1.2.3+2.3.4+...+8.9.10 vừa nãy mình nhầm
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D= 1.2.3+2.3.4+....+8.9.10
\(D=1.2.3+2.3.4+...+8.9.10\)
\(4D=1.2.3.4+2.3.4.\left(5-1\right)+...+8.9.10.\left(11-7\right)\)
\(4D=1.2.3.4+2.3.4.5-1.2.3.4+...+8.9.10.11-7.8.9.10\)
\(4D=8.9.10.11\)
\(D=\frac{8.9.10.11}{4}=\frac{7920}{4}=1980\)
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1.2.3+2.3.4+3.4.5+...+8.9.10
Đặt A=1.2.3+2.3.4+...+8.9.10
\(4A=1.2.3.4+2.3.4.4+...+8.9.10.4\)
\(4A=1.2.3.\left(4-0\right)+2.3.4.\left(5-1\right)+...+8.9.10.\left(11-7\right)\)
\(4A=1.2.3.4-0.1.2.3+2.3.4.5-1.2.3.4+...+8.9.10.11-7.8.9.10\)
\(4A=\left(1.2.3.4+2.3.4.5+...+8.9.10.11\right)-\left(0.1.2.3+1.2.3.4+...+7.8.9.10\right)\)
\(4A=8.9.10.11-0.1.2.3\)
\(4A=8.9.10.11\)
\(A=2.9.10.11\)
\(\Rightarrow A=1980\)
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B=1.3+3.5+5.7+...+97.98
C=1.2.3+2.3.4+3.4.5+4.5.6+5.6.7+7.8.9+8.9.10
D=1.2.3+2.3.4+...+99.100.101
B=1/1.2.3+1/2.3.4+...+1/8.9.10
\(B=\dfrac{1}{1.2.3}+\dfrac{1}{3.4.5}+...+\dfrac{1}{8.9.10}\)
\(B=2.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)\)
\(B=2.\left(1-\dfrac{1}{10}\right)\)
\(B=2.\dfrac{9}{10}\)
\(B=\dfrac{9}{5}\)
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anh ơi , đại học rồi mà ko giải đc bài này ạ?
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Xem thêm câu trả lời
1/1.2.3+1/2.3.4+...+1/8.9.10
\(=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}...+\frac{2}{8.9.10}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{90}\right)=\frac{11}{45}\)
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\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{90}\right)\)
\(=\frac{1}{2}.\frac{22}{45}\)
\(=\frac{11}{45}\)
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1/1.2.3+1/2.3.4+...+1/8.9.10 =?
Cho A=1.2.3+2.3.4+....+8.9.10 Tính A
<=>4S=1.2.3.4 + 2.3.4.4+3.4.5.4+.....+8.9.10.4
<=>4S =1.2.3.4 + 2.3.4.(5-1) + 3.4.5.(6-2)+......+8.9.10.(11-7)
<=>4S=1.2.3.4 + 2.3.4.5 -1.2.3.4+3.4.5.6- 2.3.4.5+......+8.9.10.11 - 7.8.9.10
<=> 4S=8.9.10.11
<=>S=1980
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S=1.2.3+2.3.4+..........+8.9.10 TÌM 4S
10 tháng 8 2021 lúc 8:31
(\(\dfrac{1}{1.2.3}\)+\(\dfrac{1}{2.3.4}\)+....\(\dfrac{1}{8.9.10}\)) . x = \(\dfrac{22}{45}\)
Tìm x nha mọi người mình đang cần gấp lắm ạ
\(\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{8.9.10}\right).x=\dfrac{22}{45}\)
=> \(\dfrac{1}{2}.\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{8.9.10}\right).x=\dfrac{22}{45}\)
=> \(\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-...-\dfrac{1}{9.10}\right).x=\dfrac{22}{45}\)
=> \(\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{9.10}\right).x=\dfrac{22}{45}\)
=> \(\dfrac{1}{2}.\dfrac{22}{45}.x=\dfrac{22}{45}\)
=> \(\dfrac{1}{2}.x=1\)
=> \(x=2\)
Vậy x = 2
Chúc bạn học tốt !!!
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