a) (2x+3)-(5x-17)=9

2x+3-5x+17=9

-3x+20=9

-3x=9-20=-11

=>x=11/3

b) \(\left(\frac{1}{5}+\frac{4}{5}x\right)-\left(\frac{2}{5}x+\frac{1}{9}\right)=\frac{3}{7}\)

=> \(\frac{1}{5}+\frac{4}{5}x-\frac{2}{5}x-\frac{1}{9}=\frac{3}{7}\)

=> \(\left(\frac{1}{5}-\frac{1}{9}\right)+\left(\frac{4}{5}x-\frac{2}{5}x\right)=\frac{3}{7}\)

\(\frac{4}{45}+\frac{2}{5}x=\frac{3}{7}\)

\(\frac{2}{5}x=\frac{3}{7}-\frac{4}{45}=\frac{107}{315}\)

x=107/315:2/5=107/126

dễ tí nữa tôi lấy nk khác của tôi làm thì và kb nhé

a) = 2x-5x+3+17=9\(\Rightarrow\)-3x+20=9\(\Rightarrow\)-3x=-11\(\Rightarrow\)x=11/3

b) = \(\frac{1}{5}-\frac{1}{9}+\frac{4}{5}x-\frac{2}{5}x=\frac{3}{7}\Rightarrow\frac{4}{45}+\frac{2}{5}x=\frac{3}{7}\Rightarrow\frac{2}{5}x=\frac{107}{315}\Rightarrow x=\frac{107}{126}\)

Các bạn giúp mk với, nhanh nhé, mk cần gấp

Câu a có 1 số 0 to thôi nhé!

mk sắp phải đi học rồi các bạn giúp mình với có đc ko mk nhớ sẽ đền đáp công ơn của bạn 

a) (5 - x) +12 = -25

<-> 5 - x + 12 = -25

<-> 17 - x = - 25

<-> x = 42

b) 12 - 4(x - 2) = -4

<-> 12 - 4x + 8 = -4

<-> 20 - 4x = -4

<-> 4x = 24

<-> x = 6

a) (5 - x) + 12 = -25

<=> -x = -25 - 12 - 5

<=> -x = -42

<=> x = 42

b) 12 - 4(x - 2) = -4

<=> 12 - 4x + 8 = -4

<=> -4x = -4 - 8 - 12

<=> -4x = -24

<=> x = 6

c) -15 - |3 - x| = -19

<=> -|3 - x| = -4

<=> 3 - x = 4 hoặc 3 - x = -4

<=> x = -1 hoặc x = 7

\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)

\(\Leftrightarrow\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)

\(\Leftrightarrow\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}2x+\frac{3}{5}=\frac{3}{5}\\2x+\frac{3}{5}=-\frac{3}{5}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x=0\\2x=-\frac{6}{5}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{3}{5}\end{cases}}\)

_Tần vũ_

\(3\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)

\(\Leftrightarrow3\left(3x-\frac{1}{2}\right)^3=-\frac{1}{9}\)

\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=-\frac{1}{27}\)

\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=\left(-\frac{1}{3}\right)^3\)

\(\Leftrightarrow3x-\frac{1}{2}=\frac{-1}{3}\)

\(\Leftrightarrow3x=\frac{1}{6}\)

\(\Leftrightarrow x=\frac{1}{18}\)

_Tần Vũ_

a) \(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)

\(\Rightarrow\left(2x+\frac{3}{5}\right)^2=0+\frac{9}{25}\)

\(\Rightarrow\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)

\(\Rightarrow\left(2x+\frac{3}{5}\right)^2=\frac{3^2}{5^2}\)

\(\Rightarrow\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)

\(\Rightarrow\hept{\begin{cases}2x+\frac{3}{5}=\frac{3}{5}\\2x+\frac{3}{5}=-\frac{3}{5}\end{cases}}\)\(\Rightarrow\hept{\begin{cases}2x=\frac{3}{5}-\frac{3}{5}\\2x=-\frac{3}{5}-\frac{3}{5}\end{cases}}\)\(\Rightarrow\hept{\begin{cases}2x=0\\2x=\frac{-6}{5}\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=0:2\\x=-\frac{6}{5}:2\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=0\\x=-\frac{3}{5}\end{cases}}\)

b) \(3\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)

\(\Rightarrow3\left(3x-\frac{1}{2}\right)^3=0-\frac{1}{9}\)

\(\Rightarrow3\left(3x-\frac{1}{2}\right)^3=-\frac{1}{9}\)

\(\Rightarrow\left(3x-\frac{1}{2}\right)^3=-\frac{1}{9}:3\)

\(\Rightarrow\left(3x-\frac{1}{2}\right)^3=-\frac{1}{27}\)

\(\Rightarrow\left(3x-\frac{1}{2}\right)^3=\left(\frac{-1^3}{3^3}\right)\)

\(\Rightarrow\left(3x-\frac{1}{2}\right)^3=\left(-\frac{1}{3}\right)^3\)

\(\Rightarrow3x-\frac{1}{2}=-\frac{1}{3}\)

\(\Rightarrow3x=-\frac{1}{3}+\frac{1}{2}\)

\(\Rightarrow3x=\frac{1}{6}\)

\(\Rightarrow x=\frac{1}{6}:3\)

\(\Rightarrow x=\frac{1}{18}\)

1,\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{3}{7}.\left(7-\frac{1}{6}\right)+\frac{1}{3}\)

  \(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{3}{7}.\frac{41}{6}+\frac{1}{3}\)

\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{41}{14}+\frac{1}{3}\)

 \(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{137}{42}\)

\(\frac{2}{9}.\left(x-\frac{9}{4}\right)=\frac{137}{42}-\frac{1}{2}\)

\(\frac{2}{9}.\left(x-\frac{9}{4}\right)=\frac{58}{21}\)

 \(\left(x-\frac{9}{4}\right)=\frac{5}{2}:\frac{2}{9}\)

\(\left(x-\frac{9}{4}\right)=\frac{45}{4}\)

\(x=\frac{45}{4}+\frac{9}{4}\)

\(x=\frac{27}{2}\)

Bước cưối 58/21 minh man viết nhầm nên sai 

\(\left(x-\frac{9}{4}\right)=\frac{58}{21}:\frac{2}{9}\)

\(\left(x-\frac{9}{4}\right)=\frac{87}{7}\)

\(x=\frac{87}{7}+\frac{9}{4}\)

\(x=\frac{411}{28}\)

1.

\(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)

\(MC:12\)

Quy đồng :

\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)

\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)

\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)

\(\Leftrightarrow6x+9-3x=-4-9+16\)

\(\Leftrightarrow-7x=3\)

\(\Leftrightarrow x=\frac{-3}{7}\)

2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)

\(MC:20\)

Quy đồng :

\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)

\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)

\(\Leftrightarrow30x+15-20=15x-2\)

\(\Leftrightarrow15x=3\)

\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)