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AN
27 tháng 10 2017
Tìm X
a) \(2x+\dfrac{3}{24}=3x-\dfrac{1}{32}\)
\(\Leftrightarrow\left(2x+\dfrac{3}{24}\right)-\left(3x-\dfrac{1}{32}\right)=0\)
\(\Leftrightarrow2x+\dfrac{3}{24}-3x+\dfrac{1}{32}=0\)
\(\Leftrightarrow\left(\dfrac{3}{24}+\dfrac{1}{32}\right)+\left(2x-3x\right)=0\)
\(\Leftrightarrow\dfrac{5}{32}-x=0\)
\(\Leftrightarrow x=\dfrac{5}{32}\)
17 tháng 12 2022
a: \(\Leftrightarrow4^{x-5}\cdot17=68\)
=>4^x-5=4
=>x-5=1
=>x=6
b: \(\Leftrightarrow\dfrac{1}{3}:\left|2x-1\right|=\dfrac{1}{3}+\dfrac{2}{3}=1\)
=>|2x-1|=1/3
=>2x-1=1/3 hoặc 2x-1=-1/3
=>x=2/3 hoặc x=1/3
c: =>|2x-2|=|3x+15|
=>3x+15=2x-2 hoặc 3x+15=-2x+2
=>x=-17 hoặc x=-13/5
DK
0
HT
1
3 tháng 8 2023
2: =>(3x-7)^2=25/144=(5/12)^2
=>3x-7=5/12 hoặc 3x-7=-5/12
=>3x=5/12+7=89/12 hoặc 3x=7-5/12=79/12
=>x=89/36 hoặc x=79/36
3:Sửa đề: |2x-3|=|x+1|
=>2x-3=x+1 hoặc 2x-3=-x-1
=>x=4 hoặc 3x=2
=>x=2/3 hoặc x=4
4: =>3x+1=5 hoặc 3x+1=-5
=>3x=4 hoặc 3x=-6
=>x=-2 hoặc x=4/3
1: =>\(2x-7=\sqrt[3]{\dfrac{26}{63}}\)
=>\(2x=\sqrt[3]{\dfrac{26}{63}}+7\)
=>\(x=\dfrac{1}{2}\cdot\left(\sqrt[3]{\dfrac{26}{63}}+7\right)\)
22 tháng 6 2016
\(a,\left(\frac{3}{8}+-\frac{3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
= \(\left(-\frac{3}{8}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
= \(\frac{5}{24}:\frac{5}{6}+\frac{1}{2}\)
= \(\frac{1}{4}+\frac{1}{2}\)
= \(\frac{3}{4}\)
b)\(-\frac{7}{3}.\frac{5}{9}+\frac{4}{9}.\left(-\frac{3}{7}\right)+\frac{17}{7}\)
=\(-\frac{35}{27}+\left(-\frac{4}{21}\right)+\frac{17}{7}\)
= \(-\frac{35}{27}+\frac{47}{21}\)
= \(\frac{178}{189}\)
c) \(\frac{117}{13}-\left(\frac{2}{5}+\frac{57}{13}\right)\)
= \(\frac{117}{13}-\frac{311}{65}\)
= \(\frac{274}{65}\)
d) \(\frac{2}{3}-0,25:\frac{3}{4}+\frac{5}{8}.4\)
= \(\frac{2}{3}-\frac{1}{4}:\frac{3}{4}+\frac{5}{8}.4\)
= \(\frac{2}{3}-\frac{1}{3}+\frac{5}{2}\)
= \(\frac{1}{3}+\frac{5}{2}\)
= \(\frac{17}{6}\)
18 tháng 7 2019
\(a,\left(x-3\right)^2=1\)
=> \(\sqrt{\left(x-3\right)^2}=\sqrt{1}\)
=> \(\left|x-3\right|=1\)
=> \(\left[{}\begin{matrix}x-3=1\\x-3=-1\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=1+3=4\\x=-1+3=2\end{matrix}\right.\)
Vậy \(x\in\left\{4;2\right\}\)
\(\left(2x+1\right)^3=-8\)
=> \(\sqrt[3]{\left(2x+1\right)^3}=\sqrt[3]{-8}\)
=> \(2x+1=-2\)
=> \(2x=-2-1=-3\)
=> \(x=-3:2=-\frac{3}{2}\)
Vậy \(x\in\left\{-\frac{3}{2}\right\}\)
\(c,\left(x-\frac{1}{4}\right)^2=\frac{1}{25}\)
=> \(\sqrt{\left(x-\frac{1}{4}\right)^2}=\sqrt{\frac{1}{25}}\)
=> \(\left|x-\frac{1}{4}\right|=\frac{1}{5}\)
=> \(\left[{}\begin{matrix}x-\frac{1}{4}=\frac{1}{5}\\x-\frac{1}{4}=-\frac{1}{5}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\frac{1}{5}+\frac{1}{4}=\frac{9}{20}\\x=-\frac{1}{5}+\frac{1}{4}=\frac{1}{20}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{9}{20};\frac{1}{20}\right\}\)
NT
0
\(\left(x+\dfrac{1}{3}\right)^2=\dfrac{1}{25}\\ \Rightarrow\left(x+\dfrac{1}{3}\right)^2=\left(\pm\dfrac{1}{5}\right)^2\\ \Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=\dfrac{1}{5}\\x+\dfrac{1}{3}=-\dfrac{1}{5}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{2}{15}\\x=-\dfrac{8}{15}\end{matrix}\right.\)
\(\left(x+\dfrac{1}{3}\right)^2=\dfrac{1}{25}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=\dfrac{1}{5}\\x+\dfrac{1}{3}=-\dfrac{1}{5}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{15}\\x=-\dfrac{8}{15}\end{matrix}\right.\)
Vậy phương trình có nghiệm \(S=\left\{-\dfrac{2}{15};-\dfrac{8}{15}\right\}\)