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\(A=1.2+2.3+3.4+4.5+...+59.60\)
\(\Rightarrow3A=1.2.3+2.3.\left(4-1\right)+...+59.60.\left(61-58\right)\)
\(\Rightarrow3A=1.2.3+2.3.4-1.2.3+...+59.60.61-58.59.60\)
\(\Rightarrow3A=59.60.61\)
\(\Rightarrow A=\frac{59.60.61}{3}\)
\(B=1+2+4+5+7+8+10+...+119+121+122\)
Ta có :
\(A=1+2+3+...+121+122\)
\(A=\left[\left(122-1\right):1+1\right]\left(1+122\right):2\)
\(A=122.\left(123\right):2=7503\)
Ta lại có :
\(C=3+6+9+...+120\)
\(C=\left[\left(120-3\right):3+1\right]\left(3+120\right):2\)
\(C=40.123:2=2460\)
Ta thấy : \(B=A-C\)
\(B=7503-2460=5043\)
Phần A sai nha phải là: 10 + 13 + ... + 79 + 82
đáp án:
SSH: A = ( 82 - 10 ) : 3 + 1 = 25
Tổng: ( 10 + 82 ) . 25 : 2 = 1150
a) A = 10+13 + ...+79 + 81
A = ( 79+10) x 24 : 2 + 81
A = 89 x 24 : 2 +81
A = 1068+ 81
A= 1149
b) chỗ " 1/12.20" phải là 1/12.13 chứ !
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{12.13}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{12}-\frac{1}{13}\)
\(=1-\frac{1}{13}\)
\(=\frac{12}{13}\)
c) \(C=\frac{3}{7}\times\frac{4}{13}+\frac{3}{7}+5\frac{4}{7}\)
\(C=\frac{3}{7}\times\left(\frac{4}{13}+1\right)+\frac{39}{7}\)
\(C=\frac{3}{7}\times\frac{17}{13}+\frac{39}{7}\)
\(C=\frac{51}{91}+\frac{39}{7}\)
\(C=\frac{558}{91}\)
c; 17\(\dfrac{2}{31}\) - (\(\dfrac{15}{17}\) + 6\(\dfrac{2}{31}\))
= 17 + \(\dfrac{2}{31}\) - \(\dfrac{15}{17}\) - 6 - \(\dfrac{2}{31}\)
= (17 - 6) - \(\dfrac{15}{17}\) + (\(\dfrac{2}{31}\) - \(\dfrac{2}{31}\))
= 11 - \(\dfrac{15}{17}\)+ 0
= \(\dfrac{172}{17}\)
b; 130\(\dfrac{25}{28}\) + 120\(\dfrac{17}{35}\)
= 130 + \(\dfrac{25}{28}\) + 120 + \(\dfrac{17}{35}\)
= (130 + 120) + (\(\dfrac{25}{28}\) + \(\dfrac{17}{35}\))
= 250 + (\(\dfrac{125}{140}\) + \(\dfrac{68}{140}\))
= 250 + \(\dfrac{193}{140}\)
= 250\(\dfrac{193}{140}\)
A = 1 + 2 + 3 + ... + 2018 (có 2018 số )
= (2018 + 1) . 2018 : 2 = 2037171
B = 1 + 3 + 5 + ... + 2017(có 1009 số )
= (2017 + 1) . 1009 : 2 = 1018081
C = 2 + 4 + 6 + ... + 2018 (Có 1009 số )
= (2018 + 2) x 1009 : 2 = 1019090
D = 72 . 153 + 27.153 + 153
= (72 + 27 + 1) . 153
= 100 . 153 = 15300
Ta có: A = 1.2 + 2.3 + 3.4 + 4.5 +.....+ 98.99
=> 3A = 1.2.(3 - 0) + 2.3.(4 - 1) + ..... +98.99.(100 - 97)
=> 3A = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ..... + 98.99.100
=> 3A = 98.99.100
=> A = 98.99.100 / 3
=> A = 323400
\(\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{19\cdot20}\right)\div x=\frac{9}{10}\)
\(\Leftrightarrow\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}\right)\div x=\frac{9}{10}\)
\(\Leftrightarrow\left(\frac{1}{1}-\frac{1}{20}\right)\div x=\frac{9}{10}\)
\(\Leftrightarrow\frac{19}{20}\div x=\frac{9}{10}\)
\(\Leftrightarrow x=\frac{19}{18}\)
Sửa đề : \(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{19.20}\right):x=\frac{9}{10}\)
\(\Leftrightarrow VT=\frac{9}{10}x\)
\(\Leftrightarrow\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}\right)=\frac{9}{10}x\)
\(\Leftrightarrow\left(1-\frac{1}{20}\right)=\frac{9}{10}x\Leftrightarrow\frac{19}{20}=\frac{9}{10}x\)
\(\Leftrightarrow\frac{19}{20}=\frac{18x}{20}\) Khử mẫu ta đc : \(\Leftrightarrow18x=19\Leftrightarrow x=\frac{19}{18}\)
c: C=1*2+2*3+3*4+...+58*59+59*60
=>3*C=1*2*3+2*3*(4-1)+3*4*(5-2)+...+58*59(60-57)+59*60(61-58)
=>3*C=1*2*3+2*3*4-1*2*3+...+58*59*60-58*59*57+59*60*61-58*59*60
=>3*C=59*60*61
=>C=59*20*61=71980