a/ \(A=\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{99.100}\)
b/ \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{20}}< 1\)
c/ \(A=\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{97.99}\)
d/ \(A=\dfrac{2015}{2016}+\dfrac{2016}{2017}+\dfrac{2017}{2018}+\dfrac{2018}{2015}>4\)
\(\dfrac{2}{3.5}+\dfrac{2}{5.7}\dfrac{2}{7.9}+.........+\dfrac{2}{99.101}\)
\(P=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}+\dfrac{2}{13.15}\)
Đặt A=\(\dfrac{2}{3.5}.\dfrac{2}{7.9}.....\dfrac{2}{99.101}\)
A=\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
A=\(\dfrac{1}{3}-\dfrac{1}{101}=\dfrac{98}{303}\)
Ta có: \(P=\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{15}\)
\(=\dfrac{1}{3}-\dfrac{1}{15}\)
\(=\dfrac{4}{15}\)
Câu 1:
\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{99.101}\)
= \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
= \(\dfrac{1}{3}-\dfrac{1}{101}\)
= \(\dfrac{98}{303}\)
Câu 2 làm tương tự ở câu 1 nhé
CMR: nếu \(\dfrac{a+104}{a-104}=\dfrac{b+105}{b-105}\) thì \(\dfrac{a}{104}=\dfrac{b}{105}\) (với a,b \(\in\) Z và a \(a\ne104;b\ne105\))
Ta có:
\(\dfrac{a+104}{a-104}=\dfrac{a-104+208}{a-104}=1+\dfrac{208}{a-104}=\dfrac{b+105}{b-105}=\dfrac{b-105+210}{b-105}=1+\dfrac{210}{b-105}\)
\(\Rightarrow\dfrac{208}{a-104}=\dfrac{210}{b-105}\)
\(\Rightarrow\dfrac{a-104}{b-105}=\dfrac{208}{210}=\dfrac{104}{105}\)
\(\Rightarrow\dfrac{a-104}{104}=\dfrac{b-105}{105}\)
\(\Rightarrow\dfrac{a}{104}-1=\dfrac{b}{105}-1\)
\(\Rightarrow\dfrac{a}{104}=\dfrac{b}{105}\left(đpcm\right)\)
a)chứng minh rằng :\(\dfrac{1}{3^2}\)+\(\dfrac{1}{4^2}\)+\(\dfrac{1}{5^2}\)+\(\dfrac{1}{6^2}\)........+\(\dfrac{1}{100^2}< \dfrac{1}{2}\)
b)tính nhanh tổng S với S= \(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+......+\dfrac{1}{61.63}\)
các cao nhân gải giúp với ạ !!! iem đang cần gấp
So sánh A và B :
A = \(\dfrac{1}{101^2}+\dfrac{1}{102^2}+\dfrac{1}{103^2}+\dfrac{1}{104^2}+\dfrac{1}{105^2}\)
B = \(\dfrac{1}{2^2.3.5^2.7}\)
Ta có: \(A=\dfrac{1}{101^2}+\dfrac{1}{102^2}+\dfrac{1}{103^2}+\dfrac{1}{104^2}+\dfrac{1}{105^2}\)
\(A>\dfrac{1}{100.101}+\dfrac{1}{101.102}+\dfrac{1}{102.103}+\dfrac{1}{103.104}+\dfrac{1}{104.105}\)\(A>\dfrac{1}{100}-\dfrac{1}{101}+\dfrac{1}{101}-\dfrac{1}{102}+\dfrac{1}{102}-\dfrac{1}{103}+\dfrac{1}{103}-\dfrac{1}{104}+\dfrac{1}{104}-\dfrac{1}{105}\)\(A>\dfrac{1}{100}-\dfrac{1}{105}\)
\(A>\dfrac{1}{2100}\)
Mà \(B=\dfrac{1}{2^2.3.5^2.7}\)=\(\dfrac{1}{2100}\)
=> \(A>B\)
Vậy \(A>B\)
\(\text{Tìm x, biết:}\)
\(a\)) \(x-\dfrac{2}{3.5}-\dfrac{2}{5.7}-\dfrac{2}{7.9}-\dfrac{2}{9.11}-\dfrac{2}{11.13}-\dfrac{2}{13.15}=\dfrac{2}{5}\)
\(b\)) \(\dfrac{1}{2.3}.x+\dfrac{1}{3.4}.x+\dfrac{1}{4.5}.x+...+\dfrac{1}{49.50}.x=1\)
\(c\)) \(x-\dfrac{20}{11.3}-\dfrac{20}{13.15}-...-\dfrac{53}{55}=\dfrac{3}{11}\)
\(d\)) \(x+\dfrac{4}{5.9}+\dfrac{4}{9.13}+...+\dfrac{4}{41.45}=\dfrac{-37}{45}\)
\(e\)) \(\left(\dfrac{11}{12}.\dfrac{11}{2.23}.\dfrac{11}{23.34}...\dfrac{11}{89.100}\right).x=\dfrac{5}{3}\)
\(f\)) \(\left(\dfrac{2}{11.13}.\dfrac{2}{13.15}.\dfrac{2}{15.17}...\dfrac{2}{19.21}\right)-x+4+\dfrac{221}{231}=\dfrac{7}{3}\)
d) Ta có: \(x+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{41\cdot45}=\dfrac{-37}{45}\)
\(\Leftrightarrow x+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}=\dfrac{-37}{45}\)
\(\Leftrightarrow x+\dfrac{1}{5}-\dfrac{1}{45}=\dfrac{-37}{45}\)
\(\Leftrightarrow x=\dfrac{-37}{45}+\dfrac{1}{45}-\dfrac{1}{5}=\dfrac{-36}{45}-\dfrac{1}{5}=\dfrac{-4}{5}-\dfrac{1}{5}=-1\)
Vậy: x=-1
Tính giá trị của biểu thức sau bằng cách thay số bởi chữ một cách hợp lí:
A = 2\(\dfrac{1}{315}\) . \(\dfrac{1}{651}\)- \(\dfrac{1}{105}\) . 3\(\dfrac{650}{651}\) - \(\dfrac{4}{315.651}\)+ \(\dfrac{4}{105}\)
Tính :
a) A = \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+....+\dfrac{2}{37.39}\)
b) B = \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+.....+\dfrac{3}{73.76}\)
a, \(A=\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{37.39}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{37}-\dfrac{1}{39}\)
\(=\dfrac{1}{3}-\dfrac{1}{39}\)
\(=\dfrac{12}{39}\)
Vậy \(A=\dfrac{12}{39}\)
b,\(B=\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{73.76}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{73}-\dfrac{1}{76}\)
\(=1-\dfrac{1}{76}\)
\(=\dfrac{75}{76}\)
Vậy \(B=\dfrac{75}{76}\)
a) Ta có :
\(A=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+....................+\dfrac{2}{37.39}\)
\(A=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...................+\dfrac{1}{37}-\dfrac{1}{39}\)
\(A=\dfrac{1}{3}-\dfrac{1}{39}=\dfrac{4}{13}\)
b) Ta có :
\(B=\dfrac{3}{1.4}+\dfrac{3}{4.7}+..................+\dfrac{3}{73.76}\)
\(B=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+..................+\dfrac{1}{73}-\dfrac{1}{76}\)
\(B=1-\dfrac{1}{76}=\dfrac{75}{76}\)
~ Học tốt ~
A=\(\dfrac{4}{3.5}-\dfrac{6}{5.7}+\dfrac{8}{7.9}-\dfrac{10}{9.11}+\dfrac{12}{11.13}-...-\dfrac{100}{99.100}\)
Tính giá trị của A
Tính nhanh tổng sau
A= \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{59.61}\)
\(A=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{59.61}\)
\(A=\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+\dfrac{9-7}{7.9}+...+\dfrac{61-59}{59.61}\)
\(A=\dfrac{5}{3.5}-\dfrac{3}{3.5}+\dfrac{7}{5.7}-\dfrac{5}{5.7}+\dfrac{9}{7.9}-\dfrac{7}{7.9}+...+\dfrac{61}{59.61}-\dfrac{59}{59.61}\)
\(A=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\)
\(A=\dfrac{1}{3}-\dfrac{1}{61}=\dfrac{61}{183}-\dfrac{3}{183}=\dfrac{58}{183}\)
\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{59.61}\)
= \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\)
= \(\dfrac{1}{3}-\dfrac{1}{61}\)
= \(\dfrac{58}{183}\)
A= \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{59.61}\)
A= \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\)
A= \(\dfrac{1}{3}-\dfrac{1}{61}\)
A= \(\dfrac{58}{183}\)
