A=\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^4}+.....+\frac{1}{3^{104}}+\frac{1}{3^{105}}\)
Những câu hỏi liên quan
2frac{1}{315}.frac{1}{651}-frac{1}{105}.3frac{650}{651}-frac{4}{315.651}+frac{4}{105}left(2+frac{1}{315}right).frac{1}{651}-frac{1}{105}.left(3+frac{650}{651}right)-frac{4}{315.651}+frac{4}{105}2.frac{1}{651}+frac{1}{315}.frac{1}{651}-frac{1}{105}.3+frac{1}{105}.frac{650}{651}-frac{4}{315.651}+frac{4}{105}frac{2}{651}+frac{1}{315.651}-frac{3}{105}-frac{650}{105.651}-frac{4}{315.651}+frac{4}{105}frac{2}{615}+left(frac{1}{315.651}-frac{4}{315.651}right)+left(frac{-3}{105}+frac{4}{105}right)-frac{6...
Đọc tiếp
\(2\frac{1}{315}.\frac{1}{651}-\frac{1}{105}.3\frac{650}{651}-\frac{4}{315.651}+\frac{4}{105}\)
\(=\left(2+\frac{1}{315}\right).\frac{1}{651}-\frac{1}{105}.\left(3+\frac{650}{651}\right)-\frac{4}{315.651}+\frac{4}{105}\)
\(=2.\frac{1}{651}+\frac{1}{315}.\frac{1}{651}-\frac{1}{105}.3+\frac{1}{105}.\frac{650}{651}-\frac{4}{315.651}+\frac{4}{105}\)
\(=\frac{2}{651}+\frac{1}{315.651}-\frac{3}{105}-\frac{650}{105.651}-\frac{4}{315.651}+\frac{4}{105}\)
\(=\frac{2}{615}+\left(\frac{1}{315.651}-\frac{4}{315.651}\right)+\left(\frac{-3}{105}+\frac{4}{105}\right)-\frac{650}{105.651}\)
\(=\frac{2}{651}-\frac{3}{315.651}+\frac{1}{105}-\frac{650}{105.651}\)
\(=\left(\frac{2}{651}+\frac{1}{105}\right)-\frac{650}{105.651}-\frac{3}{315.651}\)
\(=\left(\frac{2.105}{105.651}+\frac{651}{105.651}\right)-\frac{650}{105.651}-\frac{3}{315.651}\)
\(=\frac{211}{105.651}-\frac{3}{315.651}\)
\(=\frac{1}{651}.\left(\frac{211}{105}-\frac{3}{315}\right)\)
\(=\frac{1}{651}.\left(\frac{633}{315}-\frac{3}{315}\right)\)
\(=\frac{1}{651}.2\)
\(=\frac{2}{651}\)
Tìm số hữu tỉ x biết:\(\frac{x-1}{105}+\frac{x-2}{104}-\frac{x-3}{103}=\frac{x-4}{102}+\frac{x-5}{101}-1\)
1 tháng 3 2017 lúc 23:23
Giá trị biếu thức \(A=\frac{1+2+3+...+103+104+105}{1-2+3-4+...+103-105+105}\)
Đặt tử số là B=1+2+3+....+105
Số các số hạng của B là
(105-1):1+1=105(số)
Tổng B là:
(105+1)x105:2=5565
Đặt mẫu số là C =1-2+3-4+...+103-104+105
C=(1-2)+(3-4)+...+(103-104)+105
C=-1+(-1)+...+(-1)(52 số hạng) + 105
C=-52 + 105
C=53
Vậy A=\(\dfrac{B}{C}\)=\(\dfrac{5565}{53}=105\)
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a)\(\left(\frac{2}{3}+\frac{4}{5}+\frac{12}{7}\right)+\left(2^7.5^6-\frac{25^3}{\left(0.125\right)3}\right).\left(0.0001\right)^2\)
b) \(2\frac{1}{315}.\frac{1}{651}-\frac{1}{105}.3\frac{650}{651}-\frac{4}{315.651}+\frac{4}{105}\)
So sánh \(\frac{1}{101^2}+\frac{1}{102^2}+\frac{1}{103^2}+\frac{1}{104^2}+\frac{1}{105^2}\) với \(\frac{1}{2^2×3×5^2×7}\)
an vo cai nay la vo tra loi
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Tính \(A=2\frac{1}{315}.\frac{1}{615}-\frac{1}{105}.3\frac{650}{651}-\frac{4}{315.651}+\frac{4}{105}\)
\(Tính:\)
\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{56}\)
\(B=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{11.13}\)
\(C=\frac{3}{4}+\frac{3}{28}+\frac{3}{70}+\frac{3}{130}+\frac{3}{208}+\frac{3}{304}\)
\(D=\frac{1}{2}+\frac{1}{14}+\frac{1}{35}+\frac{1}{65}+\frac{1}{104}+\frac{1}{152}\)
A = \(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{56}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{7.8}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{7}-\frac{1}{8}\)
\(=\frac{1}{2}-\frac{1}{8}=\frac{3}{8}\)
B = \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{11.13}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{11}-\frac{1}{13}\)
\(=1-\frac{1}{13}=\frac{12}{13}\)
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\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{56}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{7.8}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{8}\)
\(=\frac{1}{2}-\frac{1}{8}=\frac{3}{8}\)
\(B=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{11.13}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{11}-\frac{1}{13}\)
\(=1-\frac{1}{13}=\frac{12}{13}\)
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A=\(\frac{1}{6}\)+\(\frac{1}{12}\)+......+\(\frac{1}{56}\)
A=\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+...+\(\frac{1}{7.8}\)
A=\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+....+\(\frac{1}{7}\)-\(\frac{1}{8}\)
A=\(\frac{1}{2}\)-\(\frac{1}{8}\)=\(\frac{3}{8}\)
B=\(\frac{2}{1.3}\)+\(\frac{2}{3.5}\)+...+\(\frac{2}{11.13}\)
B=1-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{5}\)+....+\(\frac{1}{11}\)-\(\frac{1}{13}\)
B=1-\(\frac{1}{13}\)=\(\frac{12}{13}\)
C=\(\frac{3}{4}\)+\(\frac{3}{28}\)+....+\(\frac{3}{304}\)
C=\(\frac{3}{1.4}\)+\(\frac{3}{4.7}\)+..+\(\frac{3}{16.19}\)
Rồi bạn cũng tách ra như câu A và câu B
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Rút gọn bằng cách thay số bằng chữ:
1)\(2\frac{1}{317}.\frac{3}{111}-\frac{316}{317}.\frac{1}{111}-\frac{4}{317.111}\)
2)\(2\frac{1}{15}.\frac{3}{651}-\frac{1}{105}.3\frac{650}{651}-\frac{4}{315.651}+\frac{4}{105}\)
3)999994.999999.999992-999996.999991.999998
4)444443.444448.444441-444445.444440.444447
1) Đặt \(\frac{1}{317}=a;\frac{3}{111}=b\) thế vào mà làm thôi
mấy câu sau tương tự
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Tính \(A=2\frac{1}{315}.\frac{1}{651}-\frac{1}{105}.3\frac{650}{651}-\frac{4}{315.651}+\frac{4}{105}\)
Đặt \(a=\frac{1}{315}\), \(b=\frac{1}{651}\)ta có :
\(A=\left(2+a\right)\cdot b-3a\left(3+1-b\right)-4ab+12a\)
\(\Rightarrow A=2b+ab-12a+3ab-4ab+12a\)
\(\Rightarrow A=2b=\frac{2}{651}\)
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