\(2.11^x=\left(3^2+2\right)^3:\left(5^3-2^5:2^3\right)\)
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Q
12 tháng 9 2017
(x2-1)2=9
=> x2-1 = 3
x2 = 3+1
x2 = 4
=> x2 = 4 = 22 ( x2=22 )
<=> x = 2
12:{390:[5.102-(53+x.72)]} = 4
390:[5.102-(53+x.72)] = 12:4
390:[5.102-(53+x.72)] = 3
5.102-(53+x.72) = 390 : 3
5.102-(53+x.72) = 130
=> 500-(125+x+49)=130
125+x+49 = 500-130
125+x+49 = 370
125+x = 370-49
125+x = 321
x = 321-125
x = 106
53(3x+2):13=103:(135:134)
53(3x+2):13=103:13
53(3x+2):13= 1000/13
125(3x+2):13 = 1000/13
125(3x+2) = 1000/13 . 13
125(3x+2) = 1000
3x+2 = 1000:125
3x+2 = 8
3x = 8-2
3x = 6
x = 6:3
x = 2
TC
19 tháng 4 2022
a)\(3x-\dfrac{2}{5}=0=>3x=\dfrac{2}{5}=>x=\dfrac{2}{15}\)
b)\(\left(x-3\right)\left(2x+8\right)=0=>\left[{}\begin{matrix}x-3=0\\2x=-8\end{matrix}\right.=>\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\)
c)\(3x^2-x-4=0=>3x^2+3x-4x-4=0=>\left(3x-4\right)\left(x+1\right)=0\)
\(=>\left[{}\begin{matrix}3x=4\\x+1=0\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-1\end{matrix}\right.\)
H
30 tháng 12 2021
\(\dfrac{-3}{5}-x=\dfrac{21}{10}\)
\(x=\dfrac{-3}{5}-\dfrac{21}{10}\)
\(x=\)-\(\dfrac{27}{10}\)
\(x:\dfrac{2}{9}=\dfrac{9}{2}\)
\(x.\dfrac{9}{2}=\dfrac{9}{2}\)
\(x=\dfrac{9}{2}:\dfrac{9}{2}\)
\(x=1\)
\(\dfrac{x}{9}=\dfrac{5}{3}\)
\(x.3=5.9\)
\(x.3=45\)
\(x=45:3=15\)
\(x:\left(\dfrac{2}{5}\right)^3=\left(\dfrac{5}{2}\right)^3\)
\(x:\dfrac{8}{125}=\dfrac{125}{8}\)
\(x.\dfrac{125}{8}=\dfrac{125}{8}\)
\(x=\dfrac{125}{8}:\dfrac{125}{8}=1\)
KT
28 tháng 7 2018
a) \(\left|x+\frac{1}{2}\right|=\frac{1}{3}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+\frac{1}{2}=\frac{1}{3}\\x+\frac{1}{2}=-\frac{1}{3}\end{cases}}\) \(\Leftrightarrow\)\(\orbr{\begin{cases}x=-\frac{1}{6}\\x=-\frac{5}{6}\end{cases}}\)
Vậy....
b) \(\left|x-\frac{1}{2}\right|=\frac{1}{3}-\frac{1}{2}\)
\(\Leftrightarrow\)\(\left|x-\frac{1}{2}\right|=-\frac{1}{6}\) vô lí do \(\left|a\right|\ge0\)
Vậy pt vô nghiệm
c) \(\left|x+\frac{1}{3}\right|-4=-1\)
\(\Leftrightarrow\)\(\left|x+\frac{1}{3}\right|=3\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+\frac{1}{3}=3\\x+\frac{1}{3}=-3\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\frac{8}{3}\\x=-\frac{10}{3}\end{cases}}\)
Vậy..
KT
28 tháng 7 2018
d) \(\left|x-\frac{1}{5}\right|+\frac{1}{3}=\frac{1}{4}-\left|-\frac{3}{2}\right|\)
\(\Leftrightarrow\)\(\left|x-\frac{1}{5}\right|+\frac{1}{3}=-\frac{5}{4}\)
\(\Leftrightarrow\)\(\left|x-\frac{1}{5}\right|=-\frac{19}{12}\)vô lí do \(\left|a\right|\ge0\)với mọi a
Vậy pt vô nghiệm
e) \(\left|x-\frac{5}{2}\right|=\frac{4}{3}-\left(\frac{2}{3}-\frac{1}{2}\right)\)
\(\Leftrightarrow\)\(\left|x-\frac{5}{2}\right|=\frac{7}{6}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x-\frac{5}{2}=\frac{7}{6}\\x-\frac{5}{2}=-\frac{7}{6}\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=3\frac{2}{3}\\x=\frac{4}{3}\end{cases}}\)
Vậy...
\(\Rightarrow11^x=11^3:\left(125-2^2\right)\\ \Rightarrow11^x=11^3:121=11^3:11^2=11^1\\ \Rightarrow x=1\)