a) 3 phần 1.4 + 3 phần 4.7 + 3 phần 7.10+....+3 phần 97.100
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giúp mình với
KV
27 tháng 4 2023
5/1.4 + 5/4.7 + 5/7.10 + ... + 5/97.100
= 5/3 . (3/1.4 + 3/4.7 + 3/7.10 + ... + 3/97.100)
= 5/3 . (1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + ... + 1/97 - 1/100)
= 5/3 . ( 1 - 1/100)
= 5/3 . 99/100
= 33/20
Q
9 tháng 4 2017
a) \(P=\dfrac{1}{1.2}+\dfrac{2}{2.4}+\dfrac{3}{4.7}+...\dfrac{10}{46.56}\)
\(P=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...\dfrac{1}{46}-\dfrac{1}{56}\)
\(P=1-\dfrac{1}{56}\)
\(P=\dfrac{55}{56}\)
b) \(A=\dfrac{3}{1.2}+\dfrac{3}{2.3}+\dfrac{3}{3.4}+...+\dfrac{3}{99.100}\)
\(A=3\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)\)
\(A=3\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(A=3\left(1-\dfrac{1}{100}\right)\)
\(A=3.\dfrac{99}{100}\)
\(A=\dfrac{297}{100}\)
c) \(B=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{100.103}\)
\(B=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\)
\(B=1-\dfrac{1}{103}\)
\(B=\dfrac{102}{103}\)
d) \(C=\dfrac{5}{1.4}+\dfrac{5}{4.7}+\dfrac{5}{7.10}+...+\dfrac{5}{100.103}\)
\(C=\dfrac{5}{3}\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{100.103}\right)\)
\(C=\dfrac{5}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)
\(C=\dfrac{5}{3}\left(1-\dfrac{1}{103}\right)\)
\(C=\dfrac{5}{3}.\dfrac{102}{103}\)
\(C=\dfrac{170}{103}\)
e) \(D=\dfrac{7}{1.5}+\dfrac{7}{5.9}+\dfrac{7}{9.13}+...+\dfrac{7}{101.105}\)
\(D=\dfrac{7}{4}\left(\dfrac{4}{1.5}+\dfrac{4}{5.9}+\dfrac{4}{9.13}+...+\dfrac{4}{101.105}\right)\)
\(D=\dfrac{7}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{101}-\dfrac{1}{105}\right)\)
\(D=\dfrac{7}{4}\left(1-\dfrac{1}{105}\right)\)
\(D=\dfrac{7}{4}.\dfrac{104}{105}\)
\(D=\dfrac{26}{15}\)
D
9 tháng 4 2017
a)\(P=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{46}-\frac{1}{56}\)
=\(1-\frac{1}{56}=\frac{55}{56}\)
b)\(A.\frac{1}{3}=\frac{1}{3}.\left(\frac{3}{1.2}+\frac{3}{2.3}+....+\frac{3}{99.100}\right)\)
= \(\frac{1}{3}A=\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{3}{99.100}\)
=> \(\frac{1}{3}A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
=> \(\frac{1}{3}A=1-\frac{1}{100}=\frac{99}{100}\)
=> \(A=\frac{99}{100}.3=\frac{297}{100}\)
c)\(B=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\)
=\(1-\frac{1}{103}=\frac{102}{103}\)
d) \(\frac{3}{5}C=\frac{3}{5}.\left(\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}\right)\)
=\(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\)
=\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{100}-\frac{1}{103}\)
=\(1-\frac{1}{103}=\frac{102}{103}\)
=>\(C=\frac{102}{103}.\frac{5}{3}=\frac{170}{103}\)
e) \(\frac{4}{7}D=\frac{4}{7}.\left(\frac{7}{1.5}+\frac{7}{5.9}+...+\frac{7}{101.105}\right)\)
=\(\frac{4}{1.5}+\frac{4}{5.9}+...+\frac{4}{101.105}\)
=\(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{101}-\frac{1}{105}\)
=\(1-\frac{1}{105}=\frac{104}{105}\)
=< D=\(\frac{104}{105}.\frac{7}{4}=\frac{26}{15}\)
BH
20 tháng 4 2022
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mik chỉ hỏi thui mà, mik có bắt bạn giúp âu
AF
21 tháng 9 2020
Bài giải
\(\frac{2}{3}+\frac{3}{4}+\frac{4}{5}=\frac{40}{60}+\frac{45}{60}+\frac{48}{60}=\frac{133}{60}\)
\(\frac{8}{5}+\frac{7}{6}+\frac{10}{9}+\frac{1}{2}=\frac{144}{90}+\frac{105}{90}+\frac{100}{90}+\frac{45}{90}=\frac{394}{90}\)
\(\frac{15}{17}-\frac{11}{13}+\frac{3}{26}=\frac{390}{442}+\frac{374}{442}+\frac{51}{442}=\frac{815}{442}\)
\(\frac{9}{12}\text{ x }\frac{4}{3}\text{ : }\frac{8}{5}=\frac{9}{12}\text{ x }\frac{4}{3}\text{ x }\frac{5}{8}=\frac{9\text{ x }4\text{ x }5}{12\text{ x }3\text{ x }8}=\frac{5}{8}\)
\(\frac{4}{5}\text{ x }\frac{15}{8}\text{ : }\frac{5}{7}=\frac{4}{5}\text{ x }\frac{15}{8}\text{ x }\frac{7}{5}=\frac{4\text{ x }15\text{ x }7}{5\text{ x }8\text{ x }5}=\frac{21}{10}\)
26 tháng 4 2021
\(\frac{2}{3}+\frac{3}{4}+\frac{4}{5}=\frac{40}{60}+\frac{45}{60}+\frac{48}{60}=\frac{133}{60}\)
\(\frac{8}{5}+\frac{7}{6}+\frac{10}{9}+\frac{1}{2}=\frac{144}{90}+\frac{105}{90}+\frac{100}{90}+\frac{45}{90}=\frac{197}{45}\)
\(\frac{15}{17}-\frac{11}{13}+\frac{1}{26}=\frac{390}{442}+\frac{374}{442}+\frac{51}{442}=\frac{815}{442}\)
\(\frac{9}{12}\times\frac{4}{3}:\frac{8}{5}=1:\frac{8}{5}=\frac{5}{8}\)
\(\frac{4}{5}\times\frac{15}{8}:\frac{5}{7}=\frac{3}{2}:\frac{5}{7}=\frac{21}{10}\)
BN
19 tháng 3 2017
\(A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2015.2017}\)
\(A=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2015}-\dfrac{1}{2017}\)
\(A=1-\dfrac{1}{2017}=\dfrac{2016}{2017}\)
\(B=\dfrac{3}{1.4}+\dfrac{3}{5.7}+\dfrac{3}{7.10}+...+\dfrac{3}{100.103}\)
\(B=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\)
\(B=1-\dfrac{1}{103}=\dfrac{102}{103}\)
\(C=\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{62.65}\)
\(3C=3\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{62.65}\right)\)
\(3C=\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{62.65}\)
\(3C=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{62}-\dfrac{1}{65}\)
\(3C=\dfrac{1}{2}-\dfrac{1}{65}\)
\(3C=\dfrac{63}{130}\)
\(C=\dfrac{63}{130}:3=\dfrac{21}{130}\)
NG
3
20 tháng 8 2020
Ta có: \(\left(4,5-2x\right):\frac{3}{4}=1\frac{1}{3}\)
=> \(\left(4,5-2x\right):\frac{3}{4}=\frac{4}{3}\)
=> \(4,5-2x=1\)
=> \(2x=3,5\Rightarrow x=1,75\)
Vậy x=1,75
\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\)
=\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\)
=\(1-\frac{1}{100}\)
=\(\frac{99}{100}\)