Mk nhầm nha câu đầu chỉ có 1 cái x-1 + x -2 thôi ko có cái đằng sau nhé ! giá trị tuyệt đối thì vẫn giữ nguyên !

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

\(\left|2x-3\right|=\dfrac{1}{3}+\dfrac{5}{2}=\dfrac{2}{6}+\dfrac{15}{6}\)

\(\left|2x-3\right|=\dfrac{17}{6}\)

\(+)2x-3=\dfrac{17}{6}\Rightarrow2x=\dfrac{35}{6}\Rightarrow x=\dfrac{35}{12}\)

\(+)2x-3=\dfrac{-17}{6}\Rightarrow2x=\dfrac{1}{6}\Rightarrow x=\dfrac{1}{12}\)

vậy...

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

Dấu ngoặc vuông nhé

thánh bấm nhầm

a: \(\Rightarrow\left(2x-4\right)^{x+1}\left[\left(2x-4\right)^4-1\right]=0\)

=>(2x-4)(2x-3)(2x-5)=0

hay \(x\in\left\{2;\dfrac{3}{2};\dfrac{5}{2}\right\}\)

b: \(\Leftrightarrow\left(x-3\right)^{x+4}\left(x-3-1\right)=0\)

=>(x-3)^x+4(x-4)=0

=>x=3 hoặc x=4

c: \(\Leftrightarrow\left[{}\begin{matrix}x-1>2\\x-1< -2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>3\\x< -1\end{matrix}\right.\)

d: =>-5<=2x+3<=5

=>-8<=2x<=2

=>-4<=x<=1

1,

\(\left(2x+1\right)^3=-0,001\\ \left(2x+1\right)^3=\left(-0.1\right)^3\\ \Leftrightarrow2x+1=-0.1\\ 2x=-1.1\\ x=-\dfrac{11}{10}:2\\ x=-\dfrac{11}{20}\\ Vậy...\)

2,

\(\left(2x-3\right)^4=\left(2x-3\right)^6\\ \Leftrightarrow\left(2x-3\right)^6-\left(2x-3\right)^4=0\\ \Leftrightarrow\left(2x-3\right)^4\cdot\left[\left(2x-3\right)^2-1\right]=0\\ \Rightarrow\left\{{}\begin{matrix}\left(2x-3\right)^4=0\\\left(2x-3\right)^2-1=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x-3=0\\\left(2x-3\right)^2=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x=3\\2x-3=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\x=2\end{matrix}\right.\\ Vậyx\in\left\{\dfrac{3}{2};2\right\}\)

3, Làm tương tự câu 2

5,

\(9^x:3^x=3\\ \left(9:3\right)^x=3\\ 3^x=3\\ \Rightarrow x=1\\ Vậy...\)

6,

\(3^x+3^{x+3}=756\\ 3^x+3^x\cdot3^3\\ 3^x\cdot\left(1+27\right)=756\\ 3^x\cdot28=756\\ \Leftrightarrow3^x=27\\ 3^x=3^3\\ \Rightarrow x=3\\ vậy...\)

7,

\(5^{x+1}+6\cdot5^{x+1}=875\\ 5^{x+1}\cdot\left(1+6\right)=875\\ 5^{x+1}\cdot7=875\\ \Leftrightarrow5^{x+1}=125\\ \Leftrightarrow5^{x+1}=5^3\Leftrightarrow x+1=3\\ \Rightarrow x=2\\ Vậy...\)

9,

lê thị hồng vân trả lời típ đi

#)Giải :

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

\(\Leftrightarrow2x-3-x+\frac{1}{2}=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x-3=0\\x+\frac{1}{2}=0\end{cases}}\Rightarrow\orbr{\begin{cases}2x=3\\x=-\frac{1}{2}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=-\frac{1}{2}\end{cases}}}\)

a) \(2x-3=x+\frac{1}{2}\)

\(\Leftrightarrow2x-x=\frac{1}{2}+3\)

\(\Leftrightarrow x=\frac{7}{2}\)

Vậy...

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

\(\Leftrightarrow4x-2x-1=3-\frac{1}{3}+x\)

\(\Leftrightarrow4x-2x-x=3-\frac{1}{3}+1\)

\(\Leftrightarrow x=\frac{11}{3}\)

Vậy ...

c) \(2x-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-...-\frac{1}{49.50}=7-\frac{1}{50}+x\)

\(\Leftrightarrow2x-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{49.50}\right)=\frac{349}{50}+x\)

\(\Leftrightarrow2x-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\right)=\frac{349}{50}+x\)

\(\Leftrightarrow2x-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\right)=\frac{349}{50}+x\)

\(\Leftrightarrow2x-\left(1-\frac{1}{50}\right)=\frac{349}{50}+x\)

\(\Leftrightarrow2x-\frac{49}{50}=\frac{349}{50}+x\)

\(\Leftrightarrow2x-x=\frac{349}{50}+\frac{49}{50}\)

\(\Leftrightarrow x=\frac{199}{25}\)

Vậy ...

Câu a:

2.(3\(x\) - \(\frac12\)) - 2\(x\) = \(\frac12\).(2\(x\) - 3)

6\(x\) - 1 - 2\(x\) = \(x\) - \(\frac32\)

6\(x\) - 2\(x\) - \(x\) = 1 - \(\frac32\)

4\(x\) - \(x\) = - \(\frac12\)

3\(x\) = - \(\frac12\)

\(x\) = - \(\frac12\) : 3

\(x=-\frac16\)

Vậy \(x=-\frac16\)

Câu b:

(2\(x\) - \(\frac35\))\(^2\) = \(\frac{4}{25}\)

(2\(x-\frac35\))\(^2\) = \(\left(\frac{2}{25}\right)\)\(^2\)

2\(x\) - \(\frac35\) = \(\frac25\) hoặc 2\(x\) - \(\frac35\) = - \(\frac25\)

TH: 2\(x\) - \(\frac35\) = \(\frac25\)

2\(x\) = \(\frac25+\frac35\)

2\(x\) = 1

\(x=\frac12\)

2\(x\) - \(\frac35\) = - \(\frac25\)

2\(x\) = - \(\frac25\) + \(\frac35\)

2\(x\) = \(\frac15\)

\(x\) = \(\frac{13}{25}\) : 2

\(x\) = \(\frac15\)

Vậy \(x\) ∈ {1/5; 1/2}

a: \(\Leftrightarrow2^{2x}\cdot8+2^{2x}\cdot2+2^{2x}=176\)

\(\Leftrightarrow2^{2x}=16\)

=>2x=4

=>x=2

b: \(\Leftrightarrow2^{2x}\left(2^3+2-1\right)=144\)

\(\Leftrightarrow2^{2x}=16\)

=>2x=4

=>x=2