$S=2^2+4^2+....+20^2=?$

$={\left(2.1\right)}^2+{\left(2.2\right)}^2+{\left(2.3\right)}^2+....+{\left(2.10\right)}^2$

$=2^2.1^2+2^2.2^2+2^2.2^3+...+2^2.10^2$

$=2^2.\left(1^2+2^2+3^2+...+10^2\right)$

$=2^2.385$

$=4.385$

$=1540$

a,5200>3200(Cuˋng.so^ˊ.mu~,cơ.so^ˊ:5>3)b,(−4)30=[(−4)3]10=(−64)10=6410;(−3)40=[(−3)4]10=8110Vıˋ:6410<8110(Cuˋng.so^ˊ.mu~,cơ.so^ˊ:64<81)⇒(−4)30<(−3)40

$c,{\left(-\genfrac{}{}{0ex}{}{1}{16}\right)}^{10}=\genfrac{}{}{0ex}{}{{\left(-1\right)}^{10}}{{\left(2^4\right)}^{10}}=\genfrac{}{}{0ex}{}{1}{2^{40}};{\left(\genfrac{}{}{0ex}{}{1}{64}\right)}^7=\genfrac{}{}{0ex}{}{1^7}{{\left(2^6\right)}^7}=\genfrac{}{}{0ex}{}{1}{2^{42}}V\grave{ı}:2^{40}<2^{42}\Rightarrow \genfrac{}{}{0ex}{}{1}{2^{40}}>\genfrac{}{}{0ex}{}{1}{2^{42}}Vậy:{\left(-\genfrac{}{}{0ex}{}{1}{16}\right)}^{10}>{\left(\genfrac{}{}{0ex}{}{1}{64}\right)}^7$

$d,6^{10}={\left(6^2\right)}^5=36^{5}V\grave{ı}:37^5>36^5\left(C\grave{u}ng.s\acute{\hat{o}}.m\tilde{u},cơ.s\acute{\hat{o}}:37>36\right)N\hat{e}n:37^5>6^{10}$

 Đúng(1)   POP POP CTVVIP 5 giờ trước (9:12)  

Bài 6:

$a)27.3^3.\genfrac{}{}{0ex}{}{1}{81}.3^{27}=3^3.3^3.\genfrac{}{}{0ex}{}{1}{3^4}.3^{27}=\genfrac{}{}{0ex}{}{3^{3+3+27}}{3^4}=3^{33-4}=3^{29}b,5^3.625:5^2=5^3.5^4:5^2=5^{3+4-2}=5^{5}c,\mathrm{64.125.}{3}^3.\genfrac{}{}{0ex}{}{1}{27}:25^3=64.5^3.3^3.\genfrac{}{}{0ex}{}{1}{3^3}.\genfrac{}{}{0ex}{}{1}{{\left(5^2\right)}^3}=64.\genfrac{}{}{0ex}{}{5^3}{5^{)6}}.\genfrac{}{}{0ex}{}{3^3}{3^3}=64.\genfrac{}{}{0ex}{}{1}{5^3}.1=\genfrac{}{}{0ex}{}{64}{125}d,{\left(\genfrac{}{}{0ex}{}{1}{7}\right)}^2.7^{-1}:\genfrac{}{}{0ex}{}{1}{49^2}=\genfrac{}{}{0ex}{}{1}{7^2}.\genfrac{}{}{0ex}{}{1}{7}:\genfrac{}{}{0ex}{}{1}{{\left(7^2\right)}^2}=\genfrac{}{}{0ex}{}{1}{7^3}.7^4=7^{4-3}=7^1=7$

x=5

=83

 

3600

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