A = \(\frac{1+2+3+...+105}{1-2+3-...+105}\) là
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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
\(2\frac{1}{315}.\frac{1}{651}-\frac{1}{105}.3\frac{650}{651}-\frac{4}{315.651}+\frac{4}{105}\)
\(=\left(2+\frac{1}{315}\right).\frac{1}{651}-\frac{1}{105}.\left(3+\frac{650}{651}\right)-\frac{4}{315.651}+\frac{4}{105}\)
\(=2.\frac{1}{651}+\frac{1}{315}.\frac{1}{651}-\frac{1}{105}.3+\frac{1}{105}.\frac{650}{651}-\frac{4}{315.651}+\frac{4}{105}\)
\(=\frac{2}{651}+\frac{1}{315.651}-\frac{3}{105}-\frac{650}{105.651}-\frac{4}{315.651}+\frac{4}{105}\)
\(=\frac{2}{615}+\left(\frac{1}{315.651}-\frac{4}{315.651}\right)+\left(\frac{-3}{105}+\frac{4}{105}\right)-\frac{650}{105.651}\)
\(=\frac{2}{651}-\frac{3}{315.651}+\frac{1}{105}-\frac{650}{105.651}\)
\(=\left(\frac{2}{651}+\frac{1}{105}\right)-\frac{650}{105.651}-\frac{3}{315.651}\)
\(=\left(\frac{2.105}{105.651}+\frac{651}{105.651}\right)-\frac{650}{105.651}-\frac{3}{315.651}\)
\(=\frac{211}{105.651}-\frac{3}{315.651}\)
\(=\frac{1}{651}.\left(\frac{211}{105}-\frac{3}{315}\right)\)
\(=\frac{1}{651}.\left(\frac{633}{315}-\frac{3}{315}\right)\)
\(=\frac{1}{651}.2\)
\(=\frac{2}{651}\)
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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Tính \(A=2\frac{1}{315}.\frac{1}{615}-\frac{1}{105}.3\frac{650}{651}-\frac{4}{315.651}+\frac{4}{105}\)
Tính \(A=2\frac{1}{315}.\frac{1}{651}-\frac{1}{105}.3\frac{650}{651}-\frac{4}{315.651}+\frac{4}{105}\)
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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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) \(2\frac{1}{315}.\frac{1}{651}-\frac{1}{105}.3\frac{650}{651}-\frac{4}{315.651}+\frac{4}{105}\)
Bài 1: Tính giá trị biểu thức:
\(A=2\frac{1}{315}.\frac{1}{651}-\frac{1}{105}.3\frac{650}{651}-\frac{4}{315.651}+\frac{4}{105}\)
\(A=2\frac{1}{315}.\frac{1}{651}-\frac{1}{105}.3\frac{650}{651}-\frac{4}{315.651}+\frac{4}{105}\)
\(A=\left(2+\frac{1}{315}\right).\frac{1}{651}-3.\frac{1}{315}.\left(4-\frac{1}{651}\right)-4.\frac{1}{315}.\frac{1}{651}+12.\frac{1}{315}\)
Đặt \(\frac{1}{315}=x;\frac{1}{651}=y\),khi đó:
\(A=\left(2+x\right)y-3x\left(4-y\right)-4xy+12y\)
\(A=2y+xy-12x+3xy-4xy+12y=2y=2.\frac{1}{651}=\frac{2}{651}\)
Vậy A=2/651
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A=\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^4}+.....+\frac{1}{3^{104}}+\frac{1}{3^{105}}\)
Tính nhanh
\(2\frac{1}{315}.\frac{1}{651}-\frac{1}{105}.3\frac{650}{651}-\frac{4}{315.651}+\frac{4}{105}\)
\(\left(2+\frac{1}{315}\right).\frac{1}{651}-\frac{1}{105}\left(3+1-\frac{1}{651}\right)-\frac{4}{315.651}+\frac{4}{105}\)
\(=\frac{2}{651}+\frac{1}{315.651}-\frac{4}{105}+\frac{1}{105.651}-\frac{4}{315.651}+\frac{4}{105}\)
\(=\frac{2}{651}-\frac{3}{315.651}+\frac{1}{105.651}\)
\(=\frac{2}{651}-\frac{1}{105.651}+\frac{1}{105.651}=\frac{2}{651}\)
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\(2\frac{1}{315}.\frac{1}{651}-\frac{1}{105}.3\frac{650}{651}-\frac{4}{315.651}+\frac{4}{105}\)