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Ta có : S = 22 + 42 + 62 + ...... + 202
1/4S = 12 + 22 + 32 + ..... + 102
1/4S = 1.1 + 2.2 + 3.3 + ..... + 10.10
= 1.(2 - 1) + 2(3 - 1) + 3(4 - 1) + ..... + 10.(11 - 1)
= 1.2 - 1 + 2.3 - 2 + 3.4 - 3 + ..... + 10.11 - 10
= (1.2 + 2.3 + 3.4 + ..... + 10.11) - (1 + 2 + 3 + ..... + 10)
1/4S = 385
=> S = 385 . 4 = 1540
\(A=\frac{14^{16}-21^{32}.35^{68}}{10^{16}.15^2.7^{96}}=\frac{2^{16}.7^{16}-3^{32}.7^{32}.7^{68}.5^{68}}{5^{16}.2^{16}.3^2.5^2.7^{96}}\)
\(A=\frac{2^{16}.7^{16}-3^{32}.7^{100}.5^{68}}{5^{18}.2^{16}.3^2.7^{96}}=\frac{7^{16}.\left(2^{16}-3^{32}.7^{84}.5^{68}\right)}{5^{18}.2^{16}.3^2.7^{96}}=\frac{2^{16}-3^{32}.7^{84}.5^{68}}{5^{18}.2^{16}.3^2.7^{88}}\)
\(B=\frac{4^{20}-2^{20}+6^{20}}{6^{20}-3^{20}+9^{20}}=\frac{2^{40}-2^{20}+2^{20}.3^{20}}{2^{20}.3^{20}-3^{20}+3^{40}}=\frac{2^{20}.\left(2^{20}-1+3^{20}\right)}{3^{20}.\left(2^{20}-1+3^{20}\right)}\)
\(B=\frac{2^{20}}{3^{20}}\)
\(2^2+4^2+6^2+...+20^2\)
\(=1^2.2^2+2^2.2^2+2^2.3^2+...+2^2.10^2\)
\(=2^2.\left(1^2+2^2+3^2+...+10^2\right)\)
\(=4.385=1540\)
\(=\frac{\left(2^{20}-1^{20}+3^{20}\right)\cdot2^{20}}{\left(2^{20}-1^{20}+3^{20}\right)\cdot3^{20}}\left(\text{Mình làm hơi tắt, xin bạn thông cảm}\right)\)
\(=\frac{2^{20}}{3^{20}}=\left(\frac{2}{3}\right)^{20}\)
a/\(\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9}{2^{10}.3^8+\left(2.3\right)^8.2^2.5}\)
= \(\frac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}=\frac{2^{10}.3^8.\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}=\frac{1-3}{1+5}=\frac{-2}{6}=-3\)
\(B=2^2+4^2+........+20^2\)
\(B=1^2.2^2+2^2.2^2+3^2.2^2+.......+10^2.2^2\)
\(B=2^2.\left(1^2+2^2+3^2+.......+10^2\right)\)
\(B=4.385\) (đề bài cho)
\(B=1540\)
Ta có: S=22+42+62+...202
=2 x (12+22+32+...+102)
= 2 x 385
= 770
Vậy S=770
S=(1.2)2+(2.2)2+(2.3)2+.........+(2.10)2
S=12.22+22.22+22.32+.........+22.102
S=22.(12+22+32+............+102)
S=4.385
S=1540
= 110/3 nka
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