20*3+7*20+20*88+20*2
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TD
4
28 tháng 11 2016
= 20 + 40 + 60 + 80 + 100 + 120 + 140 + 160 + 180 + 200
=(20 + 80)+ (40 + 60) + (120 +180) + (140 + 160) + 100
=100 + 100 + 300 + 300 + 100
= 200 + 300 + 300 + 100
= 500 + 300 + 100
= 800 + 100
= 900
28 tháng 6 2018
Đáp án: thiếu đề
@#@
mời bn xem xét lại đề bài.
~hok tốt~
TD
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 = ?
12
27 tháng 11 2016
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
MJ
1
KN
11 tháng 1 2019
\(20+\frac{20}{10}+87+88+90+\frac{10\times100}{10}\times\frac{800}{1000}\)
\(=20+2+175+90+100+0,8\)
\(=22+265+100,8\)
\(=287+100,8\)
\(=387,8\)
S
7 tháng 5 2017
a, Ta có: \(\frac{2001}{2002}=\frac{2002-1}{2002}=\frac{2002}{2002}-\frac{1}{2002}=1-\frac{1}{2002}\)
\(\frac{2000}{2001}=\frac{2001-1}{2001}=\frac{2001}{2001}-\frac{1}{2001}=1-\frac{1}{2001}\)
Vì \(\frac{1}{2002}< \frac{1}{2001}\Rightarrow1-\frac{1}{2002}>1-\frac{1}{2001}\Rightarrow\frac{2001}{2002}>\frac{2000}{2001}\)
b, Ta có: \(\left(\frac{1}{80}\right)^7>\left(\frac{1}{81}\right)^7=\left(\frac{1}{3^4}\right)^7=\left(\frac{1}{3}\right)^{28}=\frac{1}{3^{28}}\)
\(\left(\frac{1}{243}\right)^6=\left(\frac{1}{3^5}\right)^6=\left(\frac{1}{3^5}\right)^6=\frac{1}{3^{30}}\)
Vì \(\frac{1}{3^{28}}>\frac{1}{3^{30}}\Rightarrow\left(\frac{1}{81}\right)^7>\left(\frac{1}{243}\right)^6\Rightarrow\left(\frac{1}{80}\right)^7>\left(\frac{1}{243}\right)^6\)
c, Ta có: \(\left(\frac{3}{8}\right)^5=\frac{3^5}{\left(2^3\right)^5}=\frac{243}{2^{15}}>\frac{243}{3^{15}}>\frac{125}{3^{15}}=\frac{5^3}{\left(3^5\right)^3}=\frac{5^3}{243^3}=\left(\frac{5}{243}\right)^3\)
Vậy \(\left(\frac{3}{8}\right)^5>\left(\frac{5}{243}\right)^3\)
d, Ta có: \(\frac{2011}{2012}>\frac{2011}{2012+2013}\)
\(\frac{2012}{2013}>\frac{2012}{2012+2013}\)
\(\Rightarrow\frac{2011}{2012}+\frac{2012}{2013}>\frac{2011}{2012+2013}+\frac{2012}{2012+2013}=\frac{2011+2012}{2012+2013}\)
e, \(C=\frac{20^{10}+1}{20^{10}-1}=\frac{20^{10}-1+2}{20^{10}-1}=\frac{20^{10}-1}{20^{10}-1}+\frac{2}{2^{10}-1}=1+\frac{2}{2^{10}-1}\)
\(D=\frac{20^{10}-1}{20^{10}-3}=\frac{20^{10}-3+2}{20^{10}-3}=\frac{20^{10}-3}{20^{10}-3}+\frac{2}{2^{10}-3}=1+\frac{2}{2^{10}-3}\)
Vì \(\frac{2}{10^{10}-1}< \frac{2}{10^{10}-3}\Rightarrow1+\frac{2}{10^{10}-1}< 1+\frac{2}{10^{10}-3}\Rightarrow C< D\)
g, \(G=\frac{10^{100}+2}{10^{100}-1}=\frac{10^{100}-1+3}{10^{100}-1}=\frac{10^{100}-1}{10^{100}-1}+\frac{3}{10^{100}-1}=1+\frac{3}{10^{100}-1}\)
\(H=\frac{10^8}{10^8-3}=\frac{10^8-3+3}{10^8-3}=\frac{10^8-3}{10^8-3}+\frac{3}{10^8-3}=1+\frac{3}{10^8-3}\)
Vì \(\frac{3}{10^{100}-1}< \frac{3}{10^8-3}\Rightarrow1+\frac{3}{10^{100}-1}< 1+\frac{3}{10^8-3}\Rightarrow G< H\)
h, Vì E < 1 nên:
\(E=\frac{98^{99}+1}{98^{89}+1}< \frac{98^{99}+1+97}{98^{89}+1+97}=\frac{98^{99}+98}{98^{89}+98}=\frac{98\left(98^{98}+1\right)}{98\left(98^{88}+1\right)}=\frac{98^{98}+1}{98^{88}+1}=F\)
Vậy E = F
NT
1
H9
HT.Phong (9A5)
CTVHS
13 tháng 10 2023
\(\dfrac{1}{10}+\dfrac{1}{40}+\dfrac{1}{88}+...+\dfrac{1}{\left(x+2\right)\left(x+5\right)}=\dfrac{3}{20}\)
\(\Rightarrow\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{\left(x+2\right)\left(x+5\right)}=\dfrac{3}{20}\)
\(\Rightarrow\dfrac{1}{3}\cdot\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{\left(x+2\right)\left(x+5\right)}\right)=\dfrac{3}{20}\)
\(\Rightarrow\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+..+\dfrac{3}{\left(x+2\right)\left(x+5\right)}=\dfrac{9}{20}\)
\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{x+2}-\dfrac{1}{x+5}=\dfrac{9}{20}\)
\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{x+5}=\dfrac{9}{20}\)
\(\Rightarrow\dfrac{1}{x+5}=\dfrac{1}{2}-\dfrac{9}{20}\)
\(\Rightarrow\dfrac{1}{x+5}=\dfrac{1}{20}\)
\(\Rightarrow x+5=20\)
\(\Rightarrow x=20-5\)
\(\Rightarrow x=15\)
12 tháng 3 2023
8:
\(A=\dfrac{20^{10}-1+2}{20^{10}-1}=1+\dfrac{2}{20^{10}-1}\)
\(B=\dfrac{20^{10}-3+2}{20^{10}-3}=1+\dfrac{2}{20^{10}-3}\)
mà 20^10-1>20^10-3
nên A<B
10 tháng 5 2019
làm lại
ta có : \(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+...+\frac{1}{\left(x+2\right)\left(x+5\right)}=\frac{3}{20}\)
=>\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{\left(x+2\right)\left(x+5\right)}=\frac{3}{20}\)
=>\(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+....+\frac{3}{\left(x+2\right)\left(x+5\right)}=\frac{9}{20}\)
=>\(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{x+2}-\frac{1}{x+5}=\frac{9}{20}\)
=>\(\frac{1}{2}-\frac{1}{x+5}=\frac{9}{20}\)
=>\(\frac{1}{x+5}=\frac{1}{2}-\frac{9}{20}\)
=>\(\frac{1}{x+5}=\frac{1}{20}\)
=>\(x+5=20\)
=>\(x=20-5\)
=>\(x=15\)
10 tháng 5 2019
ta có : \(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+....+\frac{1}{\left(x+2\right)\left(x+5\right)}=\frac{3}{20}\)
=>\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{\left(x+2\right)\left(x+3\right)}=\frac{3}{20}\)
=>\(3.\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+....+\frac{1}{\left(x+2\right)\left(x+3\right)}\right)=3.\frac{3}{20}\)
=>\(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+....+\frac{3}{\left(x+2\right)\left(x+3\right)}=\frac{9}{20}\)
=>\(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{x+2}-\frac{1}{x+3}=\frac{9}{20}\)
=>\(\frac{1}{2}-\frac{1}{x+3}=\frac{9}{20}\)
=>\(\frac{1}{x+3}=\frac{1}{2}-\frac{9}{20}\)
=>\(\frac{1}{x+3}=\frac{1}{20}\)
=>\(x+3=20\)
=>\(x=20-3\)
=>\(x=17\)
=20x(3+7+88+2)=20x100=2000
Tk mk nha bn!!!
20 x 3 + 7 x 20 + 20 x 88 + 20 x 2
=20 x (3 + 7 + 88 + 2)
= 20 x 100
= 2000
k cho tớ nha bn