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LD
28 tháng 12 2017
\(\left(2x-\sqrt{\dfrac{9}{4}}\right)^2=\dfrac{1}{\sqrt{625}}\)
\(\Leftrightarrow\left(2x-\dfrac{3}{2}\right)^2=\dfrac{1}{25}\)
\(\Rightarrow\left[{}\begin{matrix}2x-\dfrac{3}{2}=\dfrac{1}{5}\\2x-\dfrac{3}{2}=-\dfrac{1}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17}{20}\\x=\dfrac{13}{20}\end{matrix}\right.\)
8 tháng 10 2017
\(a,\left(\dfrac{4}{9}\right)^x=\left(\dfrac{3}{2}\right)^{-5}\\ \Leftrightarrow\left(\dfrac{2}{3}\right)^{2x}=\left(\dfrac{2}{3}\right)^5\\ \Rightarrow x=\dfrac{5}{2}\)
Vậy....
NT
1
1 tháng 2 2022
a: =>2x+5=4
=>2x=-1
hay x=-1/2
b: \(\Leftrightarrow\left(3x-4\right)^2\cdot\left[\left(3x-4\right)^2-1\right]=0\)
=>(3x-4)(3x-5)(3x-3)=0
hay \(x\in\left\{1;\dfrac{4}{3};\dfrac{5}{3}\right\}\)
c: \(\Leftrightarrow3^{x+1}=3^{2x}\)
=>2x=x+1
=>x=1
d: \(\Leftrightarrow2^{2x+3}=2^{2x-10}\)
=>2x+3=2x-10
=>0x=-13(vô lý)
NT
1
23 tháng 5 2022
a: =>|7x-9|=5x-3
\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{3}{5}\\\left(7x-9-5x+3\right)\left(7x-9+5x-3\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{3}{5}\\\left(2x-6\right)\left(12x-12\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{3;1\right\}\)
b: =>|17x-5|=|17x+5|
=>17x-5=17x+5(vô lý) hoặc 17x-5=-17x-5
=>34x=0
hay x=0
c: =>|3x+4|=|4x-18|
=>4x-18=3x+4 hoặc 4x-18=-3x-4
=>x=22 hoặc 7x=14
=>x=22 hoặc x=2
DT
0
30 tháng 9 2018
a)\(2^{3x+2}=4^{x+5}\)
\(2^{3x+2}=2^{2x+10}\)
\(\Rightarrow3x+2=2x+10\)
\(\Rightarrow x=8\)
Vậy \(x=8\)
b) \(3^{x+1}=9^x\)
\(3^{x+1}=3^{2x}\)
\(\Rightarrow x+1=2x\)
\(\Rightarrow x=1\)
Vậy \(x=1\)
c) \(\left(\frac{4}{5}\right)^{2x+7}=\frac{625}{256}\)
\(\left(\frac{4}{5}\right)^{2x+7}=\frac{5^4}{4^4}\)
\(\frac{4^{2x+7}}{5^{2x+7}}=\frac{5^4}{4^4}\)
\(\Rightarrow5^{2x+7}.5^4=4^{2x+7}.4^4\)
\(\Leftrightarrow5^{2x+11}=4^{2x+11}\)
\(\Leftrightarrow5=4\)( vô lý )
\(\Rightarrow\)x không có giá trị
Vậy không tìm được giá trị của x
Fan alibaba nguyễn~
13 tháng 9 2017
\(\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}\)
\(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{4\left(\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{3}{4}\)
\(=\dfrac{1}{4}+\dfrac{3}{4}\)
\(=1\)
13 tháng 9 2017
\(\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}\)
\(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{4\left(\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}=\dfrac{1}{4}+\dfrac{3}{4}=1\)
AT
30 tháng 6 2017
\(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{4\left(\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}\)
\(=\dfrac{1}{4}+\dfrac{3}{4}=\dfrac{4}{4}=1\)
\(\left(2x-\sqrt{\dfrac{9}{4}}\right)^2=\dfrac{1}{\sqrt{625}}\)
\(\Leftrightarrow\left(2x-\dfrac{3}{2}\right)^2=\dfrac{1}{25}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{3}{2}=\sqrt{\dfrac{1}{25}}\\2x-\dfrac{3}{2}=-\sqrt{\dfrac{1}{25}}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{3}{2}=\dfrac{1}{5}\\2x-\dfrac{3}{2}=-\dfrac{1}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{17}{10}\\2x=\dfrac{13}{10}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17}{20}\\x=\dfrac{13}{20}\end{matrix}\right.\)
Vậy ..
\(\left(2x-\sqrt{\dfrac{9}{4}}\right)^2=\dfrac{1}{\sqrt{625}}\\ \Leftrightarrow\left(2x-\dfrac{3}{2}\right)^2=\dfrac{1}{25}\\ \Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{3}{2}=\dfrac{1}{5}\\2x-\dfrac{3}{2}=-\dfrac{1}{5}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17}{20}\\x=\dfrac{13}{20}\end{matrix}\right.\)