bạn viết rõ đề ra nhé

b, \(\left|4x-8\right|=1-x\)ĐK : \(x\le1\)

TH1 : \(4x-8=1-x\Leftrightarrow5x=9\Leftrightarrow x=\dfrac{9}{5}\)( ktm )

TH2 : \(4x-8=x-1\Leftrightarrow3x=7\Leftrightarrow x=\dfrac{7}{3}\)( ktm )

b) Ta có: \(\left|4x-8\right|=1-x\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-8=1-x\left(x\ge2\right)\\4x-8=x-1\left(x< 2\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x+x=1+8\\4x-x=-1+8\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}5x=9\\3x=7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{5}\left(loại\right)\\x=\dfrac{7}{3}\left(loại\right)\end{matrix}\right.\)

a) \(\dfrac{13}{20}+\dfrac{3}{5}+x=\dfrac{5}{6}\)

\(\Rightarrow\dfrac{5}{4}+x=\dfrac{5}{6}\)

\(\Rightarrow x=\dfrac{5}{6}-\dfrac{5}{4}\)

\(\Rightarrow x=\dfrac{-5}{12}\)

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

\(\Rightarrow x+\dfrac{1}{3}=\dfrac{11}{15}\)

\(\Rightarrow x=\dfrac{11}{15}-\dfrac{1}{3}\)

\(\Rightarrow x=\dfrac{2}{5}\)

c)\(\dfrac{-5}{8}-x=\dfrac{-3}{20}-\dfrac{-1}{6}\)

\(\dfrac{-5}{8}-x=\dfrac{1}{60}\)

\(\Rightarrow x=\dfrac{-5}{8}-\dfrac{1}{60}\)

\(\Rightarrow x=\dfrac{-77}{120}\)

d) \(\dfrac{3}{5}-x=\dfrac{1}{4}+\dfrac{7}{10}\)

\(\Rightarrow\dfrac{3}{5}-x=\dfrac{19}{20}\)

\(\Rightarrow x=\dfrac{3}{5}-\dfrac{19}{20}\)

\(\Rightarrow x=\dfrac{-7}{20}\)

e) \(\dfrac{-3}{7}-x=\dfrac{4}{5}+\dfrac{-2}{3}\)

\(\Rightarrow\dfrac{-3}{7}-x=\dfrac{2}{15}\)

\(\Rightarrow x=\dfrac{-3}{7}-\dfrac{2}{15}\)

\(\Rightarrow x=\dfrac{-59}{105}\)

g) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)

\(\Rightarrow\dfrac{-5}{6}-x=\dfrac{1}{4}\)

\(\Rightarrow x=\dfrac{-5}{6}-\dfrac{1}{4}\)

\(\Rightarrow x=\dfrac{-13}{12}\)

`x-1/9 =8/3`

`=>x=8/3 +1/9`

`=> x= 24/9 +1/9`

`=>x= 25/9`

Vậy `x=25/9`

__

`x-2/20=-5/2-x`

`=>x+x=-5/2 +2/20`

`=> 2x= -50/20 +2/20`

`=> 2x= -48/20`

`=> x= -12/5:2`

`=>x=-12/5 xx1/2`

`=>x= -12/10`

`=>x= -6/5`

Vậy `x=-6/5`

`@` `\text {Ans}`

`\downarrow`

`a)`

Đề là \(\dfrac{x-1}{9}=\dfrac{8}{3}\) phải hongg bạn?

\(\dfrac{x-1}{9}=\dfrac{8}{3}\)

`=>` `(x-1)3 = 8*9`

`=> (x-1)3=72`

`=> x-1=72 \div 3`

`=> x-1=24`

`=> x=25`

`b)`

\(\dfrac{x-2}{20}=\dfrac{-5}{2-x}\)

`=>` `(x-2)(2-x)=20*(-5)`

`=> (x-2)(2-x)=-100`

`=> -[(x-2)(x-2)]=-100`

`=> -(x-2)^2 = -100`

`=> (x-2)^2 = 100`

`=> (x-2)^2 = (+-10)^2`

`=>`\(\left[{}\begin{matrix}x-2=10\\x-2=-10\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=12\\x=-8\end{matrix}\right.\)

Giải:

a) \(2\dfrac{17}{20}-1\dfrac{15}{11}+6\dfrac{9}{20}:3\)

\(=\dfrac{57}{20}-\dfrac{26}{11}+\dfrac{129}{20}:3\) 

\(=\dfrac{107}{220}+\dfrac{43}{20}\)

\(=\dfrac{29}{11}\)

b) \(4\dfrac{3}{7}:\left(\dfrac{7}{5}.4\dfrac{3}{7}\right)\) 

\(=\dfrac{31}{7}:\left(\dfrac{7}{5}.\dfrac{31}{7}\right)\) 

\(=\dfrac{31}{7}:\dfrac{31}{5}\) 

\(=\dfrac{5}{7}\) 

c) \(\left(3\dfrac{2}{9}.\dfrac{15}{23}.1\dfrac{7}{29}\right):\dfrac{5}{23}\) 

\(=\left(\dfrac{29}{9}.\dfrac{15}{23}.\dfrac{36}{29}\right):\dfrac{5}{23}\) 

\(=\dfrac{60}{23}:\dfrac{5}{23}\) 

\(=12\)

\(a,\dfrac{x}{7}=\dfrac{6}{12}\\ x\cdot12=7\cdot6=42\\ x=42:12\\ x=\dfrac{7}{2}\\ b,\dfrac{-5}{x}=\dfrac{20}{28}\\ x\cdot20=\left(-5\right)\cdot28=-140\\ x=\left(-140\right):20\\ x=-7\\ c,\dfrac{x-2}{8}=\dfrac{3}{4}\\ \left(x-2\right)4=8\cdot3=24\\ x-2=24:4\\ x-2=6\\ x=6+2\\ x=8\\ d,\dfrac{x}{-5}=\dfrac{-5}{x}\\ x^2=\left(-5\right)\cdot\left(-5\right)=25\\ x=5\)

b) (3.x - 24) . 7³ = 2 . 7⁴

(3.x - 24) . 343 = 4802

3.x - 24 = 4802 : 343

3.x - 24 = 14

3.x = 14 + 24

3.x = 38

Tự giải :))

c) x = 1 

Đề hơi khó hiểu bn nhé

\(\left(\dfrac{1}{4}x-\dfrac{1}{8}\right).\dfrac{3}{4}=\dfrac{1}{4}\)

\(\dfrac{1}{4}x-\dfrac{1}{8}=\dfrac{1}{4}:\dfrac{3}{4}=\dfrac{1}{3}\)

\(\dfrac{1}{4}x=\dfrac{1}{3}+\dfrac{1}{8}=\dfrac{11}{24}\)

\(x=\dfrac{11}{24}:\dfrac{1}{4}=\dfrac{11}{6}\)

\(\left(\dfrac{1}{4}x\times\dfrac{-1}{8}\right)\times\dfrac{3}{4}=\dfrac{1}{4}< =>\dfrac{-x}{32}\times\dfrac{3}{4}=\dfrac{1}{4}< =>\dfrac{-x}{32}=\dfrac{1}{4}:\dfrac{3}{4}< =>\dfrac{-x}{32}=\dfrac{1}{3}< =>x=\dfrac{1}{3}\times\left(-32\right)< =>x=\dfrac{-32}{3}\)

\(a,78-\left(x-3\right)=16\)

\(78-x+3=16\)

\(51-x=16\)

\(x=51-16=35\)

\(b,\left(2x+3\right):4=60\)

\(2x+3=60\cdot4=240\)

\(2x=240-3=237\)

\(x=\frac{237}{2}\)

\(c,\left(2x-8\right)=\left(2x-8\right)^7\)

\(\left(2x-8\right)-\left(2x-8\right)^7=0\)

Để mik nghĩ thêm tí r làm tiép

Để tui làm nốt ý C cho !!!

( 2x - 8 ) = ( 2x - 8 )⁷

=> ( 2x - 8 ) . 1 - ( 2x - 8 ) . ( 2x - 8 )⁶ = 0

=> ( 2x - 8 ) . [ 1 - ( 2x - 8 )⁶ ] = 0

=> 1 - ( 2x - 8 )⁶ = 0

=> ( 2x - 8 )⁶ = 1

=> ( 2x - 8 )⁶ = ( 1⁶ )

\(\Rightarrow\orbr{\begin{cases}2x-8=1\\2x-8=-1\end{cases}\Rightarrow\orbr{\begin{cases}2x=9\\2x=7\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{9}{2}\\x=\frac{7}{2}\end{cases}}}\)

Vậy ........

c) \(\left(2x-8\right)=\left(2x-8\right)^7\)

\(\Leftrightarrow\left(2x-8\right)^7-\left(2x-8\right)=0\)

\(\Leftrightarrow\left(2x-8\right)\left[\left(2x-8\right)^6-1\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x-8=0\\\left(2x-8\right)^6-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=4\\\left(2x-8\right)^6=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=4\\x\in\left\{\frac{7}{2};\frac{9}{2}\right\}\end{cases}}\)

Vậy \(x\in\left\{4;\frac{7}{2};\frac{9}{2}\right\}\)