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Lời giải:
$(1.2+2.3+3.4+...+2012.2013)-(2^2+3^2+...+2013^2)$
$=[(2-1).2+(3-1).3+(4-1).4+...+(2013-1).2013]-(2^2+3^2+...+2013^2)$
$=(2^2+3^2+4^2+...+2013^2)-(2+3+4+...+2013)-(2^2+3^2+...+2013^2)$
$=-(2+3+4+...+2013)$
$=1-(1+2+3+...+2013)$
$=1-2013.2014:2=1-2027091=-2027090$
(1.2+2.3+3.4+.....+2012.2013)-(22+32+42+......+20132)
= 1.2 + 2.3 + 3.4 +...+ 2012.2013 - 22 -32 - 42 -....-20132
=1.2 + 2.3 + 3.4 + ...+2012.2013 - 2.2 -3.3 - 4.4 -...- 2013.2013
=(1.2 - 2.2) + (2.3 - 3.3) + (3.4 - 4.4) + ...+(2012.2013 - 2013.2013)
=2.(1-2) + 3.(2-3) + 4.(3-4) +...+2013.(2012-2013)
=2.(-1) + 3.(-1) + 4.(-1) + ...+2013.(2012-2013)
= -2 - 3 - 4 -...- 2013
= -(2+3+4+...+2013)
= -[(2013+2).2012:2]
=-2027090
BÀI 1:
\(N=\frac{2}{2.5}+\frac{2}{5.8}+...+\frac{2}{17.20}\)
\(N=2.\left(\frac{1}{2.5}+\frac{1}{5.8}+...+\frac{1}{17.20}\right)\)
\(N=2.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\right)\)
\(N=2.\left(\frac{1}{2}-\frac{1}{20}\right)\)
\(N=2.\frac{9}{20}\)
\(N=\frac{9}{10}\)
BÀI 2:
\(C=1.2+2.3+3.4+...+99.100\)
\(\Rightarrow3B=1.2.3+2.3.3+3.4.3+...+99.100.3\)
\(3B=1.2\left(3-0\right)+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+99.100.\left(101-98\right)\)
\(3B=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100\)
\(3B=\left(1.2.3+2.3.4+3.4.5+...+99.100.101\right)-\left(1.2.3+2.3.4+...+98.99.100\right)\)
\(3B=99.100.101\)
\(3B=999900\)
\(\Rightarrow B=999900:3\)
\(B=333300\)
CHÚC BN HỌC TỐT!!!!
a. 4.52.5.25.2
=(4.25).(5.2).52
=100.10.52
=1000.52
=52000
b. 2.32 + 4.32
=32 . (2+4)
=9.6
=54
a)\(\frac{2}{3}+\frac{3}{4}+\frac{5}{6}\)
\(=\frac{8+9+10}{12}\)
\(=\frac{27}{12}=\frac{9}{4}\)
b)\(\frac{15}{8}-\frac{7}{12}+\frac{5}{6}\)
\(=\frac{45-14+20}{24}\)
\(=\frac{51}{24}=\frac{17}{8}\)
2)
a)\(\frac{2}{5}+\frac{7}{13}+\frac{3}{5}+\frac{1}{7}\)
\(=\frac{2}{5}+\frac{3}{5}+\frac{7}{13}+\frac{1}{7}\)
\(=1+\frac{7}{13}+\frac{1}{7}\)
\(=\frac{20}{13}+\frac{1}{7}\)
\(=\frac{153}{91}\)
Tí tớ trả lời tiếp
(1.2 + 2.3 + 3.4 + ... + 2012.2013) - (22 + 32 + 42 + 52 + ... + 20132)
= [(2 - 1).2 + (3 - 1).3 + (4 - 1).4 + ... + (2013 - 1).2013] - (22 + 32 + 42 + 52 + ... + 20132)
= (22 - 2 + 32 - 3 + 42 - 4 + ... + 20132 - 2013) - (22 + 32 + 42 + 52 + ... + 20132)
= 22 - 2 + 32 - 3 + 42 - 4 + ... + 20132 - 2013 - 22 - 32 - 42 - 52 - ... - 20132
= (22 - 22) + (32 - 32) + (42 - 42) + ... + (20132 - 20132) - (2 + 3 + 4 + ... + 2013)
= 0 - (2 + 3 + 4 + ... + 2013)
= 0 - (1 + 2 + 3 + 4 + ... + 2013) + 1
= 0 - \(\dfrac{2013.\left(2013+1\right)}{2}\) + 1
= 0 - 2027091 + 1
= (-2027091) + 1
= -2027090