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\(=4\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2014}-\dfrac{1}{2015}\right)=4\cdot\dfrac{2014}{2015}=\dfrac{8056}{2015}\)
\(\dfrac{4}{1.2}+\dfrac{4}{2.3}+...+\dfrac{4}{2014.2015}\\ =4\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2014.2015}\right)\\ =4\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2014}-\dfrac{1}{2015}\right)\\ =4\left(1-\dfrac{1}{2015}\right)\\ =4.\dfrac{2014}{2015}\\ =\dfrac{8056}{2015}\)
a) A = \(\dfrac{1^2}{1.2}.\dfrac{2^2}{2.3}.\dfrac{3^2}{3.4}.\dfrac{4^2}{4.5}\)
A = \(\dfrac{1.1}{1.2}.\dfrac{2.2}{2.3}.\dfrac{3.3}{3.4}.\dfrac{4.4}{4.5}\)
A = \(\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}\)= \(\dfrac{1}{5}\)
b) B = \(\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}.\dfrac{5^2}{4.6}\)
B = \(\dfrac{2.3.4.5}{1.2.3.4}.\dfrac{2.3.4.5}{3.4.5.6}\)= \(\dfrac{5}{3}\)
\(A=\frac{4}{1.2}+\frac{4}{2.3}+...+\frac{4}{2014.2015}\)
\(A=4\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2014.2015}\right)\)
\(A=4\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2014}-\frac{1}{2015}\right)\)
\(A=4\left(\frac{1}{1}-\frac{1}{2015}\right)\)
\(A=4\left(\frac{2015-1}{2015}\right)\)
\(A=4.\frac{2014}{2015}\)
... BẠN TỰ LÀM NỐT NHÉ!
A = \(\dfrac{9}{1.2}\)+ \(\dfrac{9}{2.3}\)+\(\dfrac{9}{3.4}\)+......+\(\dfrac{99}{99.100}\)
A = 9( \(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+.......+\(\dfrac{1}{99.100}\))
A = 9( 1-\(\dfrac{1}{2}\)+\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+........+\(\dfrac{1}{99}\)-\(\dfrac{1}{100}\))
A = 9 ( 1 - \(\dfrac{1}{100}\))
A = 9 . \(\dfrac{99}{100}\)
A = \(\dfrac{891}{100}\)
\(A=\dfrac{9}{1\cdot2}+\dfrac{9}{2\cdot3}+\dfrac{9}{3\cdot4}+...+\dfrac{9}{98\cdot99}+\dfrac{9}{99\cdot100}\)
\(=9\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{98\cdot99}+\dfrac{1}{99\cdot100}\right)\)
\(=9\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=9\left(1-\dfrac{1}{100}\right)\)
\(=9\left(\dfrac{100}{100}-\dfrac{1}{100}\right)\)
\(=9\cdot\dfrac{99}{100}\)
\(=\dfrac{891}{100}\)
1)Tính
a)\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+..........+\dfrac{1}{9.10}\)
=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.....+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=1-\dfrac{1}{10}\)
\(=\dfrac{9}{10}\)
b)\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.........+\dfrac{1}{99.100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+..............+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}\)
\(=\dfrac{99}{100}\)
2) tìm x
\(a\)) \(\dfrac{2}{5}+\dfrac{4}{5}x-\dfrac{7}{5}\)\(=\dfrac{9}{5}\)
\(\dfrac{4}{5}x+\dfrac{7}{5}=\dfrac{9}{5}-\dfrac{2}{5}\)
\(\dfrac{4}{5}x+\dfrac{7}{5}=\dfrac{7}{5}\)
\(\dfrac{4}{5}x=\dfrac{7}{5}-\dfrac{7}{5}\)
\(\dfrac{4}{5}x=0\)
\(x=0:\dfrac{4}{5}\)
\(x=0\)
b)\(\dfrac{2}{5}x-\dfrac{6}{4}=\dfrac{8}{5}\)
\(\dfrac{2}{5}x=\dfrac{8}{5}+\dfrac{6}{4}\)
\(\dfrac{2}{5}x=\dfrac{31}{10}\)
\(x=\dfrac{31}{10}:\dfrac{2}{5}\)
\(x=\dfrac{31}{4}\)
1. Tính:
a. \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{9.10}\)
= \(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
= \(\dfrac{1}{1}-\dfrac{1}{10}\)
= \(\dfrac{10}{10}-\dfrac{1}{10}\)
= \(\dfrac{9}{10}\)
b. \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
= \(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
= \(\dfrac{1}{1}-\dfrac{1}{100}\)
= \(\dfrac{100}{100}-\dfrac{1}{100}\)
= \(\dfrac{99}{100}\)
2. Tìm x, biết:
a. \(\dfrac{2}{5}+\dfrac{4}{5}x-\dfrac{7}{5}=\dfrac{9}{5}\)
\(\dfrac{4}{5}x-\dfrac{7}{5}=\dfrac{9}{5}-\dfrac{2}{5}\)
\(\dfrac{4}{5}x-\dfrac{7}{5}=\dfrac{7}{5}\)
\(\dfrac{4}{5}x=\dfrac{7}{5}+\dfrac{7}{5}\)
\(\dfrac{4}{5}x=\dfrac{14}{5}\)
\(x=\dfrac{14}{5}:\dfrac{4}{5}\)
\(x=\dfrac{14}{5}.\dfrac{5}{4}\)
\(x=14.\dfrac{1}{4}\)
\(x=\dfrac{14}{4}\)
Vậy \(x=\dfrac{14}{4}\)
b. \(\dfrac{2}{5}x-\dfrac{6}{4}=\dfrac{8}{5}\)
\(\dfrac{2}{5}x=\dfrac{8}{5}+\dfrac{6}{4}\)
\(\dfrac{2}{5}x=\dfrac{32}{20}+\dfrac{30}{20}\)
\(\dfrac{2}{5}x=\dfrac{62}{20}\)
\(\dfrac{2}{5}x=\dfrac{31}{10}\)
\(x=\dfrac{31}{10}:\dfrac{2}{5}\)
\(x=\dfrac{31}{10}.\dfrac{5}{2}\)
\(x=\dfrac{31}{2}.\dfrac{2}{2}\)
\(x=\dfrac{31}{2}.1\)
\(x=\dfrac{31}{2}\)
Vậy \(x=\dfrac{31}{2}\)
bài này mk tự làm ko sao chép trên mạng
nếu thấy đúng thì tick đúng cho mk nha
A = \(\dfrac{3}{4}\).\(\dfrac{8}{9}\).\(\dfrac{15}{16}.\)\(\dfrac{24}{25}\)...\(\dfrac{9800}{9801}\)
A = \(\dfrac{1.3}{2.2}\).\(\dfrac{2.4}{3.3}\).\(\dfrac{3.5}{4.4}\)...\(\dfrac{98.100}{99.99}\)
A = \(\dfrac{1}{2}.\dfrac{100}{99}\)
A = \(\dfrac{50}{99}\)
B = \(\dfrac{1.2+2.3+3.4+...+98.99}{98.99.100}\)
Đặt tử số là C Thì
C = 1.2 + 2.3 + 3.4 +...+ 98.99
C = \(\dfrac{1}{3}\).(1.2.3 + 2.3.3 + 3.4.3 + ...+ 98.99.3)
C = \(\dfrac{1}{3}\).[1.2.3 + 2.3.(4-1) + 3.4.(5-2) +...+ 98.99.(100-97)]
C = \(\dfrac{1}{3}\).[1.2.3 -1.2.3+2.3.4- 2.3.4 + 2.4.5 - .... - 97.98.99 + 98.99.100]
C = \(\dfrac{1}{3}\).98.99.100
B = \(\dfrac{\dfrac{1}{3}.98.99.100}{98.99.100}\)
B = \(\dfrac{1}{3}\) = \(\dfrac{33}{99}\) < \(\dfrac{50}{99}\) = A
Vậy B < A
\(A=\dfrac{1}{1.2}-\dfrac{1}{1.2.3}+\dfrac{1}{2.3}-\dfrac{1}{2.3.4}+\dfrac{1}{3.4}-\dfrac{1}{3.4.5}+\dfrac{1}{99.100}-\dfrac{1}{99.100.101}\)
\(A=\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)-\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{99.100.101}\right)\)
\(A=\left(1-\dfrac{1}{100}\right)-\left(\dfrac{\dfrac{1}{1.2}-\dfrac{1}{100.101}}{2}\right)\)
Bấm máy nha
\(B=\dfrac{5}{1.2.3.4}+\dfrac{5}{2.3.4.5}+\dfrac{5}{3.4.5.6}+...+\dfrac{5}{98.99.100.101}\)
\(B=\dfrac{5}{3}.\left(\dfrac{3}{1.2.3.4}+\dfrac{3}{2.3.4.5}+...+\dfrac{3}{98.99.100.101}\right)\)
\(B=\dfrac{5}{3}.\left(\dfrac{4-1}{1.2.3.4}+\dfrac{5-2}{2.3.4.5}+...+\dfrac{101-98}{98.99.100.101}\right)\)
\(B=\dfrac{5}{3}.\left(\dfrac{4}{1.2.3.4}-\dfrac{1}{1.2.3.4}+\dfrac{5}{2.3.4.5}-\dfrac{2}{2.3.4.5}+...+\dfrac{101}{98.99.100.101}-\dfrac{98}{98.99.100.101}\right)\)
\(B=\dfrac{5}{3}.\left(\dfrac{1}{1.2.3}-\dfrac{1}{99.100.101}\right)\)
\(B=\dfrac{5}{3}.\dfrac{166649}{999900}\approx0,3\)
\(P=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{99}{100}\)
Áp dụng tính chất phân phối, rồi tính giá trị biểu thức.
Chẳng hạn,
Với , thì
ĐS. ; C = 0.
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Mình k ghi lại đề nhé!~~
\(A=\dfrac{4}{1}.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2014.2015}\right)\)
\(A=\dfrac{4}{1}.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2014}-\dfrac{1}{2015}\right)\)
\(A=\dfrac{4}{1}.\left(1-\dfrac{1}{2015}\right)\)
\(A=\dfrac{4}{1}.\left(\dfrac{2015-1}{2015}\right)\)
\(A=\dfrac{4}{1}.\dfrac{2014}{2015}\)
\(A=3,998014888\)
\(A\approx4\)
đúng òi ó bn