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1: \(=\dfrac{1}{4}:\dfrac{-1}{4}-2\cdot\dfrac{-1}{8}+5-4\)
\(=-1+1+\dfrac{1}{4}=\dfrac{1}{4}\)
2: \(=5^{20}\cdot\dfrac{1}{5^{20}}+\left(\dfrac{3}{8}\cdot\dfrac{4}{3}\right)^8-1=1-1+\dfrac{1}{2}^8=\dfrac{1}{2^8}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =>10căn x=30
=>căn x=3
=>x=9
c: =>|1/2x-3/4|=1/2
=>1/2x-3/4=1/2 hoặc 1/2x-3/4=-1/2
=>1/2x=5/4 hoặc 1/2x=1/2
=>x=1 hoặc x=5/2
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\(a,\left(\frac{3}{8}+-\frac{3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
= \(\left(-\frac{3}{8}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
= \(\frac{5}{24}:\frac{5}{6}+\frac{1}{2}\)
= \(\frac{1}{4}+\frac{1}{2}\)
= \(\frac{3}{4}\)
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) \(\frac{2}{3}-0,25:\frac{3}{4}+\frac{5}{8}.4\)
= \(\frac{2}{3}-\frac{1}{4}:\frac{3}{4}+\frac{5}{8}.4\)
= \(\frac{2}{3}-\frac{1}{3}+\frac{5}{2}\)
= \(\frac{1}{3}+\frac{5}{2}\)
= \(\frac{17}{6}\)
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a) 2x.3/5 - 5x = 3/2 - 7x
6x/5 - 5x = 3/2 - 7x
-19x/5 = 3/2 - 7x
-19x = 15/2 - 35x
-19x + 35x = 15/2
16x = 15/2
x = 15/2 : 16
x = 15/32
b) (x - 1/5)2 = 4/25
(x - 1/5)2 = (+-2/5)2
x - 1/5 = +-2/5
x - 1/5 = 2/5 hoặc x - 1/5 = -2/5
x = 2/5 + 1/5 x = -2/5 + 1/5
x = 3/5 x = -1/5
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\left(\dfrac{2}{3}\right)^x=\left(\dfrac{4}{9}\right)^4.\)
\(\left(\dfrac{2}{3}\right)^x=\left[\left(\dfrac{2}{3}\right)^2\right]^4.\)
\(\left(\dfrac{2}{3}\right)^x=\left(\dfrac{2}{3}\right)^8\Rightarrow x=8.\)
Vậy.....
\(b,\left(2x-1\right)^2=25.\)
\(\left(2x-1\right)^2=\left(\pm5\right)^2.\)
\(\Rightarrow\left(2x-1\right)=\pm5.\)
+) Xét \(2x-1=5\), ta có:
\(2x-1=5.\)
\(\Rightarrow2x=6.\)
\(\Rightarrow x=3.\)
+) Xét \(2x-1=-5\), ta có:
\(2x-1=-5.\)
\(\Rightarrow2x=-4.\)
\(\Rightarrow x=-2.\)
Vậy.....
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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
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\left(x-3\right)^2=1\)
=> \(\sqrt{\left(x-3\right)^2}=\sqrt{1}\)
=> \(\left|x-3\right|=1\)
=> \(\left[{}\begin{matrix}x-3=1\\x-3=-1\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=1+3=4\\x=-1+3=2\end{matrix}\right.\)
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 \(x\in\left\{\frac{9}{20};\frac{1}{20}\right\}\)
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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: =>\(2x-7=\sqrt[3]{\dfrac{26}{63}}\)
=>\(2x=\sqrt[3]{\dfrac{26}{63}}+7\)
=>\(x=\dfrac{1}{2}\cdot\left(\sqrt[3]{\dfrac{26}{63}}+7\right)\)
\(\left(2x-3\right)^2=\dfrac{4}{25}\\ \Rightarrow\left(2x-3\right)^2=\left(\pm\dfrac{2}{5}\right)^2\\ \Rightarrow\left[{}\begin{matrix}2x-3=\dfrac{2}{5}\\2x-3=-\dfrac{2}{5}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=\dfrac{17}{5}\\2x=\dfrac{13}{5}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{17}{10}\\x=\dfrac{13}{10}\end{matrix}\right.\)
\(\left(2x-3\right)^2=\dfrac{4}{25}\\ \Leftrightarrow\left(2x-3\right)^2=\left(\dfrac{2}{5}\right)^2\\ \Leftrightarrow\left[{}\begin{matrix}2x-3=\dfrac{2}{5}\\-2x+3=\dfrac{2}{5}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{17}{5}\\-2x=-\dfrac{13}{5}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17}{10}\\x=\dfrac{13}{10}\end{matrix}\right.\)
Vậy...