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\(2\dfrac{2}{5}-y:2\dfrac{3}{4}=1\dfrac{1}{2}\\ \dfrac{12}{5}-y:\dfrac{11}{4}=\dfrac{3}{2}\\ y:\dfrac{11}{4}=\dfrac{12}{5}-\dfrac{3}{2}\\ y:\dfrac{11}{4}=\dfrac{9}{10}\\ y=\dfrac{9}{10}\times\dfrac{11}{4}=\dfrac{99}{40}\\ b,1\dfrac{1}{4}+2\dfrac{1}{5}\times y=2\dfrac{3}{5}\\ \dfrac{5}{4}+\dfrac{11}{5}\times y=\dfrac{13}{5}\\ \dfrac{11}{5}\times y=\dfrac{13}{5}-\dfrac{5}{4}\\ \dfrac{11}{5}\times y=\dfrac{27}{20}\\ y=\dfrac{27}{20}:\dfrac{11}{5}=\dfrac{27}{44}\)
\(c,2\dfrac{4}{5}-2\dfrac{1}{4}:y=\dfrac{3}{4}\\ \dfrac{14}{5}-\dfrac{9}{4}:y=\dfrac{3}{4}\\ \dfrac{9}{4}:y=\dfrac{14}{5}-\dfrac{3}{4}\\ \dfrac{9}{4}:y=\dfrac{41}{20}\\ y=\dfrac{9}{4}:\dfrac{41}{20}=\dfrac{45}{41}\\ c2,x:3\dfrac{1}{3}=2\dfrac{2}{5}+\dfrac{7}{10}\\ x:\dfrac{10}{3}=\dfrac{12}{5}+\dfrac{7}{10}\\ x:\dfrac{10}{3}=\dfrac{31}{10}\\ x=\dfrac{31}{10}\times\dfrac{10}{3}=\dfrac{31}{3}\)
a: \(y+\dfrac{1}{2}\cdot\dfrac{1}{3}=\dfrac{3}{4}\)
=>\(y+\dfrac{1}{6}=\dfrac{3}{4}\)
=>\(y=\dfrac{3}{4}-\dfrac{1}{6}=\dfrac{9}{12}-\dfrac{2}{12}=\dfrac{7}{12}\)
b: \(y+\dfrac{1}{2}\cdot\dfrac{1}{5}=\dfrac{3}{10}\)
=>\(y+\dfrac{1}{10}=\dfrac{3}{10}\)
=>\(y=\dfrac{3}{10}-\dfrac{1}{10}=\dfrac{2}{10}=\dfrac{1}{5}\)
1/2-2y=9/20
=>2y=1/2-9/20=1/20
=>y=1/20:2=1/40
b,3/5:4/3:y=2+7/10=9/20:y=27/10
=>y=9/20:27/10=1/6
c,y+y*3/2-y*1/2=1/10
=>y(1+3/2-1/2)=1/10
=>2y=1/10
=>y=1/10:2=1/20
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 = \(\dfrac{7}{6}\)
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
\(3\dfrac{1}{2}+4\dfrac{5}{7}-5\dfrac{5}{14}\)
= \(\dfrac{7}{2}+\dfrac{33}{7}-\dfrac{75}{14}\)
= \(\dfrac{49}{14}+\dfrac{66}{14}-\dfrac{75}{14}\)
= \(\dfrac{40}{14}=\dfrac{20}{7}\)
\(4\dfrac{1}{2}+\dfrac{1}{2}\div5\dfrac{1}{2}\)
=\(\dfrac{9}{2}+\dfrac{1}{2}\div\dfrac{11}{2}\)
=\(\dfrac{9}{2}+\dfrac{1}{2}\times\dfrac{2}{11}\)
=\(\dfrac{9}{2}+\dfrac{1}{11}\)
=\(\dfrac{101}{22}\)
\(x\times3\dfrac{1}{3}=3\dfrac{1}{3}\div4\dfrac{1}{4}\)
\(x\times\dfrac{10}{3}=\dfrac{10}{3}\div\dfrac{17}{4}\)
\(x\times\dfrac{10}{3}=\dfrac{10}{3}\times\dfrac{4}{17}\)
\(x\times\dfrac{10}{3}=\dfrac{40}{51}\)
\(x=\dfrac{40}{51}\div\dfrac{10}{3}\)
\(x=\dfrac{40}{51}\times\dfrac{3}{10}\)
\(x=\dfrac{120}{510}=\dfrac{12}{51}=\dfrac{4}{7}\)
\(5\dfrac{2}{3}\div x=3\dfrac{2}{3}-2\dfrac{1}{2}\)
\(\dfrac{17}{3}\div x=\dfrac{11}{3}-\dfrac{5}{2}\)
\(\dfrac{17}{3}\div x=\dfrac{7}{6}\)
\(x=\dfrac{17}{3}\div\dfrac{7}{6}\)
\(x=\dfrac{17}{3}\times\dfrac{6}{7}\)
\(x=\dfrac{102}{21}=\dfrac{34}{7}\)
a) ....
=> 14/3 : x = 4/3 + 3/5 = 29/15
=> x = 14/3 : 29/15 = 70/29
b) ...
=> 4/3 + 5/2x = 2
=> 5/2x = 2 - 4/3 = 2/3
=> x = 2/3 : 5/2 = 4/15
c)...
=> x : 10/3 = 31/10
=> x = 31/10 x 10/3 = 31/3
\(A=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{55}\)
\(\dfrac{A}{2}=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{110}=\)
\(=\dfrac{1}{2x3}+\dfrac{1}{3x4}+\dfrac{1}{4x5}+...+\dfrac{1}{10x11}=\)
\(=\dfrac{3-2}{2x3}+\dfrac{4-3}{3x4}+\dfrac{5-4}{4x5}+...+\dfrac{11-10}{10x11}=\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{10}-\dfrac{1}{11}=\)
\(=\dfrac{1}{2}-\dfrac{1}{11}=\dfrac{9}{22}\Rightarrow A=\dfrac{9}{11}\)