\(\dfrac{3}{5}\)\(x\) - \(\dfrac{11}{5}\) = \(\dfrac{-3}{14}\) : \(\dfrac{5}{7}\) 

\(\dfrac{3}{5}\)\(x\) - \(\dfrac{11}{5}\) =  - \(\dfrac{3}{10}\)

\(\dfrac{3}{5}\)\(x\)         = - \(\dfrac{3}{10}\) + \(\dfrac{11}{5}\)

\(\dfrac{3}{5}\)\(x\)       = \(\dfrac{19}{10}\) 

\(x\)         = \(\dfrac{19}{10}\) : \(\dfrac{3}{5}\)

\(x\)        = \(\dfrac{19}{6}\)

\(\dfrac{3}{5}x-\dfrac{11}{5}=-\dfrac{3}{14}:\dfrac{5}{7}\)

\(\Rightarrow\dfrac{3}{5}x-\dfrac{11}{5}=-\dfrac{3}{14}\cdot\dfrac{7}{5}\)

\(\Rightarrow\dfrac{3}{5}x-\dfrac{11}{5}=-\dfrac{3}{10}\)

\(\Rightarrow\dfrac{3}{5}x=-\dfrac{3}{10}+\dfrac{11}{5}\)

\(\Rightarrow\dfrac{3}{5}x=\dfrac{19}{10}\)

\(\Rightarrow x=\dfrac{19}{10}:\dfrac{3}{5}\)

\(\Rightarrow x=\dfrac{19}{6}\)

\(a,\dfrac{7}{12}+\dfrac{3}{4}\times\dfrac{2}{9}=\dfrac{7}{12}+\dfrac{1}{6}=\dfrac{7}{12}+\dfrac{2}{12}=\dfrac{9}{12}=\dfrac{3}{4}\)

\(b,\dfrac{8}{9}-\dfrac{4}{15}:\dfrac{2}{5}=\dfrac{8}{9}-\dfrac{4}{15}\times\dfrac{5}{2}=\dfrac{8}{9}-\dfrac{2}{3}=\dfrac{8}{9}-\dfrac{6}{9}=\dfrac{2}{9}\)

đùa thì kệ con mẹ mày xàm loz

Đáp án a là 0,75 

Đáp án b: là 0.86222222222

=>x-1/2=3+1/14-5-1/3-4/7+11/21

=>x-1/2=-97/42

=>x-21/42=-97/42

=>x=-97/42+21/42=-76/42=-38/21

\(\left\{{}\begin{matrix}\dfrac{12}{x-1}+\dfrac{7}{y+3}=19\\\dfrac{2x+6}{x-1}+\dfrac{3y+14}{y+3}=18\end{matrix}\right.\left(x\ne1;y\ne-3\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{12}{x-1}+\dfrac{7}{y+3}=19\\\dfrac{2x-2+8}{x-1}+\dfrac{3y+9+5}{y+3}=18\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{12}{x-1}+\dfrac{7}{y+3}=19\\\dfrac{2\left(x-1\right)}{x-1}+\dfrac{8}{x-1}+\dfrac{3\left(y+3\right)}{y+3}+\dfrac{5}{y+3}=18\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{12}{x-1}+\dfrac{7}{y+3}=19\\2+\dfrac{8}{x-1}+3+\dfrac{5}{y+3}=18\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{12}{x-1}+\dfrac{7}{y+3}=19\\\dfrac{8}{x-1}+\dfrac{5}{y+3}=13\end{matrix}\right.\) (I) 

Đặt: \(\left\{{}\begin{matrix}u=\dfrac{1}{x-1}\\v=\dfrac{1}{y+3}\end{matrix}\right.\)

Hệ (I) trở thành: 

\(\Leftrightarrow\left\{{}\begin{matrix}12u+7v=19\\8u+5v=13\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}24u+14v=38\\24u+15v=39\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}12u+7=19\\v=1\end{matrix}\right.\) 

\(\Leftrightarrow\left\{{}\begin{matrix}12u=12\\v=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}u=1\\v=1\end{matrix}\right.\) 

Trả ẩn phụ: 

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x-1}=1\\\dfrac{1}{y+3}=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y+3=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-2\end{matrix}\right.\left(tm\right)\)

Vậy hệ pt có 1 cặp nghiệm duy nhất là: (2;-2) 

⇔⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩12x−1+7y+3=192x−2+8x−1+3y+9+5y+3=18⇔{12�−1+7�+3=192�−2+8�−1+3�+9+5�+3=18

⇔⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩12x−1+7y+3=192+8x−1+3+5y+3=18⇔{12�−1+7�+3=192+8�−1+3+5�+3=18

⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩u=1x−1v=1y+3{�=1�−1�=1�+3

Hệ (I) trở thành: 

⇔{12u+7v=198u+5v=13⇔{12�+7�=198�+5�=13

⇔{24u+14v=3824u+15v=39⇔{24�+14�=3824�+15�=39

⇔{12u+7=19v=1⇔{12�+7=19�=1 

⇔{12u=12v=1⇔{12�=12�=1

⇔{u=1v=1⇔{�=1�=1 

Trả ẩn phụ: 

`7 xx 3/14 -1/14`

`= 21/14 -1/14`

`= 20/14`

`=10/7`

__

`3/2 + 7/4 xx 2/5`

`= 3/2 + 14/20`

`= 3/2 + 7/10`

`= 15/10 +7/10`

`= 23/10`

__

`9/8 : 3 + 7/3 : 3`

`= 9/8 xx 1/3 + 7/3 xx 1/3`

`=1/3 xx ( 9/8 + 7/3)`

`= 1/3 xx 83/24`

`= 83/72`

__

`2 : 3/4 - 5/8 xx 4/3`

`= 2 xx 4/3 - 5/8 xx 4/3`

`= 4/3 xx ( 2-5/8)`

`= 4/3 xx ( 16/8 -5/8)`

`= 4/3 xx 11/8`

`= 44/24`

`=11/6`

` @ \color{Red}{sushiteam}`

a) Ta có: \(\dfrac{2}{3}x-1=\dfrac{3}{2}\)

\(\Leftrightarrow x\cdot\dfrac{2}{3}=\dfrac{5}{2}\)

hay \(x=\dfrac{5}{2}:\dfrac{2}{3}=\dfrac{5}{2}\cdot\dfrac{3}{2}=\dfrac{15}{4}\)

b) Ta có: \(\left|5x-\dfrac{1}{2}\right|-\dfrac{2}{7}=25\%\)

\(\Leftrightarrow\left|5x-\dfrac{1}{2}\right|=\dfrac{1}{4}+\dfrac{2}{7}=\dfrac{15}{28}\)

\(\Leftrightarrow\left[{}\begin{matrix}5x-\dfrac{1}{2}=\dfrac{15}{28}\\5x-\dfrac{1}{2}=\dfrac{-15}{28}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{29}{28}\\5x=\dfrac{-1}{28}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{29}{140}\\x=\dfrac{-1}{140}\end{matrix}\right.\)

c) Ta có: \(\dfrac{x-3}{4}=\dfrac{16}{x-3}\)

\(\Leftrightarrow\left(x-3\right)^2=64\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=8\\x-3=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-5\end{matrix}\right.\)

d) Ta có: \(\dfrac{-8}{13}+\dfrac{7}{17}+\dfrac{21}{31}\le x\le\dfrac{-9}{14}+4-\dfrac{5}{14}\)

\(\Leftrightarrow\dfrac{3246}{6851}\le x\le3\)

\(\Leftrightarrow x\in\left\{1;2;3\right\}\)

ko biết =))

b: \(\Leftrightarrow\left(x-\dfrac{1}{2}\right):\dfrac{1}{3}=9+\dfrac{5}{7}-\dfrac{5}{7}=9\)

=>x-1/2=27

hay x=55/2

c: =>1/2x-3/4=42/63=2/3

=>1/2x=17/12

hay x=17/6

D

\(B=\dfrac{1+\dfrac{1}{7}+\dfrac{1}{7^2}-\dfrac{1}{7^3}}{4+\dfrac{4}{7}+\dfrac{4}{7^2}-\dfrac{4}{7^3}}\cdot\dfrac{858585}{313131}\cdot\left(-1\dfrac{14}{17}\right)\)

\(=\dfrac{1}{4}\cdot\dfrac{85}{31}\cdot\dfrac{-31}{17}\)

\(=\dfrac{-5}{4}\)

Ta có: \(B=\dfrac{1+\dfrac{1}{7}+\dfrac{1}{7^2}-\dfrac{1}{7^3}}{4+\dfrac{4}{7}+\dfrac{4}{7^2}-\dfrac{4}{7^3}}.\dfrac{858585}{313131}.\left(-1\dfrac{14}{17}\right)\)

             \(=\dfrac{1+\dfrac{1}{7}+\dfrac{1}{7^2}-\dfrac{1}{7^3}}{4\left(1+\dfrac{1}{7}+\dfrac{1}{7^2}-\dfrac{1}{7^3}\right)}.\dfrac{85.10101}{32.10101}.\dfrac{-31}{17}=\dfrac{1}{4}.\dfrac{85}{31}.\dfrac{-31}{17}=-\dfrac{5}{4}\)

TK

https://lazi.vn/edu/exercise/giai-phuong-trinh-4x-5-x-1-2-x-x-1-7-x-2-3-x-5

a: \(\Leftrightarrow4x-5=2x-2+x\)

=>4x-5=3x-2

=>x=3(nhận)

b: =>7x-35=3x+6

=>4x=41

hay x=41/4(nhận)

c: \(\Leftrightarrow\dfrac{14}{3\left(x-4\right)}-\dfrac{x+2}{x-4}=\dfrac{-3}{2\left(x-4\right)}-\dfrac{5}{6}\)

\(\Leftrightarrow\dfrac{28}{6\left(x-4\right)}-\dfrac{6\left(x+2\right)}{6\left(x-4\right)}=\dfrac{-9}{6\left(x-4\right)}-\dfrac{5\left(x-4\right)}{6\left(x-4\right)}\)

\(\Leftrightarrow28-6x-12=-9-5x+20\)

=>-6x+16=-5x+11

=>-x=-5

hay x=5(nhận)

d: \(\Leftrightarrow x^2+2x+1-\left(x^2-2x+1\right)=16\)

\(\Leftrightarrow4x=16\)

hay x=4(nhận)