2012.(2011-x)+25=5 mũ 5 :5 mũ 3
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a) \(5\left(x+7\right)-10=2^3\cdot5\)
\(\Rightarrow5\left(x+7\right)-10=40\)
\(\Rightarrow5\left(x+7\right)=40+10\)
\(\Rightarrow x+7=\dfrac{50}{5}\)
\(\Rightarrow x+7=10\)
\(\Rightarrow x=10-7\)
\(\Rightarrow x=3\)
b) \(9x-2\cdot3^2=3^4\)
\(\Rightarrow9x-18=81\)
\(\Rightarrow9x=81+18\)
\(\Rightarrow9x=99\)
\(\Rightarrow x=\dfrac{99}{9}\)
\(\Rightarrow x=11\)
c) \(5^{25}\cdot5^{x-1}=5^{25}\)
\(\Rightarrow5^{x-1}=5^{25}:5^{25}\)
\(\Rightarrow5^{x-1}=1\)
\(\Rightarrow5^{x-1}=5^0\)
\(\Rightarrow x-1=0\)
\(\Rightarrow x=1\)
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\(\left(x+1\right)^3=27\)
\(\left(x+1\right)^3=3^3\)
\(\Rightarrow x+1=3\)
\(x=2\)
\(\left(x+1\right)^3=27\)
\(< =>\left(x+1\right)^3=3.3.3=3^3\)
\(< =>x+1=3< =>x=3-1=2\)
\(\left(2x+3\right)^3=9.81\)
\(< =>\left(2x+3\right)^3=9.9.9\)
\(< =>\left(2x+3\right)^3=9^3\)
\(< =>2x+3=9< =>2x=6\)
\(< =>x=\frac{6}{2}=3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, (-0,2)2 \(\times\) 5 - \(\dfrac{2^{13}\times27^3}{4^6\times9^5}\)
= 0,04 \(\times\) 5 - \(\dfrac{2^{13}\times3^9}{2^{12}\times3^{10}}\)
= 0,2 - \(\dfrac{2}{3}\)
= \(\dfrac{2}{10}\) - \(\dfrac{2}{3}\)
= - \(\dfrac{7}{15}\)
b, \(\dfrac{5^6+2^2.25^3+2^3.125^2}{26.5^6}\)
= \(\dfrac{5^6+4.5^6+8.5^6}{26.5^6}\)
= \(\dfrac{5^6.\left(1+4+8\right)}{26.5^6}\)
= \(\dfrac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(\left(-17\right)+5+8+17+\left(-3\right)\)
\(=\left(-17+17\right)+\left[5+\left(-3\right)\right]+8\)
\(=0+8+8=8+8=16\)
b, \(\left(5^{19}:5^{17}+3\right):7=\left(5^2+3\right):7\)
\(=\left(25+3\right):7=28:7=4\)
c, \(|-8|+\left(-5\right)+9+\left(-7\right)+|-4|\)
\(=8-5+9-7+4=3+2+4=5+4=9\)
ý d mk ko biết nha.
thông cảm cho mk nha.
k mk nha.
#mon
2012.(2011-x) + 25 = 55:53 = 25
2012.(2011-x) = 25 - 25 = 0
2011 - x = 0
x = 2011 - 0 = 2011
2012.(2011-x)+25=55:53
=>2012.(2011-x)+25=55-3
=>2012.(2011-x)+25=52
=>2012.(2011-x)+25=25
=>2012.(2011-x)=25-25
=>2012.(2011-x)=0
=>2011-x=0:2012
=>2011-x=0
=>x=0+2011
=>x=2011