Tính :
1.2010+2.2009+3.2008+.....+2010.1/(1+2+3+...+2010)+(1+2+3+....+2009)+....+(1+2)+1
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Tính : 1.2010+2.2009+3.2008+.........+2010.1/(1+2+3 +.....+2010)+(1+2+3+........+2009)+.......(1+2)+1
1.2010+2.2009+3.2008+...+2010.1 / [(1+2+3+...+2010)+(1+2+3+...+2009)+...+(1+2)+1]
Để tui nhận xét, đầu tiên là đề bài đã đủ dễ thấy sai rùi vì đây là tính chia theo ý bạn nhưng người ta sẽ làm tưởng là cộng riêng ra, mặc khác bạn lại tụ giải thiếu dấu ngoặc của biểu thức chia là 1, cộng các số hạng là số chia mà ko có số số hạng là bao nhiu là 2 làm người ta phân vân bao nhiu số hạng.
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1.2010+2.2009+3.2008+...+2010.1/(1+2+3+...+2010)+(1+2+3+…+2009)+….+(1+2)+1
=1.2010+2.2009+3.2008+...+2010.1/(1+1+...+1)+(2+2+...+2)+(3+3+...+3)+...+(2009+2009)+2010
=1.2010+2.2009+3.2008+...+2010.1/1.2010+2.2009+3.2008+...+2010.1
=1
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11 tháng 2 2022 lúc 21:26
Tính:
\(\dfrac{1.2010+2.2009+3.2008+...+2010.1}{\left(1+2+3+...+2010\right)+\left(1+2+3+...+2009\right)+...\left(1+2\right)+1}\)
\(=\dfrac{1\cdot2010+2\cdot2009+3\cdot2008+...+2010\cdot1}{\left(1+1+...+1\right)+\left(2+2+...+2\right)+.....+\left(2009+2009\right)+2010}\\ =\dfrac{1\cdot2010+2\cdot2009+3\cdot2008+...+2010\cdot1}{1\cdot2010+2\cdot2009+3\cdot2008+...+2010\cdot1}\\ =1\)
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1.2010 + 2.2009 + 3.2008 + ... + 2010.1
( 1 + 2 + 3 + ... + 2010 ) + ( 1+ 2 + 3 + ... + 2009 ) + ... + ( 1 + 2 ) +1
Giúp mik nha bài này khó quá!
A=1.2010+2.2009+3.2008+...........+2010.1
Tính
1.2010 + 2.2009 + 3.2008 + ... + 2010.1
( 1 + 2 + 3 + ... + 2010 ) + ( 1+ 2 + 3 + ... + 2009 ) + ... + ( 1 + 2 ) +1
Giúp mik nha bài này khó quá, mik ko lm đc là bị phạt đó! :)
Xem thêm câu trả lời
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Đọc tiếp
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CMR: A= 3/1.2^2+5/2^2.3^2+...+4019/2010^2.2009^2<1
a) 2010/1+2009/2+2008/3+ ... +1/2010+2010 : 1+1/2+1/3+ ... +1/2010=
b) 1/2011+1/2010+1/2009+ ... +1/3+1/2 : 2010/1+2009/2+2008/3+ ... +1/2010=