\(\frac{\frac{4000}{1}+\frac{3999}{2}+\frac{3998}{3}+...+\frac{1}{4000}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{4001}}\)=?
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28 tháng 2 2016 lúc 18:03
Rút gọn
\(B=\frac{\frac{4000}{1}+\frac{3999}{2}+\frac{3998}{3}+...+\frac{1}{4000}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{4001}}\)
\(y=\frac{\frac{4000}{1}+\frac{3999}{2}+\frac{3998}{3}+...+\frac{1}{4000}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{4001}}=?\)
Đặt A=\(\frac{4000}{1}+\frac{3999}{2}+\frac{3998}{3}+........+\frac{1}{4000}\)
A=\(1+\left(1+\frac{3999}{2}\right)+\left(1+\frac{3998}{3}\right)+........+\left(1+\frac{1}{4000}\right)\)
A=\(\frac{4001}{4001}+\frac{4001}{2}+\frac{4001}{3}+...........+\frac{4001}{4000}\)
A=\(4001.\left(\frac{1}{2}+\frac{1}{3}+........+\frac{1}{4000}+\frac{1}{4001}\right)\)
=>\(y=\frac{4001.\left(\frac{1}{2}+\frac{1}{3}+........+\frac{1}{4001}\right)}{\frac{1}{2}+\frac{1}{3}+.........+\frac{1}{4001}}\)
=>\(y=4001\)
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\(C=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{4000}}{\frac{3999}{1}+\frac{3998}{2}+\frac{3997}{3}+...+\frac{1}{3999}}\) = ?
\(C=\frac{T}{M}\)
\(M=\left(1+\frac{3998}{2}\right)+\left(1+\frac{3997}{3}\right)+.....+\left(1+\frac{1}{3999}\right)+\frac{4000}{4000}\)
\(=\frac{4000}{2}+\frac{4000}{3}+......+\frac{4000}{3999}+\frac{4000}{4000}=4000.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{4000}\right)\)
\(=4000.T\)
\(C=\frac{T}{M}=\frac{T}{4000T}=\frac{1}{4000}\)
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Tính nhanh:\(\frac{1+3+5+...+4001}{2+4+6+...+4000}\)
Rút gọn: 4000/1+3999/2+3998/3+...+1/4000 / 1/2+1/3+1/4+...+1/4001
A=[(3999/2+1)+(3998/3+1)+...+(1/4000+1)+1]/(1/2+1/3+...+1/4001)
A=(4001/2+4001/3+...+4001/4001)/(1/2+1/3+...+1/4001)
A=[4001(1/2+1/3+...+1/4001)]/(1/2+1/3+...+1/4001)
A=4001
Vậy A=4001
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Tìm x, biết:
\(\left(\frac{1999}{2}+\frac{1998}{3}+\frac{1997}{4}+.......+\frac{1}{2000}+4000\right)x=1+\frac{1}{2}+\frac{1}{3}\)\(\frac{1}{3}\)
Ta có:(1+1999/2)+(1+1998/3)+...(2/1999)(có 1998 tổng<=>1998 số 1)+(2000 - 1998)+400
= 2001/2+2001/3+...+2001/1999+402
=2001.(1/2+1/3+...+1/1999)+402(1)
Thay (1) vào biểu thức trên và tính(tự tính nha!,tk cho mk!!!)
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1) Tính A/B biết:
A = 4000/1+3999/2+3998/3+...+1/4000
B = 1/2+1/3+1/4+...+1/4001
2) So sánh
A=(2014/2015)+(2015/2014) ; B=666665/333333
! ) A = (3999 /2 +1 ) + ( 3998/ 3 + 1 ) + ( 3997 / 4 + 1 ) +...+ ( 1/ 4000 + 1 ) + 1
(Ta lấy 4000/1 = 4000 rải đều 1, 1 ,1 cho 3999 phân số và dư lại 1 = 4001/4001 )
= 4001 /2 + 4001 / 3 + 4001 /4 + ...+ 4001 /4000 + 4001 / 4001
= 4001 ( 1/2 + 1/3 + 1/4 +..+ 1/ 4001 ) vay A: B = 4001
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Cho Afrac{frac{1}{2001}+frac{1}{2002}+frac{1}{2003}+...+frac{1}{4000}}{frac{1}{1.2}+frac{1}{3.4}+frac{1}{5.6}+...+frac{1}{3999.4000}}. Bfrac{left(17+1right).left(frac{17}{2}+1right).left(frac{17}{3}+1right).....left(frac{17}{19}+1right)}{left(1+frac{19}{17}right).left(1+frac{19}{16}right).left(1+frac{19}{15}right).....left(1+19right)}So sánh A-Bvới 0.LÀM ĐÚNG ĐƯỢC CHO TICK
Đọc tiếp
Cho \(A=\frac{\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}+...+\frac{1}{4000}}{\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{3999.4000}}\).
\(B=\frac{\left(17+1\right).\left(\frac{17}{2}+1\right).\left(\frac{17}{3}+1\right).....\left(\frac{17}{19}+1\right)}{\left(1+\frac{19}{17}\right).\left(1+\frac{19}{16}\right).\left(1+\frac{19}{15}\right).....\left(1+19\right)}\)
So sánh \(A-B\)với \(0\).
LÀM ĐÚNG ĐƯỢC CHO TICK
Ta có:
\(A=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{3999.4000}}\)
\(=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{3999}-\frac{1}{4000}}\)
\(=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\left(1+\frac{1}{3}+...+\frac{1}{3999}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{4000}\right)}\)
\(=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{3999}+\frac{1}{4000}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{4000}\right)}\)
\(=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{3999}+\frac{1}{4000}\right)-\left(1+\frac{1}{2}+...+\frac{1}{2000}\right)}\)
\(=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}=1\)
Ta lại có:
\(B=\frac{\left(17+1\right)\left(\frac{17}{2}+1\right)...\left(\frac{17}{19}+1\right)}{\left(1+\frac{19}{17}\right)\left(1+\frac{19}{16}\right)...\left(1+19\right)}\)
\(=\frac{\frac{18}{1}.\frac{19}{2}.\frac{20}{3}...\frac{36}{19}}{\frac{36}{17}.\frac{35}{16}.\frac{34}{15}...\frac{20}{1}}\)
\(=\frac{1.2.3...36}{1.2.3...36}=1\)
Từ đây ta suy ra được
\(A-B=1-1=0\)
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xin lỗi, mk mới học lớp 5 nên ko giúp đc gì, mong cậu thông cảm, chúc cậu học giỏi nha
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