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a/|x|-2,5=27,5
=>|x|=27,5+2,5=30
=>x=30 hoặc x=-30
b/\(\dfrac{3}{4}+\dfrac{2}{5}.x=\dfrac{29}{60}\)
=>\(\dfrac{2}{5}.x\)=\(\dfrac{29}{60}-\dfrac{3}{4}\)=\(\dfrac{-4}{15}\)
=>x=\(\dfrac{-4}{15}:\dfrac{2}{5}\)=\(\dfrac{-2}{3}\)
c/(x-1)\(^5\)=-32
=>x-1=-2 vì (-2)\(^5\)=-32
=>x=-2+1=-1
d/\(\dfrac{4}{5}.x+0,5=4.5\)
=>\(\dfrac{4}{5}.x+0,5=20\)
=>\(\dfrac{4}{5}.x=20-0,5=19,5\)
=>\(x=19,5:\dfrac{4}{5}\)=\(\dfrac{195}{8}\)
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Ta có: \(3\cdot5^{x+2}+4\cdot5^{x-3}=19\cdot5^{10}\)
\(\Leftrightarrow5^{x-3}\cdot\left(3\cdot5+4\right)=19\cdot5^{10}\)
\(\Leftrightarrow5^{x-3}\cdot19=19\cdot5^{10}\)
\(\Leftrightarrow5^{x-3}=5^{10}\)
\(\Rightarrow x-3=10\)
\(\Rightarrow x=13\)
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a) 3x + 1 + 3x + 3 = 270
=> 3x .3 + 3x . 27 = 270
=> 3x .(3 + 27) = 270
=> 3x .30 = 270
=> 3x = 270 : 30
=> 3x = 9
=> 3x = 32
=> x = 2
b) 5x + 4.5x + 1 = 2625
=> 5x + 4.5x.5 =2625
=> 5x.(1 +20) = 2625
=> 5x .21 = 2625
=> 5x = 2625 : 21
=> 5x = 125
=> 5x = 53
=> x = 3
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1/3.4+1/4.5+1/5.6+1/6.7+....+1/x(x+1)=3/10
<=> \(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{\left(x+1\right)x}=\frac{1}{3}-\frac{1}{x+1}=\frac{3}{10}\)
<=> \(\frac{1}{x+1}=\frac{1}{3}-\frac{3}{10}=\frac{1}{30}\)=> x+1=30=>x=29
\(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{x\left(x+1\right)}=\frac{3}{10}\)
\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{3}{10}\)
\(\frac{1}{3}-\frac{1}{x+1}=\frac{3}{10}\)
\(\frac{1}{x+1}=\frac{1}{3}-\frac{3}{10}\)
\(\frac{1}{x+1}=\frac{1}{30}\)
\(\Rightarrow x+1=30\)
\(x=30-1=29\)
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\(5^{x+3}-4\cdot5^{x+1}=21\cdot5\)
\(\Rightarrow5^{x+1}\cdot5^2-4\cdot5^{x+1}=105\)\(\Rightarrow5^{x+1}\cdot\left(25-4\right)=105\)
\(\Rightarrow5^{x+1}\cdot21=105\) \(\Rightarrow5^{x+1}=5\) \(\Rightarrow x+1=1\) \(\Rightarrow x=0\)
5x+3 - 4 . 5x+1 = 21 . 5
=> 5x+1 . ( 52 - 4 ) = 21 . 5
=> 5x+1 . ( 25 - 4 ) = 21 . 51
=> 5x+1 . 21 = 21 . 51
=> 5x+1 = 51
=> x + 1 = 1
=> x = 0