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a)\(5^x+5^{x+2}=650\Rightarrow5^x+5^2.5^x=650\Rightarrow5^x+25.5^x=650\Rightarrow26.5^x=650\)\(5^x=25\Rightarrow5^x=5^2\Rightarrow x=2\)
b) \(3^{x-1}+5.3^{x-1}=162\Rightarrow6.3^{x-1}=162\Rightarrow3^{x-1}=27=3^3\)x-1=3 nên x=4

ta có: 99^20=99^(2.10)=9801^10
mà 9801<9999 nên 9801^10<9999^10
vậy 99^20<9999^10

= \(3^n\left(3^3+3\right)+2^n\left(2^3+2^2\right)\)
= \(3^n.30+2^n.12\)
= \(6.\left(3^n.5+2^n.2\right)⋮6\)
Ok nha bn :D

\(\left(1-\frac{1}{15}\right)\left(1-\frac{1}{21}\right)\left(1-\frac{1}{28}\right)...\left(1-\frac{1}{225}\right)\)
\(=\frac{14}{15}.\frac{20}{21}.\frac{27}{28}...\frac{224}{225}\)
\(=\frac{2.7}{3.5}.\frac{5.4}{7.3}.\frac{3.9}{4.7}...\frac{16.14}{15.15}\)
\(=\frac{2}{3}.\frac{14}{15}\) ( rút gọn )
\(=\frac{28}{45}\)

a) \(\left(x-\dfrac{1}{2}\right)^2=0\)
\(\Rightarrow x-\dfrac{1}{2}=0\)
\(\Rightarrow x=\dfrac{1}{2}\)
b) \(\left(x-2\right)^2=1\)
\(\Rightarrow x-2=1\)
\(\Rightarrow x=3\)
c) \(\left(2x-1\right)^3=-8\)
\(\Rightarrow\left(2x-1\right)^3=\left(-2\right)^3\)
\(\Rightarrow2x-1=-2\)
\(\Rightarrow2x=-1\)
\(\Rightarrow x=\dfrac{-1}{2}\)
d) \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)
\(\Rightarrow\left(x+\dfrac{1}{2}\right)^2=\left(\dfrac{1}{4}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{4}\\x+\dfrac{1}{2}=-\dfrac{1}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1}{4}\\x=\dfrac{-3}{4}\end{matrix}\right.\).
a , \(\left(x-\dfrac{1}{2}\right)^2=0\)
<=> \(x-\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{2}\)
b , \(\left(x-2\right)^2=1\Rightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
c , \(\left(2x-1\right)^3=-8\Rightarrow2x-1=-2\Rightarrow x=\dfrac{-1}{2}\)
d , \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{4^2}\)
<=> \(\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{4}\\x+\dfrac{1}{2}=\dfrac{-1}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{4}\\x=\dfrac{-3}{4}\end{matrix}\right.\)

\(3^{x-1}+5\times3^{x-1}=162\)
\(3^{x-1}\left(5+1\right)=162\)
\(3^{x-1}\times6=162\Rightarrow3^{x-1}=162\div6\)
\(3^{x-1}=27\Rightarrow3^{x-1}=3^3\)
\(\Rightarrow x-1=3\)
\(\Rightarrow x=4\)


\(1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+...+\frac{1}{x}.\left(1+2+3+...+x\right)=115\)
\(\Rightarrow1.\left(\frac{1.2}{2}\right)+\frac{1}{2}.\left(\frac{2.3}{2}\right)+\frac{1}{3}.\left(\frac{3.4}{2}\right)+....+\frac{1}{x}.\left[\frac{x\left(x+1\right)}{2}\right]=115\)
\(\Rightarrow\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+....+\frac{x+1}{2}=115\Rightarrow2+3+...+\left(x+1\right)=230\)
\(\frac{\Rightarrow\left[\frac{\left(x+1-2\right)}{1}+1\right].\left(x+1+2\right)}{2}=\frac{x.\left(x+3\right)}{2}=230\Rightarrow x.\left(x+3\right)=460\)
vì x và x+3 là 2 số tự nhiên cách nhau 3 đơn vị => \(x.\left(x+3\right)=460=20.23\Rightarrow x=20\)
Vậy x=20
\(3^{x-1}+5.3^{x-1}=162\)
`=>` \(6.3^{x-1}=162\)
`=>` \(3^{x-1}=162:6\)
`=>` \(3^{x-1}=27\)
`=>` \(3^{x-1}=3^3\)
`=> x-1 = 3`
`=> x = 4`
Vậy `x = 4`