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\(S=\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+...+\frac{2}{110}\)
\(S=\frac{2}{4\cdot5}+\frac{2}{5\cdot6}+...+\frac{2}{10\cdot11}\)
\(S=2\left(\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+...+\frac{1}{10\cdot11}\right)\)
\(S=2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{10}-\frac{1}{11}\right)\)
\(S=2\left(\frac{1}{4}-\frac{1}{11}\right)=2\cdot\frac{7}{44}=\frac{7}{22}\)
\(\dfrac{5}{2}+\dfrac{5}{6}+\dfrac{5}{12}+\dfrac{5}{20}+\dfrac{5}{30}+\dfrac{5}{42}+...+\dfrac{5}{110}\)
\(=\dfrac{5}{1.2}+\dfrac{5}{2.3}+\dfrac{5}{3.4}+\dfrac{5}{4.5}+\dfrac{5}{5.6}+\dfrac{5}{6.7}+...+\dfrac{5}{10.11}\)
\(=\dfrac{5}{1}-\dfrac{5}{2}+\dfrac{5}{2}-\dfrac{5}{3}+\dfrac{5}{3}-\dfrac{5}{4}+\dfrac{5}{4}-\dfrac{5}{5}+...+\dfrac{5}{10}-\dfrac{5}{11}\)
\(=\dfrac{5}{1}-\dfrac{5}{11}=\dfrac{50}{11}\)
=5(1/2+1/6+1/12+1/20+1/30+1/42+...+1/110)
=5(1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/10.11)
=5(1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+...+1/10-1/11)
=5(1-1/11)=5.10/11=50/11
1,a,\(\left|x+2\right|=x+3\Leftrightarrow\orbr{\begin{cases}x+2=x+3\\x+2=-x-3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-x=3-2\\x+x=-3-2\end{cases}\Leftrightarrow\orbr{\begin{cases}0=1\left(voly\right)\\x=\frac{-5}{2}\end{cases}}}\)
b, \(|x-2|=2-x\Leftrightarrow\orbr{\begin{cases}x-2=2-x\\x-2=x-2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x+x=2+2\\x-x=-2+2\end{cases}\Rightarrow x=2}\)
c,\(\left|2x-1\right|=3\Leftrightarrow\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=4\\2x=-2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}}\)
d,\(\left|x-12\right|=x\Leftrightarrow\orbr{\begin{cases}x-12=x\\x-12=-x\end{cases}\Leftrightarrow\orbr{\begin{cases}0=12\left(voly\right)\\x=6\end{cases}}}\)
(12x2 +12x4 - 12x6) : (2+4+6+....+18+20)
=12 x (2+4-6) : (2+4+6+...+18+20)
=12 x 0 : (2+4+6+...+18+20)
= 0
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{90}+\frac{1}{110}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{9.10}+\frac{1}{10.11}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)\(=\frac{1}{2}-\frac{1}{11}=\frac{9}{22}\)
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}+\frac{1}{110}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{9.10}+\frac{1}{10.11}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)
\(=\frac{1}{2}-\frac{1}{11}=\frac{9}{22}\)
Bài làm
a) 11 - 12 + 13 - 14 + 15 - 16 + 17 - 18 + 19 - 20
= ( 11 - 12 ) + ( 13 - 14 ) + ( 15 - 16 ) + ( 17 - 18 ) + ( 19 - 20 )
= ( -1 ) + ( -1 ) + ( -1 ) + ( -1 ) + ( -1 )
= -5
b) 101 - 102 - ( - 103 ) - 104 - ( - 105 ) - 106 - ( - 107 ) - 108 - ( - 109 ) - 110
= 101 - 102 + 103 - 104 + 105 - 106 + 107 - 108 + 109 - 110
= ( 101 - 102 ) + ( 103 - 104 ) + ( 105 - 106 ) + ( 107 - 108 ) + ( 109 - 110 )
= ( -1 ) + ( -1 ) + ( -1 ) + ( -1 ) + ( -1 )
= -5
# Học tốt #
a)11 - 12 + 13 - 14 + 15 - 16 + 17 - 18 + 19 - 20
= (11 - 12) + (13 - 14) + (15 - 16) + (17 - 18) + ( 19 - 20)
= (-1) + (-1) + (-1) + (-1) + (-1) = (-1) . 5 = - 5
b)101-102-(-103)-104-(-105)-106-(-107)-108-(-109)-110
= 101-102+103-104+105-106+107-108+109-110
= (101-102)+(103-104)+(105-106)+(107-108)+(109-110)
= (-1) + (-1) + (-1) + (-1) + (-1)
= (-1) . 5
= -5
2;6;12;20;30;42;56;72;90;110
moi so tu 2 den 6 la 4 don vi moi so tiep theo la cong them 2 don vi nhu 2+4=6;6+6=12....
nhu vay 5 so tiep theo cong them vao 110 la ra : so dau tien la: 22 ;24;26;28;30
vay 5 so do la 110 +22=132 ;132 +24 =156 ;156+26 =182;182+28=210 ;210+30=240
tong cac so :2+6+12+20+30+42+56+72+90+110+132+156+182+210+240=1328
ket qua la 1328
chuc ban hoc gioi nha
neu ban cam thay mh dung thi k cho mk nha cac ban !!!!!!! :))) ^^^
\(\dfrac{2}{1^2}.\dfrac{6}{2^2}.\dfrac{12}{3^2}.\dfrac{20}{4^2}.....\dfrac{110}{10^2}.x=-20\)
\(\dfrac{1.2}{1^2}.\dfrac{2.3}{2^2}.\dfrac{3.4}{3^2}.\dfrac{4.5}{4^2}.....\dfrac{10.11}{10^2}.x=-20\)
\(\dfrac{1.2.2.3.3.4.4.5.5.....10.10.11}{1.1.2.2.3.3.4.4.5.5.....10.10}.x=-20\)
\(11.x=-20\)
\(x=-20:11=-\dfrac{20}{11}\)
\(\dfrac{2}{1^2}\cdot\dfrac{6}{2^2}\cdot\dfrac{12}{3^2}\cdot\dfrac{20}{4^2}\cdot...\cdot\dfrac{110}{10^2}\cdot x=-20\)
\(\Leftrightarrow\dfrac{1\cdot2}{1\cdot1}\cdot\dfrac{2\cdot3}{2\cdot2}\cdot\dfrac{3\cdot4}{3\cdot3}\cdot\dfrac{4\cdot5}{4\cdot4}\cdot...\cdot\dfrac{10\cdot11}{10\cdot10}\cdot x=-20\)
\(\Leftrightarrow\dfrac{1\cdot2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot5\cdot...\cdot10\cdot11}{1\cdot1\cdot2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot...\cdot10\cdot10}\cdot x=-20\)
\(\Rightarrow11\cdot x=-20\)
\(\Rightarrow x=\dfrac{-20}{11}\)
Vậy \(x=\dfrac{-20}{11}\).