Giúp mình nhanh với ạ! Thank,s

a) `1/4 x + 7/4 x = -6`

`=> x(1/4 + 7/4) = -6`

`=> x. 2 = -6`

`=> x= -6/2`

`=> x= -3`

Vậy `x= -3`

b) `1/4 x +2x = 9/2`

`=> x(1/4 +2) =9/2`

`=> x(1/4 + 8/4) = 9/2`

`=>x. 9/4 =9/2`

`=> x= 9/2 : 9/4`

`=> x= 2`

Vậy `x=2`

Bạn ơi, bạn viết lại đề đi. Khó nhìn quá

a) \(\Leftrightarrow2\left|3x-1\right|=\dfrac{4}{5}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=\dfrac{2}{5}\\3x-1=-\dfrac{2}{5}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{15}\\x=\dfrac{1}{5}\end{matrix}\right.\)

b)TH1:  \(x\ge3\)

\(\Leftrightarrow x+5+x-3=9\Leftrightarrow2x=7\Leftrightarrow x=\dfrac{7}{2}\left(tm\right)\)

TH2: \(-5\le x< 3\)

\(\Leftrightarrow x+5-x+3=9\Leftrightarrow8=9\left(VLý\right)\)

TH3: \(x< -5\)

\(\Leftrightarrow-x-5-x+3=9\Leftrightarrow2x=-11\Leftrightarrow x=-\dfrac{11}{2}\left(tm\right)\)

\(a,2.|3x-1|-\dfrac{3}{4}=\dfrac{1}{20}\)

\(2.|3x-1|=\dfrac{1}{20}+\dfrac{3}{4}\)

\(2.|3x-1|=\dfrac{4}{5}\)

\(|3x-1|=\dfrac{4}{5}:2\)

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

\(\Rightarrow3x-1=\pm\dfrac{2}{5}\)

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

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

\(3x=\dfrac{7}{5}\)

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

\(x=\dfrac{7}{15}\)

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

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

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

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

\(x=\dfrac{1}{5}\)

\(a,\Rightarrow\left|3x-1\right|=\dfrac{4}{5}:2=\dfrac{2}{5}\\ \Rightarrow\left[{}\begin{matrix}3x-1=\dfrac{2}{5}\left(x\ge\dfrac{1}{3}\right)\\1-3x=\dfrac{2}{5}\left(x< \dfrac{1}{3}\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=\dfrac{7}{5}\left(x\ge\dfrac{1}{3}\right)\\3x=\dfrac{3}{5}\left(x< \dfrac{1}{3}\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{15}\left(tm\right)\\x=\dfrac{1}{5}\left(tm\right)\end{matrix}\right.\)

\(b,\Rightarrow\left[{}\begin{matrix}-x-5+3-x=9\left(x< -5\right)\\x+5+3-x=9\left(-5\le x< 3\right)\\x+5+x-3=9\left(x\ge3\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{11}{2}\left(tm\right)\\0x=1\left(ktm\right)\\x=\dfrac{7}{2}\left(tm\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{11}{2}\\x=\dfrac{7}{2}\end{matrix}\right.\)

a, \(\left(\dfrac{7}{2}-2x\right).\dfrac{10}{3}=\dfrac{22}{3}\Leftrightarrow\dfrac{7}{2}-2x=\dfrac{22}{10}=\dfrac{11}{5}\)

\(\Leftrightarrow2x=\dfrac{13}{10}\Leftrightarrow x=\dfrac{13}{20}\)

b, \(\dfrac{4x}{9}=\dfrac{9}{8}-\dfrac{125}{1000}=1\Leftrightarrow x=\dfrac{9}{4}\)

c, \(-\dfrac{x}{21}=\dfrac{60}{21}\Rightarrow x=-60\)

1

a)103/35

b)-3/13

Bài 13: 

a: =>20-x=15-8+13=20

hay x=0

a)

\(\begin{array}{l}\frac{2}{9}:x + \frac{5}{6} = 0,5\\\frac{2}{9}:x = \frac{1}{2} - \frac{5}{6}\\\frac{2}{9}:x = \frac{3}{6} - \frac{5}{6}\\\frac{2}{9}:x = \frac{{ - 2}}{6}\\x = \frac{2}{9}:\frac{{ - 2}}{6}\\x = \frac{2}{9}.\frac{{ - 6}}{2}\\x = \frac{{ - 2}}{3}\end{array}\)                        

Vậy \(x = \frac{{ - 2}}{3}\).

b)

\(\begin{array}{l}\frac{3}{4} - \left( {x - \frac{2}{3}} \right) = 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - \frac{4}{3}\\x - \frac{2}{3} = \frac{9}{{12}} - \frac{{16}}{{12}}\\x - \frac{2}{3} = \frac{{ - 7}}{{12}}\\x = \frac{{ - 7}}{{12}} + \frac{2}{3}\\x = \frac{{ - 7}}{{12}} + \frac{8}{{12}}\\x = \frac{1}{12}\end{array}\)

Vậy\(x = \frac{1}{12}\).

c)

\(\begin{array}{l}1\frac{1}{4}:\left( {x - \frac{2}{3}} \right) = 0,75\\\frac{5}{4}:\left( {x - \frac{2}{3}} \right) = \frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}:\frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}.\frac{4}{3}\\x - \frac{2}{3} = \frac{5}{3}\\x = \frac{5}{3} + \frac{2}{3}\\x = \frac{7}{3}\end{array}\)               

Vậy \(x = \frac{7}{3}\).

d)

\(\begin{array}{l}\left( { - \frac{5}{6}x + \frac{5}{4}} \right):\frac{3}{2} = \frac{4}{3}\\ - \frac{5}{6}x + \frac{5}{4} = \frac{4}{3}.\frac{3}{2}\\ - \frac{5}{6}x + \frac{5}{4} = 2\\ - \frac{5}{6}x = 2 - \frac{5}{4}\\ - \frac{5}{6}x = \frac{8}{4} - \frac{5}{4}\\ - \frac{5}{6}x = \frac{3}{4}\\x = \frac{3}{4}:\left( { - \frac{5}{6}} \right)\\x = \frac{3}{4}.\frac{{ - 6}}{5}\\x = \frac{{ - 9}}{{10}}\end{array}\)

Vậy \(x = \frac{{ - 9}}{{10}}\).

-29-9(2x-1)\(^2\)= -110

(=) 9(2x-1)² = (-29) +110

(=) 9(2x-1)² = 81

(=) (2x-1)² =81: 9

(=) (2x-1)² =9

(=) (2x-1)² = 3² =(-3)²

\(\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}}\)

\(\orbr{\begin{cases}2x=4\\2x=-2\end{cases}}\)

\(\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)

vậy : ........

a,\(-29-9\left(2x-1\right)^2=-110\)

\(=>-29+110=9.\left(2x-1\right)^2\)

\(=>81=9.\left(2x-1\right)^2\)

\(=>\left(2x-1\right)^2=9\)

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

b,\(-2.\left(-x-5\right)+28=20-3.4\)

\(=>2x+10+28=20-12=8\)

\(=>2x=8-28-10=-30\)

\(=>x=\frac{-30}{2}=-15\)

Giải:

a) \(\dfrac{-1}{5}\le\dfrac{x}{8}\le\dfrac{1}{4}\) 

\(\Rightarrow\dfrac{-8}{40}\le\dfrac{5x}{40}\le\dfrac{10}{40}\) 

\(\Rightarrow5x\in\left\{0;\pm5;10\right\}\) 

\(\Rightarrow x\in\left\{0;\pm1;2\right\}\) 

b) \(\dfrac{4}{x-6}=\dfrac{y}{24}=\dfrac{-12}{18}\) 

\(\Rightarrow\dfrac{4}{x-6}=\dfrac{-12}{18}\) 

\(\Rightarrow-12.\left(x-6\right)=4.18\) 

\(\Rightarrow-12x+72=72\) 

\(\Rightarrow-12x=72-72\) 

\(\Rightarrow-12x=0\) 

\(\Rightarrow x=0:-12\) 

\(\Rightarrow x=0\) 

\(\Rightarrow\dfrac{y}{24}=\dfrac{-12}{18}\) 

\(\Rightarrow y=\dfrac{-12.24}{18}=-16\) 

c) \(\dfrac{x+46}{20}=x.\dfrac{2}{5}\) 

\(\dfrac{x+46}{20}=\dfrac{2x}{5}\) 

\(\Rightarrow5.\left(x+46\right)=2x.20\) 

\(\Rightarrow5x+230=40x\) 

\(\Rightarrow5x-40x=-230\) 

\(\Rightarrow-35x=-230\) 

\(\Rightarrow x=-230:-35\) 

\(\Rightarrow x=\dfrac{46}{7}\) 

Chúc bạn học tốt!