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`a, 2/3 +3/4 = (8+9)/12=17/12.`
`1 1/3+4/5 = 4/3 + 4/5 = (20+12)/15=32/15`.
`=> x=2.`
`b, 5/6-1/4=(20-6)/24=7/12`.
`2 1/3-2/5= 7/3-2/5 = (35-6)/15=29/15`.
`=> x=1`.
\(1,\\ =\dfrac{2-1}{1\times2}+\dfrac{3-2}{2\times3}+\dfrac{4-3}{3\times4}+\dfrac{5-4}{4\times5}+.....+\dfrac{99-98}{98\times99}+\dfrac{100-99}{99\times100}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+....+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}=\dfrac{100-1}{100}=\dfrac{99}{100}\)
\(2,=\dfrac{13-11}{11\times13}+\dfrac{15-13}{13\times15}+....+\dfrac{21-19}{19\times21}+\dfrac{23-21}{21\times23}\\ =\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+....+\dfrac{1}{19}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{23}\\ =\dfrac{1}{11}-\dfrac{1}{23}\\ =\dfrac{23-11}{11\times23}=\dfrac{12}{253}\)
@seven
a: 1/1*2+1/2*3+...+1/99*100
=1-1/2+1/2-1/3+...+1/99-1/100
=1-1/100
=99/100
b: 2/11*13+2/13*15+...+2/21*23
=1/11-1/13+1/13-1/15+...+1/21-1/23
=1/11-1/23
=12/253
a,a+1/4=2 3/4-1 1/2
a+1/2=5/4
a=5/4-1/2
a=3/4
b,a-7/4=13/4-7/9
a-7/4=89/36
a= 89/36+7/4
a=152/36
c,3/2-a=17/6-1/6
3/2-a=8/3
a= 3/2-8/3
a= -7/6
a) \(...\dfrac{11}{4}-a+\dfrac{1}{4}=\dfrac{3}{2}\)
\(\dfrac{11}{4}+\dfrac{1}{4}-a=\dfrac{3}{2}\)
\(3-a=\dfrac{3}{2}\)
\(a=3-\dfrac{3}{2}\)
\(a=\dfrac{6}{2}-\dfrac{3}{2}\)
\(a=\dfrac{3}{2}\)
b) \(...\dfrac{13}{4}-a-\dfrac{13}{4}=\dfrac{7}{8}\)
\(\dfrac{13}{4}-\dfrac{13}{4}-a=\dfrac{7}{8}\)
\(0-a=\dfrac{7}{8}\)
\(a=-\dfrac{7}{8}\) (ra số âm lớp 5 chưa học nên bạn xem lại đề)
c) \(...\dfrac{17}{6}-\dfrac{3}{2}-a=\dfrac{1}{6}\)
\(\dfrac{17}{6}-\dfrac{9}{6}-a=\dfrac{1}{6}\)
\(\dfrac{8}{6}-a=\dfrac{1}{6}\)
\(a=\dfrac{8}{6}-\dfrac{1}{6}\)
\(a=\dfrac{7}{6}\)
a, 2\(\dfrac{3}{4}\) - a + \(\dfrac{1}{4}\) = 1\(\dfrac{1}{2}\)
a = 2 + \(\dfrac{3}{4}\) + \(\dfrac{1}{4}\) - 1 - \(\dfrac{1}{2}\)
a = 2 + 1 - 1 - \(\dfrac{1}{2}\)
a = 2 - \(\dfrac{1}{2}\)
a = \(\dfrac{3}{2}\)
b, 3\(\dfrac{1}{4}\) - a - 3\(\dfrac{1}{4}\) = \(\dfrac{7}{8}\)
(3\(\dfrac{1}{4}\) - 3\(\dfrac{1}{4}\)) - a = \(\dfrac{7}{8}\)
a = - \(\dfrac{7}{8}\)
c, 2\(\dfrac{5}{6}\) - 1\(\dfrac{1}{2}\) - a = \(\dfrac{1}{6}\)
a = 2 + \(\dfrac{5}{6}\) - 1 - \(\dfrac{1}{2}\) - \(\dfrac{1}{6}\)
a = (2-1) + (\(\dfrac{5}{6}\) - \(\dfrac{1}{6}\)) - \(\dfrac{1}{2}\)
a = 1 + \(\dfrac{2}{3}\) - \(\dfrac{1}{2}\)
a = \(\dfrac{7}{6}\)
\(3\dfrac{1}{2}+4\dfrac{5}{7}-5\dfrac{5}{14}\)
= \(\dfrac{7}{2}+\dfrac{33}{7}-\dfrac{75}{14}\)
= \(\dfrac{49}{14}+\dfrac{66}{14}-\dfrac{75}{14}\)
= \(\dfrac{40}{14}=\dfrac{20}{7}\)
\(4\dfrac{1}{2}+\dfrac{1}{2}\div5\dfrac{1}{2}\)
=\(\dfrac{9}{2}+\dfrac{1}{2}\div\dfrac{11}{2}\)
=\(\dfrac{9}{2}+\dfrac{1}{2}\times\dfrac{2}{11}\)
=\(\dfrac{9}{2}+\dfrac{1}{11}\)
=\(\dfrac{101}{22}\)
\(x\times3\dfrac{1}{3}=3\dfrac{1}{3}\div4\dfrac{1}{4}\)
\(x\times\dfrac{10}{3}=\dfrac{10}{3}\div\dfrac{17}{4}\)
\(x\times\dfrac{10}{3}=\dfrac{10}{3}\times\dfrac{4}{17}\)
\(x\times\dfrac{10}{3}=\dfrac{40}{51}\)
\(x=\dfrac{40}{51}\div\dfrac{10}{3}\)
\(x=\dfrac{40}{51}\times\dfrac{3}{10}\)
\(x=\dfrac{120}{510}=\dfrac{12}{51}=\dfrac{4}{7}\)
\(5\dfrac{2}{3}\div x=3\dfrac{2}{3}-2\dfrac{1}{2}\)
\(\dfrac{17}{3}\div x=\dfrac{11}{3}-\dfrac{5}{2}\)
\(\dfrac{17}{3}\div x=\dfrac{7}{6}\)
\(x=\dfrac{17}{3}\div\dfrac{7}{6}\)
\(x=\dfrac{17}{3}\times\dfrac{6}{7}\)
\(x=\dfrac{102}{21}=\dfrac{34}{7}\)
\(2\dfrac{2}{5}-y:2\dfrac{3}{4}=1\dfrac{1}{2}\\ \dfrac{12}{5}-y:\dfrac{11}{4}=\dfrac{3}{2}\\ y:\dfrac{11}{4}=\dfrac{12}{5}-\dfrac{3}{2}\\ y:\dfrac{11}{4}=\dfrac{9}{10}\\ y=\dfrac{9}{10}\times\dfrac{11}{4}=\dfrac{99}{40}\\ b,1\dfrac{1}{4}+2\dfrac{1}{5}\times y=2\dfrac{3}{5}\\ \dfrac{5}{4}+\dfrac{11}{5}\times y=\dfrac{13}{5}\\ \dfrac{11}{5}\times y=\dfrac{13}{5}-\dfrac{5}{4}\\ \dfrac{11}{5}\times y=\dfrac{27}{20}\\ y=\dfrac{27}{20}:\dfrac{11}{5}=\dfrac{27}{44}\)
\(c,2\dfrac{4}{5}-2\dfrac{1}{4}:y=\dfrac{3}{4}\\ \dfrac{14}{5}-\dfrac{9}{4}:y=\dfrac{3}{4}\\ \dfrac{9}{4}:y=\dfrac{14}{5}-\dfrac{3}{4}\\ \dfrac{9}{4}:y=\dfrac{41}{20}\\ y=\dfrac{9}{4}:\dfrac{41}{20}=\dfrac{45}{41}\\ c2,x:3\dfrac{1}{3}=2\dfrac{2}{5}+\dfrac{7}{10}\\ x:\dfrac{10}{3}=\dfrac{12}{5}+\dfrac{7}{10}\\ x:\dfrac{10}{3}=\dfrac{31}{10}\\ x=\dfrac{31}{10}\times\dfrac{10}{3}=\dfrac{31}{3}\)
a) \(\dfrac{2}{3}+\dfrac{3}{4}< x< 1\dfrac{1}{3}+\dfrac{4}{5}\)
\(\dfrac{2\times4}{3\times4}+\dfrac{3\times3}{4\times3}< x< \dfrac{\left(1\times3+1\right)\times5}{3\times5}+\dfrac{4\times3}{5\times3}\)
\(\dfrac{8}{12}+\dfrac{9}{12}< x< \dfrac{20}{15}+\dfrac{12}{15}\\ \dfrac{17}{12}< x< \dfrac{32}{15}\)
Ước tính: \(\dfrac{17}{12}=1,4\) và \(\dfrac{32}{15}=2,1\). Vậy số tự nhiên x = 2 sẽ thõa mãn 1,4 < x < 2,1
b)
\(\dfrac{5}{6}-\dfrac{1}{4}< x< 2\dfrac{1}{3}-\dfrac{2}{5}\\ \dfrac{5\times4}{6\times4}-\dfrac{1\times6}{4\times6}< x< \dfrac{\left(2\times3+1\right)\times5}{3\times5}-\dfrac{2\times3}{5\times3}\\ \dfrac{20}{24}-\dfrac{6}{24}< x< \dfrac{35}{15}-\dfrac{6}{15}\\ \dfrac{14}{24}< x< \dfrac{29}{15}\)
Ước tính \(\dfrac{14}{24}=0,5\) và \(\dfrac{29}{15}=1,9\)
Vậy với x là số tự nhiên x = 1 sẽ thõa mãn 0,5 < x < 1,9
a) \(2\dfrac{3}{4}-a+\dfrac{1}{4}=1\dfrac{1}{2}\)
=> \(\dfrac{11}{4}\) \(-a+\dfrac{1}{4}=\dfrac{3}{2}\)
=> \(\dfrac{11}{4}-a\) = \(\dfrac{3}{2}-\dfrac{1}{4}\)
=> a = \(\dfrac{11}{4}-\dfrac{5}{4}\) =\(\dfrac{3}{2}\)
Vậy a = \(\dfrac{3}{2}\)
b) \(3\dfrac{1}{4}-a-1\dfrac{3}{4}=\dfrac{7}{8}\)
=> \(\dfrac{13}{4}-a-\dfrac{7}{4}=\dfrac{7}{8}\)
=> \(\dfrac{13}{4}-a=\dfrac{21}{8}\)
=> \(a=\dfrac{13}{4}-\dfrac{21}{8}=\dfrac{5}{8}\)
Vậy a = \(\dfrac{5}{8}\)
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trả lời nha <3