a) \(\left(\frac{1}{7}x-\frac{2}{7}\right)\cdot\left(-\frac{1}{5}x+\frac{3}{5}\right)\cdot\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)
\(\Rightarrow\)TH1 : \(\frac{1}{7}x-\frac{2}{7}=0\)         TH2 : \(-\frac{1}{5}x+\frac{3}{5}=0\)                TH3 : \(\frac{1}{3}x+\frac{4}{3}=0\)

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

\(x=\frac{2}{7}\cdot7\)                                                                      \(x=\frac{3}{5}\cdot-5\)                             \(x=\frac{4}{3}\cdot3\)

\(x=2\)                                                                               \(x=-3\)                                     \(x=4\)
Vậy x = 2 hoặc x = -3 hoặc x = 4
b) \(\frac{1}{6}x+\frac{1}{10}x-\frac{4}{5}x+1=0\)
 

\(x\cdot\left(\frac{1}{6}+\frac{1}{10}-\frac{4}{5}\right)=1\)

\(x\cdot\frac{5+3-24}{30}=1\)

\(x\cdot\frac{-8}{15}=1\)

\(x=1\cdot\frac{-15}{8}=\frac{-15}{8}\)
Vậy x = \(\frac{-15}{8}\)

d,

\(|x-\frac{1}{3}|=\frac{5}{6}\Rightarrow \left[\begin{matrix} x-\frac{1}{3}=\frac{5}{6}\\ x-\frac{1}{3}=-\frac{5}{6}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{7}{6}\\ x=\frac{-1}{2}\end{matrix}\right.\)

e,

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

\(\Leftrightarrow 2|2x-\frac{2}{3}|=\frac{3}{4}-2=\frac{-5}{4}\)

\(\Leftrightarrow |2x-\frac{2}{3}|=-\frac{5}{8}<0\) (vô lý vì trị tuyệt đối của 1 số luôn không âm)

Vậy không tồn tại $x$ thỏa mãn đề bài.

f, 

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

\(\Leftrightarrow 6x-3=10+6x\)

\(\Leftrightarrow 13=0\) (vô lý)

Vậy không tồn tại $x$ thỏa mãn đề bài.

a,

$0-|x+1|=5$

$|x+1|=0-5=-5<0$ (vô lý do trị tuyệt đối của một số luôn không âm)

Do đó không tồn tại $x$ thỏa mãn điều kiện đề.

b,

\(2-|\frac{3}{4}-x|=\frac{7}{12}\)

\(|\frac{3}{4}-x|=2-\frac{7}{12}=\frac{17}{12}\)

\(\Rightarrow \left[\begin{matrix} \frac{3}{4}-x=\frac{17}{12}\\ \frac{3}{4}-x=\frac{-17}{12}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{-2}{3}\\ x=\frac{13}{6}\end{matrix}\right.\)

c, 

\(2|\frac{1}{2}x-\frac{1}{3}|-\frac{3}{2}=\frac{1}{4}\)

\(2|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{4}\)

\(|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{8}\)

\(\Rightarrow \left[\begin{matrix} \frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\ \frac{1}{2}x-\frac{1}{3}=-\frac{7}{8}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{29}{12}\\ x=\frac{-13}{12}\end{matrix}\right.\)

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

\(\Leftrightarrow\dfrac{x-1}{x}-\dfrac{1}{x+1}=\dfrac{2x-1}{x\left(x+1\right)}\)

ĐKXĐ : \(\left\{{}\begin{matrix}x\ne0\\x+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne-1\end{matrix}\right.\)

Ta có : `(x-1)/x -1/(x+1) =(2x-1)/(x(x+1))`

\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}-\dfrac{x}{x\left(x+1\right)}=\dfrac{2x-1}{x\left(x+1\right)}\)

`=> x^2 +x -x-1 -x-2x+1=0`

`<=> x^2 -3x =0`

`<=> x(x-3)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=3\end{matrix}\right.\)

__

`(x+2)(5-3x)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\5-3x=0\end{matrix}\right.\)

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

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

__

\(\dfrac{5\left(1-2x\right)}{3}+\dfrac{x}{2}=\dfrac{3\left(x-5\right)}{4}-2\)

\(\Leftrightarrow\dfrac{20\left(1-2x\right)}{12}+\dfrac{6x}{12}=\dfrac{9\left(x-5\right)}{12}-\dfrac{24}{12}\)

`<=> 2x- 40x + 6x = 9x - 45 -24`

`<=>  2x- 40x + 6x-9x + 45 +24=0`

`<=>-41x+69=0`

`<=>-41x=-69`

`<=> x=69/41`

Cậu tách 2 câu 1 lượt mn trl nhanh hơn đó ạ

a:=>x^2-1-x=2x-1

=>x^2-x-1=2x-1

=>x^2-3x=0

=>x=0(loại) hoặc x=3(nhận)

b:=>x+2=0 hoặc 5-3x=0

=>x=-2 hoặc x=5/3

c:=>20(1-2x)+6x=9(x-5)-24

=>20-40x+6x=9x-45-24

=>-34x+20=9x-69

=>-43x=-89

=>x=89/43

d: =>x^2+4x+4-x^2-2x+3=2x^2+8x-4x-16-3

=>2x^2+4x-19=-2x+7

=>2x^2+6x-26=0

=>x^2+3x-13=0

=>\(x=\dfrac{-3\pm\sqrt{61}}{2}\)

e: =>(2x-3)(2x-3-x-1)=0

=>(2x-3)(x-4)=0

=>x=4 hoặc x=3/2

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Việt Nam đất nước anh hùng.....^^ 
Trung Quốc là nước nửa khùng nửa điên. 
Việt Nam đang sống bình yên. 
Trung Quốc đừng có làm phiền Việt Nam. 
Trung Quốc đông dân toàn cỏ rác. 
Việt Nam lác đác toàn siêu nhân. 
Việt Nam cưỡi rồng bay trong gió. 
Trung Quốc cưỡi chó sủa:"gâu" "gâu". 
Thái Lan hỏi nó đi đâu. 
Nó cười, nó bảo:" đi hầu Việt Nam

a:=>x^2-1-x=2x-1

=>x^2-x-1=2x-1

=>x^2-3x=0

=>x=0(loại) hoặc x=3(nhận)

b:=>x+2=0 hoặc 5-3x=0

=>x=-2 hoặc x=5/3

c:=>20(1-2x)+6x=9(x-5)-24

=>20-40x+6x=9x-45-24

=>-34x+20=9x-69

=>-43x=-89

=>x=89/43

d: =>x^2+4x+4-x^2-2x+3=2x^2+8x-4x-16-3

=>2x^2+4x-19=-2x+7

=>2x^2+6x-26=0

=>x^2+3x-13=0

=>\(x=\dfrac{-3\pm\sqrt{61}}{2}\)

e: =>(2x-3)(2x-3-x-1)=0

=>(2x-3)(x-4)=0

=>x=4 hoặc x=3/2

1) Tìm x:

a) \(\frac{11}{12}-\frac{5}{12}.\left(\frac{2}{5}+x\right)=\frac{2}{3}\)

\(\Leftrightarrow\frac{5}{12}.\left(\frac{2}{5}+x\right)=\frac{11}{12}-\frac{2}{3}=\frac{1}{4}\)

\(\Leftrightarrow\frac{2}{5}+x=\frac{1}{4}:\frac{5}{12}=\frac{3}{5}\)

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

b) \(\frac{3}{4}+\frac{1}{4}:x=\frac{2}{5}\)

\(\Leftrightarrow\frac{1}{4}:x=\frac{2}{5}-\frac{3}{4}=-\frac{7}{20}\)

\(\Leftrightarrow x=-\frac{7}{20}:\frac{1}{4}=\frac{-7}{5}\)

a) \(\frac{11}{12}-\frac{5}{12}\left(\frac{2}{5}+x\right)=\frac{2}{3}\)

\(\Leftrightarrow\frac{11}{12}-\frac{5}{12}.\frac{2}{5}-\frac{5}{12}x=\frac{2}{3}\)

\(\Leftrightarrow\frac{11}{12}-\frac{1}{6}-\frac{5}{12}x=\frac{2}{3}\)

\(\Leftrightarrow\frac{-5}{12}x=\frac{2}{3}-\frac{11}{12}+\frac{1}{6}\)

\(\Leftrightarrow-\frac{5}{12}x=\frac{8}{12}-\frac{11}{12}+\frac{2}{12}=-\frac{1}{12}\)

\(\Leftrightarrow x=\frac{-1}{12}:\left(-\frac{5}{12}\right)=-\frac{1}{12}.\left(-\frac{12}{5}\right)=\frac{1}{5}\)

Vậy x = 1/5

b) \(\frac{3}{4}+\frac{1}{4}:x=\frac{2}{5}\)

\(\Leftrightarrow\frac{1}{4}:x=\frac{2}{5}-\frac{3}{4}=\frac{8}{20}-\frac{15}{20}=-\frac{7}{20}\)

\(\Leftrightarrow x=\frac{1}{4}:\left(-\frac{7}{20}\right)=\frac{1}{4}.\left(-\frac{20}{7}\right)=-\frac{5}{7}\)

Vậy x = -5/7

c) \(2x\left(x-\frac{1}{7}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x-\frac{1}{7}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\frac{1}{7}\end{matrix}\right.\)

d) \(\left(x+1\right)\left(x-2\right)< 0\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1< 0\\x-2>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+1>0\\x-2< 0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< -1\\x>2\end{matrix}\right.\\\left\{{}\begin{matrix}x>-1\\x< 2\end{matrix}\right.\end{matrix}\right.\)

Ta thấy x <-1 và x >2 vô lí

Do đó: x >-1 và x <2

Vậy -1 < x <2

e) \(\left(x-2\right)\left(x+\frac{2}{3}\right)>0\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2>0\\x+\frac{2}{3}>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2< 0\\x+\frac{2}{3}< 0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>2\\x>-\frac{2}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}x< 2\\x< -\frac{2}{3}\end{matrix}\right.\end{matrix}\right.\)

Vậy x > 2 hoặc x < -2/3

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