(\frac{x}{15}=\frac{-60}{x}\)

\(=>x.x=-60.15=>x^2=-900=>x=-30\)

cậu có thể tham khảo bài alfm trên đây ạ, chúc học tốt:>

\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)

\(\Leftrightarrow\left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\)

\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)

\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)

\(\Leftrightarrow x+2020=0\)vì \(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\ne0\)

\(\Leftrightarrow x=-2020\)

khó lắm

bây h thì bạn giải đc chưa

Cảm ơn bạn rất nhiều mình đã hiểu rồi 

Chúc bạn học tốt nhé

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

c) \(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)+\frac{5}{9}=\frac{23}{27}\)

\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)=\frac{23}{27}-\frac{5}{9}\)

\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)=\frac{23}{27}-\frac{15}{27}\)

\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)=\frac{8}{27}\)

\(2+\frac{3}{4}x=\frac{7}{9}\div\frac{8}{27}\)

\(2+\frac{3}{4}x=\frac{7}{9}.\frac{27}{8}\)

\(2+\frac{3}{4}x=\frac{21}{8}\)

\(\frac{3}{4}x=\frac{21}{8}-2\)

\(\frac{3}{4}x=\frac{21}{8}-\frac{16}{8}\)

\(\frac{3}{4}x=\frac{5}{8}\)

\(x=\frac{5}{8}\div\frac{3}{4}\)

\(x=\frac{5}{8}.\frac{4}{3}\)

\(x=\frac{5}{6}\)

Vậy \(x=\frac{5}{6}\).

d) \(\left|x-\frac{1}{3}\right|-\frac{3}{4}=\frac{5}{3}\)

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

\(\left|x-\frac{1}{3}\right|=\frac{20}{12}+\frac{9}{12}\)

\(\left|x-\frac{1}{3}\right|=\frac{29}{12}\)

\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{3}=\frac{29}{12}\\x-\frac{1}{3}=-\frac{29}{12}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{11}{4}\\x=-\frac{25}{12}\end{cases}}\)

Vậy \(x\in\left\{\frac{11}{4};-\frac{25}{12}\right\}\).