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3^x+3^(x+2)=90
3^x+3^x.3^2=90
3^x.(1+3^2)=90
3^x.10=90
3^x=9
3^x=3^2
=>x=2
Vậy x=2
3^x + 3^(x+2) = 90
=> 3^x + 3^x x 3^2 = 90
=> 3^x (9 + 1) = 90
=> 3^x x 10 = 90
=> 3^x = 9
=> 3^x = 3^2
=> x = 2
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{3}{11}\times\frac{7}{19}+\frac{3}{11}\times\frac{8}{19}-\frac{4}{9}:\frac{11}{3}\)
\(=\frac{3}{11}\times\frac{7}{9}+\frac{3}{11}\times\frac{8}{9}-\frac{4}{9}\times\frac{3}{11}\)
\(=\frac{3}{11}\times\left(\frac{7}{9}+\frac{8}{9}-\frac{4}{9}\right)\)
\(=\frac{3}{11}\times\frac{11}{9}\)
\(=\frac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=1+2+...+2^{2017}\)
\(2S=2+2^2+...+2^{2018}\)
\(2S-S=2+2^2+...+2^{2018}-1-2-...-2^{2017}\)
\(S=2^{2018}-1\)
\(S=3+3^2+...+3^{2017}\)
\(3S=3^2+3^3+...+3^{2018}\)
\(3S-S=3^2+3^3+...+3^{2018}-3-3^2-...-3^{2017}\)
\(2S=3^{2018}-3\)
\(S=\dfrac{3^{2018}-3}{2}\)
\(S=4+4^2+...+4^{2017}\)
\(4S=4^2+4^3+...+4^{2018}\)
\(4S-S=4^2+4^3+...+4^{2018}-4-4^2-...-4^{2017}\)
\(3S=4^{2018}-4\)
\(S=\dfrac{4^{2018}-4}{3}\)
\(S=5+5^2+...+5^{2017}\)
\(5S=5^2+5^3+...+5^{2018}\)
\(5S-S=5^2+5^3+...+5^{2018}-5-5^2-...-5^{2017}\)
\(4S=5^{2018}-5\)
\(S=\dfrac{5^{2018}-5}{4}\)
a) S=1+2+22+...+22017
=> 2S=2.(1+2+22+...+22017)
=>2S=2+22+23+...+22018
=>S=(2+22+23+ ..+22018) - (1+2+22+ ....+22017 )
=> S =22018-1
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hửm :)))
chết mình gửi nhầm đề bài là thế này
tính : 11 + 13 + 15 +17 +........+97 + 99 = .......