a: =>5/42-x=11/13-15/28+11/13=421/364

=>x=-1193/1092

b: =>\(\dfrac{7}{2}-2x=7+\dfrac{6}{5}-3-\dfrac{2}{5}-1-\dfrac{4}{5}=3\)

=>2x=1/2

=>x=1/4

c: =>|2x-1/3|=-1/3(vô lý)

d: =>2x-1=-3

=>2x=-2

hay x=-1

e: =>2x=16

hay x=8

|2x-1|=1,5

TH(1)2x-1=1,5

2x =1,5+1

2x =2,5

x =2,5 :2

x =1,25

TH(2) 2x-1=-1,5

2x =-1,5+1

2x =-0,5

x =-0,5:2

x =-0,25

các câu khác cứ tương tự bạn nhé

b) \(7,5-\left|5-2x\right|=-4,5\)

\(\left|5-2x\right|=7,5+4,7\)

\(\left|5-2x\right|=12\)

th1 :\(5-2x=12\)

\(2x=5-12\)

\(2x=-7\)

\(x=-7:2\)

\(x=-3,5\)

th2: \(5-2x=-12\)

\(2x=5+12\)

\(2x=17\)

\(x=17:2\)

\(x=8,5\)

c) \(-3+\left|x\right|=-1\)

\(\left|x\right|=-1+3\)

\(\left|x\right|=2\)

th1: \(x=-2\)

th2 : \(x=2\)

d)\(\left|2\dfrac{1}{3}-x\right|=\dfrac{1}{6}\)

\(\left|\dfrac{7}{3}-x\right|=\dfrac{1}{6}\)

th1 :\(\dfrac{7}{3}-x=\dfrac{1}{6}\)

\(x=\dfrac{7}{3}-\dfrac{1}{2}\)

\(x=\dfrac{11}{6}\)

th2: \(\dfrac{7}{3}-x=\dfrac{-1}{6}\)

\(x=\dfrac{7}{3}+\dfrac{1}{6}\)

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

e) \(\dfrac{5}{7}-\left|x+1\right|=\dfrac{1}{14}\)

\(\left|x+1\right|=\dfrac{5}{7}-\dfrac{1}{14}\)

\(\left|x+1\right|=\dfrac{9}{14}\)

th1 :\(x+1=\dfrac{9}{14}\)

\(x=\dfrac{9}{14}-1\)

\(x=\dfrac{-5}{14}\)

th2 : \(x+1=\dfrac{-9}{14}\)

\(x=\dfrac{-9}{14}-1\)

\(x=\dfrac{-5}{14}\)

a/dễ --> tự lm

b/ \(\left(x-\dfrac{4}{7}\right)\left(1\dfrac{3}{5}+2x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{5}=0\\1\dfrac{3}{5}+2x=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\2x=\dfrac{8}{5}\Rightarrow x=\dfrac{4}{5}\end{matrix}\right.\)

Vậy...............

c/ \(\left(x-\dfrac{4}{7}\right):\left(x+\dfrac{1}{2}\right)>0\)

TH1: \(\left\{{}\begin{matrix}x-\dfrac{4}{7}>0\\x+\dfrac{1}{2}>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x>\dfrac{4}{7}\\x>-\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow x>\dfrac{4}{7}\)

TH2: \(\left\{{}\begin{matrix}x-\dfrac{4}{7}< 0\\x+\dfrac{1}{2}< 0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x< \dfrac{4}{7}\\x< -\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow x< -\dfrac{1}{2}\)

Vậy \(x>\dfrac{4}{7}\) hoặc \(x< -\dfrac{1}{2}\) thì thỏa mãn đề

d/ \(\left(2x-3\right):\left(x+1\dfrac{3}{4}\right)< 0\)

TH1: \(\left\{{}\begin{matrix}2x-3>0\\x+1\dfrac{3}{4}< 0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x>1,5\\x< -\dfrac{7}{4}\end{matrix}\right.\)(vô lý)

TH2: \(\left\{{}\begin{matrix}2x-3< 0\\x+1\dfrac{3}{4}>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x< 1,5\\x>-\dfrac{7}{4}\end{matrix}\right.\)\(\Rightarrow-\dfrac{7}{4}< x< 1,5\)

Vậy...................

x< -7/4(vô lí ) vì sao bạn

b: =>(3x-1)(3x+1)(2x+3)=0

hay \(x\in\left\{\dfrac{1}{3};-\dfrac{1}{3};-\dfrac{3}{2}\right\}\)

c: \(\Leftrightarrow\left|2x-\dfrac{1}{3}\right|=\dfrac{5}{6}+\dfrac{3}{4}=\dfrac{19}{12}\)

=>2x-1/3=19/12 hoặc 2x-1/3=-19/12

=>2x=23/12 hoặc 2x=-15/12=-5/4

=>x=23/24 hoặc x=-5/8

d: \(\Leftrightarrow-\dfrac{5}{6}\cdot x+\dfrac{3}{4}=-\dfrac{3}{4}\)

=>-5/6x=-3/2

=>x=3/2:5/6=3/2*6/5=18/10=9/5

e: =>2/5x-1/2=3/4 hoặc 2/5x-1/2=-3/4

=>2/5x=5/4 hoặc 2/5x=-1/4

=>x=5/4:2/5=25/8 hoặc x=-1/4:2/5=-1/4*5/2=-5/8

f: =>14x-21=9x+6

=>5x=27

=>x=27/5

h: =>(2/3)^2x+1=(2/3)^27

=>2x+1=27

=>x=13

i: =>5^3x*(2+5^2)=3375

=>5^3x=125

=>3x=3

=>x=1

\(\left|x+\dfrac{1}{1.5}\right|+\left|x+\dfrac{1}{5.9}\right|+\left|x+\dfrac{1}{9.14}\right|+...+\left|x+\dfrac{1}{397.401}\right|\ge0\)

\(\Rightarrow101x\ge0\)

\(\Rightarrow x\ge0\)

\(\Rightarrow x+\dfrac{1}{1.5}+x+\dfrac{1}{5.9}+...+x+\dfrac{1}{397.401}=101x\)

\(\Rightarrow101x+\left(\dfrac{1}{1.5}+\dfrac{1}{5.9}+...+\dfrac{1}{397.401}\right)=x\)

\(\Rightarrow\dfrac{1}{4}\left(\dfrac{4}{1.5}+\dfrac{4}{5.9}+...+\dfrac{4}{397.401}\right)=x\)

\(\Rightarrow x=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+....+\dfrac{1}{397}-\dfrac{1}{401}\right)\)

\(\Rightarrow x=\dfrac{1}{4}\left(1-\dfrac{1}{401}\right)\)

\(\Rightarrow x=\dfrac{1}{4}.\dfrac{400}{401}\)

\(\Rightarrow x=\dfrac{100}{401}\)

a,

\(\left(\dfrac{3}{5}x-\dfrac{2}{3}x-x\right)\cdot\dfrac{1}{7}=-\dfrac{5}{21}\)

\(\Rightarrow\dfrac{-16}{15}x\cdot\dfrac{1}{7}=-\dfrac{5}{21}\)

\(\Rightarrow\dfrac{-16}{15}x=\dfrac{-\dfrac{5}{21}}{\dfrac{1}{7}}=-\dfrac{5}{3}\)

\(\Rightarrow x=\dfrac{-\dfrac{5}{3}}{-\dfrac{16}{15}}=\dfrac{25}{16}\)

b,

\(\left(5x-1\right)\left(2x+\dfrac{1}{3}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}5x-1=0\\2x+\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{6}\end{matrix}\right.\)

c,

\(\dfrac{5\left|x+1\right|}{2}=\dfrac{90}{\left|x+1\right|}\)

\(\Rightarrow5\left|x+1\right|^2=180\)

\(\Rightarrow\left|x+1\right|^2=36\)

Mà \(\left|x+1\right|\ge0\)

=> x + 1 = 6 <=> x = 7

a) ( x + 5 )³ = -64

x + 5 = - 4

x = - 4 - 5

x = -9

b) (2x - 3)²=9

2x - 3 = 3

2x = 3+3

2x = 6

x = 6 : 2

x = 3

e) \(\dfrac{8}{2x}=4\)

=> 4 . 2x = 8

8x =8

x = 8 : 8

x = 1

g) \(\left(\dfrac{1}{2}\right)^{2x-1}=\dfrac{1}{8}\)

\(\left(\dfrac{1}{2}\right)^{2x}:\left(\dfrac{1}{2}\right)^1=\dfrac{1}{8}\)

\(\left(\dfrac{1}{2}\right)^{2x}:\dfrac{1}{2}=\dfrac{1}{8}\)

\(\left(\dfrac{1}{2}\right)^{2x}=\dfrac{1}{8}.\dfrac{1}{2}\)

\(\left(\dfrac{1}{2}\right)^{2x}=\dfrac{1}{16}\)

\(\left(\dfrac{1}{2}\right)^{2x}=\left(\dfrac{1}{2}\right)^{2.2}\)

=> x = 2

h) \(\left(\dfrac{1}{2}\right)^2.x=\left(\dfrac{1}{2}\right)^5\)

\(\dfrac{1}{4}.x=\dfrac{1}{32}\)

x = \(\dfrac{1}{32}:\dfrac{1}{4}\)

x = \(\dfrac{1}{8}\)

i) \(\left(\dfrac{-1}{3}\right)x=\dfrac{1}{81}\)

\(x=\dfrac{1}{81}:\left(\dfrac{-1}{3}\right)\)

\(x=\dfrac{-1}{27}\)

a) (x + 5)³ = -64

=> (x + 5)³ = (-4)³

x + 5 = -4

x = -4 - 5

x = -9

b) (2x - 3)² = 9

=> (2x - 3)² = (\(\pm\)3)²

=> 2x - 3 = 3 hoặc 2x - 3 = -3

*2x - 3 = 3

2x = 3 + 3

2x = 9

x = \(\dfrac{9}{2}\)

*2x - 3 = -3

2x = -3 + 3

2x = 0

x = 0 : 2

x = 0

Vậy x \(\in\left\{\dfrac{9}{2};0\right\}\)

c) \(\dfrac{x}{\dfrac{4}{2}}=\dfrac{4}{\dfrac{x}{2}}\)

=> \(x.\dfrac{x}{2}=4.\dfrac{4}{2}\)

\(\dfrac{x}{2}=8\)

x = 8 : 2

x = 4

d) \(\dfrac{-32}{\left(-2\right)^n}=4\)

\(\Rightarrow\dfrac{\left(-2\right)^5}{\left(-2\right)^n}=\left(-2\right)^2\)

=> (-2)ⁿ . (-2)²= (-2)⁵

(-2)ⁿ = (-2)⁵ : (-2)²

(-2)ⁿ = (-2)³

Vậy n = 3

e) \(\dfrac{8}{2x}=4\)

=> 2x . 4 = 8

2x = 8 : 4

2x = 2

x = 1

g) \(\left(\dfrac{1}{2}\right)^{2x-1}=\dfrac{1}{8}\)

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

2x - 1 = 3

2x = 3 + 1

2x = 4

x = 4 : 2

x = 2

h) \(\left(\dfrac{1}{2}\right)^2.x=\left(\dfrac{1}{2}\right)^5\)

\(x=\left(\dfrac{1}{2}\right)^5:\left(\dfrac{1}{2}\right)^2\)

\(x=\left(\dfrac{1}{2}\right)^3\)

\(x=\dfrac{1}{8}\)

i) \(\left(\dfrac{-1}{3}\right)x=\dfrac{1}{81}\)

\(x=\dfrac{1}{81}:\left(\dfrac{-1}{3}\right)\)

\(x=\left(\dfrac{-1}{3}\right)^4:\left(\dfrac{-1}{3}\right)\)

\(x=\left(\dfrac{-1}{3}\right)^3\)

\(x=\dfrac{-1}{27}\).

Mấy câu này dễ mà,động não lên chứ bạn:v

Link______________Link

h) \(\left|x-1\right|+\left|x-3\right|=\left|x-1\right|+\left|3-x\right|\)

\(\ge\left|x-1+3-x\right|=2\)

\(\Rightarrow x+1>2\Leftrightarrow x>1\)

Vậy: \(\left\{{}\begin{matrix}x>1\\x\in R\end{matrix}\right.\)

Câu b xét khoảng tương tự với cái link t đưa thôi

hơi bức xúc rồi đó

tau chỉ muốn kiểm tra lại thôi

a: \(\Leftrightarrow3^x\cdot3+2x\cdot3^x-18x-27=0\)

\(\Leftrightarrow3^x\left(2x+3\right)-9\left(2x+3\right)=0\)

\(\Leftrightarrow\left(2x+3\right)\left(3^x-9\right)=0\)

=>x=2 hoặc x=-3/2

b: \(\Leftrightarrow\left|2x+5\right|\cdot\dfrac{1}{2}-\dfrac{5}{4}\cdot2\cdot\left|2x+5\right|+\dfrac{7}{3}\cdot4\cdot\left|2x+5\right|=\dfrac{1}{6}\)

\(\Leftrightarrow\left|2x+5\right|=\dfrac{1}{44}\)

=>2x+5=1/44 hoặc 2x+1=-1/44

=>x=-219/88 hoặc x=-221/88

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

\(\Rightarrow\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{10}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{3}=-\dfrac{7}{10}\\\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{10}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{11}{15}\\x=\dfrac{31}{15}\end{matrix}\right.\)

b, \(\left|\dfrac{1}{4}x-2\dfrac{1}{5}\right|=\left|0,6-\dfrac{2}{3}x\right|\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{4}x-2\dfrac{1}{5}=\dfrac{2}{3}x-0,6\\\dfrac{1}{4}x-2\dfrac{1}{5}=0,6-\dfrac{2}{3}x\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{4}x-\dfrac{2}{3}x=-0,6+2\dfrac{1}{5}\\\dfrac{1}{4}x+\dfrac{2}{3}x=0,6+2\dfrac{1}{5}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}-\dfrac{5}{12}x=1,6\\\dfrac{11}{12}x=2,8\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-3,84\\x=\dfrac{168}{55}\end{matrix}\right.\)

c, \(\left|2x-\dfrac{1}{3}\right|=x+\dfrac{1}{2}\)

+, Xét \(x\ge\dfrac{1}{6}\) thì \(2x-\dfrac{1}{3}\ge0\Rightarrow\left|2x-\dfrac{1}{3}\right|=2x-\dfrac{1}{3}\)

Thay vào ta có:

\(2x-\dfrac{1}{3}=x+\dfrac{1}{2}\Rightarrow x=\dfrac{5}{6}\)(chọn vì thoả mãn điều kiện \(x\ge\dfrac{1}{6}\))

+, Xét \(x< \dfrac{1}{6}\) thì \(2x-\dfrac{1}{3}< 0\Rightarrow\left|2x-\dfrac{1}{3}\right|=\dfrac{1}{3}-2x\)

Thay vào ta có:

\(\dfrac{1}{3}-2x=x+\dfrac{1}{2}\Rightarrow3x=-\dfrac{1}{6}\Rightarrow x=-\dfrac{1}{18}\)(chọn vì thoả mãn điều kiện \(x< \dfrac{1}{6}\))

Vậy.............

Chúc bạn học tốt!!!