Y=3^20+6^20+8^20
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có:\(2^{30}=\left(2^3\right)^{10}=8^{10}< 9^{10}=\left(3^2\right)^{10}=3^{20}\)
\(3^{30}=3^{20}.3^{10}< 3^{20}.4^{10}=3^{20}.\left(2^2\right)^{10}=3^{20}.2^{20}=\left(3.2\right)^{20}=6^{20}\)
\(4^{30}=4^{20}.4^{10}=4^{20}.\left(2^2\right)^{10}=4^{20}.2^{20}=\left(4.2\right)^{20}=8^{20}\)
nên \(2^{30}+3^{30}+4^{30}< 3^{20}+6^{20}+8^{20}\)
20 x 1 + 20 x 2 + 20 x 3 + 20 x 4 + 20 x 5 + 20 x 6 = ?
1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x ... x 100 = ?
20x1+20x2+20x3+20x4+20x5+20x6= 20x(1+2+3+4+5+6)= 20 x 21 = 420
Còn câu 2 mik thấy khó quá cho mik xin lỗi nha
Xét \(A=2^{30}+3^{30}+4^{30}=\left(2^3\right)^{10}+\left(3^3\right)^{10}+\left(2^2\right)^{30}=8^{10}+27^{10}+2^{60}\)
\(B=3^{20}+6^{20}+8^{20}=\left(3^2\right)^{10}+\left(6^2\right)^{10}+\left(2^3\right)^{20}=9^{10}+36^{10}+2^{60}\)
Vì \(8^{10}< 9^{10},27^{10}< 36^{10}\)nên A<B
230 = 23.10= 810
330=33.10=2710
430=43.10=6410
Vế trái = 810 + 2710 + 6410
320=32.10=910
620=62.10=3610
820=82.10=6410
vế phải = 910 + 3610 + 6410
Vì 6410=6410 ; 3610 > 2710 ; 910 > 810
=> vế phải > vế trái
\(\dfrac{45^{10}\cdot5^{20}}{75^{15}}=\dfrac{\left(3^2\cdot5\right)^{10}\cdot5^{20}}{\left(3\cdot5^2\right)^{15}}=\dfrac{3^{20}\cdot5^{10}\cdot5^{20}}{3^{15}\cdot5^{30}}=3^5=243\\ \dfrac{6^6+6^3+3^3+3^6}{-73}=\dfrac{46656+216+27+729}{-73}=-\dfrac{47628}{73}\\ \dfrac{27^7+3^{15}}{9^9-27}=\dfrac{\left(3^3\right)^7+3^{15}}{\left(3^2\right)^9-3^3}=\dfrac{3^{21}+3^{15}}{3^{18}-3^3}=\dfrac{3^{15}\left(3^6+1\right)}{3^3\left(3^{15}-1\right)}=\dfrac{3^5\cdot730}{3^{15}-1}\\ \dfrac{8^{20}+4^{20}}{4^{25}+64^5}=\dfrac{\left(2^3\right)^{20}+\left(2^2\right)^{20}}{\left(2^2\right)^{25}+\left(2^6\right)^5}=\dfrac{2^{60}+2^{40}}{2^{50}+2^{30}}=\dfrac{2^{40}\left(2^{20}+1\right)}{2^{30}\left(2^{20}+1\right)}=2^{10}=1024\)