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19 tháng 12 2022

2x-0,5=x+\(\dfrac{1}{4}\)

<=> 2x-x=\(\dfrac{1}{4}+0,5\)

<=> x= 0,75

24 tháng 5 2023

(5 - \(x\))(9\(x^2\) - 4) =0

\(\left[{}\begin{matrix}5-x=0\\9x^2-4=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=5\\9x^2=4\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=5\\x^2=\dfrac{4}{9}\end{matrix}\right.\)

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

Vậy \(x\) \(\in\) { - \(\dfrac{2}{3}\)\(\dfrac{2}{3}\)\(5\)}

 

24 tháng 5 2023

72\(x\)  + 72\(x\) + 3 = 344

72\(x\)  \(\times\) ( 1 + 73) = 344

72\(x\)  \(\times\) (1 + 343) = 344

72\(x\)  \(\times\) 344        = 344

72\(x\)                    = 344 : 344

72\(x\)                  = 1

72\(x\)                 =  70

\(2x\)                  = 0

\(x\)                   = 0

Kết luận: \(x\) = 0

25 tháng 12 2023

a)
=\(2x-\dfrac{1}{2}=\dfrac{12}{9}\cdot\dfrac{3}{4}=1\)
=\(2x=1+\dfrac{1}{2}=1.5\)
=\(x=1.5:2=0.75\)
b)
=\(x^2=0+2=2\)
TH1:\(x=2\)
TH2:\(x=-2\)

31 tháng 12 2023

Bài 1:

a: \(2x-\dfrac{1}{2}:\dfrac{3}{4}=\dfrac{12}{9}\)

=>\(2x-\dfrac{1}{2}\cdot\dfrac{4}{3}=\dfrac{4}{3}\)

=>\(2x=\dfrac{4}{3}+\dfrac{2}{3}=\dfrac{6}{3}=2\)

=>x=2/2=1

b: \(x^2-2=0\)

=>\(x^2=2\)

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

Bài 2:

a: \(A=\dfrac{1,11+0,19-13\cdot2}{2,06+0,54}-\left(\dfrac{1}{2}+\dfrac{1}{4}\right):2\)

\(=\dfrac{1,3-26}{2,6}-\dfrac{3}{4}:2\)

\(=-9,5-\dfrac{3}{8}=-\dfrac{79}{8}\)

\(B=\left(5\dfrac{7}{8}-2\dfrac{1}{4}-0,5\right):\left(2\dfrac{23}{26}\right)\)

\(=\left(5+\dfrac{7}{8}-2-\dfrac{1}{4}-\dfrac{1}{2}\right):\dfrac{75}{26}\)

\(=\left(3+\dfrac{1}{8}\right)\cdot\dfrac{26}{75}=\dfrac{25}{8}\cdot\dfrac{26}{75}=\dfrac{13}{12}\)

b: A<x<B

=>\(-\dfrac{79}{8}< x< \dfrac{13}{12}\)

mà \(x\in Z\)

nên \(x\in\left\{-9;-8;...;0;1\right\}\)

22 tháng 12 2021

\(\Leftrightarrow\left(x-0.5\right)\cdot\dfrac{-4}{x-0.5}=-1\cdot\left(-4\right)\)

=>-4=4(loại)

23 tháng 11 2021

\(a,\Rightarrow\left[{}\begin{matrix}5x+1=\dfrac{6}{7}\\5x+1=-\dfrac{6}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}5x=\dfrac{1}{7}\\5x=-\dfrac{13}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{35}\\x=-\dfrac{13}{35}\end{matrix}\right.\\ b,\Rightarrow\left(-\dfrac{1}{8}\right)^x=\dfrac{1}{64}=\left(-\dfrac{1}{8}\right)^2\Rightarrow x=2\\ c,\Rightarrow\left(x-2\right)\left(2x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{3}{2}\end{matrix}\right.\\ d,\Rightarrow\left(x+1\right)^{x+10}-\left(x+1\right)^{x+4}=0\\ \Rightarrow\left(x+1\right)^{x+4}\left[\left(x+1\right)^6-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\\left(x+1\right)^6=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x+1=1\\x+1=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=0\\x=-2\end{matrix}\right.\\ e,\Rightarrow\dfrac{3}{4}\sqrt{x}=\dfrac{5}{6}\left(x\ge0\right)\\ \Rightarrow\sqrt{x}=\dfrac{10}{9}\Rightarrow x=\dfrac{100}{81}\)

21 tháng 6 2021

a)

TH1: \(x< \dfrac{-2}{3}\)

<=> \(\left\{{}\begin{matrix}\left|0,5x-2\right|=2-0,5x\\\left|x+\dfrac{2}{3}\right|=-x-\dfrac{2}{3}\end{matrix}\right.\)

PT <=> \(2-0,5x+x+\dfrac{2}{3}=0< =>x=\dfrac{-16}{3}\left(c\right)\)

TH2: \(\dfrac{-2}{3}\le x< 4\)

<=> \(\left\{{}\begin{matrix}\left|0,5x-2\right|=2-0,5x\\\left|x+\dfrac{2}{3}\right|=x+\dfrac{2}{3}\end{matrix}\right.\)

PT <=> \(2-0,5x-x-\dfrac{2}{3}=0< =>x=\dfrac{8}{9}\left(c\right)\)

TH3: \(x\ge4\)

<=> \(\left\{{}\begin{matrix}\left|0,5x-2\right|=0,5x-2\\\left|x+\dfrac{2}{3}\right|=x+\dfrac{2}{3}\end{matrix}\right.\)

PT <=> \(0,5x-2-x-\dfrac{2}{3}=0< =>x=\dfrac{-16}{3}\left(l\right)\)

KL: x \(\left\{\dfrac{-16}{3};\dfrac{8}{9}\right\}\)

b) TH1: \(x\ge-1< =>\left|x+1\right|=x+1\)

PT <=> 2x - x -1 = \(\dfrac{-1}{2}\)

<=> x = \(\dfrac{1}{2}\) (c)

TH2: x < -1 <=> \(\left|x+1\right|=-x-1\)

PT <=> 2x + x + 1 = \(\dfrac{-1}{2}\)

<=> x = \(\dfrac{-1}{2}\) (l)

KL: x \(\in\left\{\dfrac{1}{2}\right\}\)

a: =>|x-1/4|=3/4

=>x-1/4=3/4 hoặc x-1/4=-3/4

=>x=1 hoặc x=-1/2

b: \(\left|x+\dfrac{1}{2}\right|=\dfrac{1}{2}-\dfrac{9}{4}=\dfrac{2-9}{4}=-\dfrac{7}{4}\)(vô lý)

c: \(\Leftrightarrow\left[{}\begin{matrix}2x+5=1-x\\2x+5=x-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-4\\x=-6\end{matrix}\right.\Leftrightarrow x\in\left\{-\dfrac{4}{3};-6\right\}\)

e: =>|3/2-x|=0

=>3/2-x=0

hay x=3/2

=>x^2+4x+3=x^2+4x+0,5x+0,5

=>4x+3=4,5x+0,5

=>-0,5x=-2,5

=>x=5

18 tháng 4 2018

\(\dfrac{x+2}{0,5}=\dfrac{2x+1}{2}\)

\(\Leftrightarrow\left(x+2\right).2=\left(2x+1\right).0,5\)

\(\Leftrightarrow2x+4=x+0,5\)

\(\Leftrightarrow x=-3,5\)

Vậy...

18 tháng 4 2018

\(\dfrac{x+2}{0,5}=\dfrac{2x+1}{2}\)

\(\Leftrightarrow\dfrac{4.\left(x+2\right)}{2}=\dfrac{2x+1}{2}\)

\(\Rightarrow4x+8=2x+1\)

\(\Leftrightarrow4x-2x=1-8\)

\(\Leftrightarrow2x=-7\)

\(\Leftrightarrow x=\dfrac{-7}{2}\)

Vậy \(x=\dfrac{-7}{2}\)