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\(\text{Ta có : }2009^{20}=(2009^2)^{10}=(2009\cdot2009)^{10}\)
\(20092009^{10}=(2009\cdot10001)^{10}\)
\(\text{Vì }(2009\cdot2009)^{10}< (2009\cdot10001)^{10}\text{nên }2009^{20}< 20092009^{10}\)
Chúc bạn học tốt ~
1. a) Ta thấy: 230=23.10=(23)10=810
320=32.10=(32)10=910
810<910 => 230<320
b) 5202=52.101=(52)101=25101
2505=25.101=(25)101=32101
Mà 25101<32101 =. 5202<2505
2. a) Ta có: 2008100+200899=200899.2008+200899=200899.(2008+1)=200899.2009
200899.2009 chia hết cho 2009 => 2008100+200899 chia hết cho 2009.
b) 12345678-12345677=12345677.12345 - 12345677=12345677.(12345-1)=12345677.12344
=> 12345677.12344 chia hết cho 12344 =. 12345678-12345677 chia hết cho 12344.
k mình nha.
1. a) Ta thấy: 2 30=2 3.10=(2 3 )10=8 10 3 20=3 2.10=(3 2 )10=9 10 8 10<9 10
=> 2 30<3 20 b) 5 202=5 2.101=(5 2 )101=25 101 2 505=2 5.101=(2 5 )101=32 101
Mà 25 101<32 101 =. 5 202<2 505 2. a) Ta có: 2008 100+2008 99=2008 99 .2008+2008 99=2008 99 .(2008+1)=2008 99 .2009
2008 99 .2009 chia hết cho 2009
=> 2008 100+2008 99 chia hết cho 2009.
b) 12345 678 -12345 677=12345 677 .12345 - 12345 677=12345 677 .(12345-1)=12345 677 .12344
=> 12345 677 .12344 chia hết cho 12344 =. 12345 678 -12345 677 chia hết cho 12344.
k mình nha.
\(\left(2a+1\right)^2+\left(b+3\right)^4+\left(5c-6\right)^2\le0\left(1\right)\)
Ta có:\(\hept{\begin{cases}\left(2a+1\right)^2\ge0;\forall a,b,c\\\left(b+3\right)^4\ge0;\forall a,b,c\\\left(5c-6\right)^2\ge0;\forall a,b,c\end{cases}}\)\(\Rightarrow\left(2a+1\right)^2+\left(b+3\right)^4+\left(5c-6\right)^2\ge0;\forall a,b,c\left(2\right)\)
Từ \(\left(1\right)\)và \(\left(2\right)\)\(\Rightarrow\left(2a+1\right)^2+\left(b+3\right)^4+\left(5c-6\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}\left(2a+1\right)^2=0\\\left(b+3\right)^4=0\\\left(5c-6\right)^2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}a=\frac{-1}{2}\\b=-3\\c=\frac{6}{5}\end{cases}}\)
Vậy \(\left(a,b,c\right)=\left(\frac{-1}{2};-3;\frac{6}{5}\right)\)
Xét: \(\frac{\left(17^{2017}+16^{2017}\right)^{2018}}{17^{2017.2018}}=\left(\frac{17^{2017}+16^{2017}}{17^{2017}}\right)^{2018}=\left(1+\left(\frac{16}{17}\right)^{2017}\right)^{2018}\)
\(\frac{\left(17^{2018}+16^{2018}\right)^{2017}}{17^{2017.2018}}=\left(\frac{17^{2018}+16^{2018}}{17^{2018}}\right)^{2017}=\left(1+\left(\frac{16}{17}\right)^{2018}\right)^{2017}\)
Ta có: \(0< \frac{16}{17}< 1\)
=> \(\left(\frac{16}{17}\right)^{2017}>\left(\frac{16}{17}\right)^{2018}\)
=> \(1+\left(\frac{16}{17}\right)^{2017}>1+\left(\frac{16}{17}\right)^{2018}>1\)
=> \(\left(1+\left(\frac{16}{17}\right)^{2017}\right)^{2018}>\left(1+\left(\frac{16}{17}\right)^{2018}\right)^{2017}\)
=> \(\left(17^{2017}+16^{2017}\right)^{2018}>\left(17^{2018}+16^{2018}\right)^{2017}\)
=>81.(x+3)=125.729
=>81.(x+3)=91125
=>x+3=91125:81
=>x+3=1125
=>x=1125:3
=>x=375
vậy x=375
a) >
b) <
Sai thì thông cảm cho mik nha ( #Huyen )
Giải chi tiết nhé mn