Bài 5:
a) 4\(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) : 5\(\dfrac{1}{2}\)
b) \(\dfrac{7}{6}\) : (\(\dfrac{1}{2}\) x \(\dfrac{3}{4}\)) - \(\dfrac{5}{8}\)
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TC
8 tháng 5 2022
bài 1
a)\(=\dfrac{16}{40}+\dfrac{15}{40}=\dfrac{31}{40}\)
b)\(=\dfrac{7}{6}-\dfrac{4}{6}=\dfrac{3}{6}=\dfrac{1}{2}\)
c)\(=\dfrac{30}{9}=\dfrac{10}{3}\)
d)\(=\dfrac{8}{5}\times\dfrac{7}{4}=\dfrac{56}{20}=\dfrac{14}{5}\)
20 tháng 12 2020
a) Ta có: \(\dfrac{1}{7}+x=-\dfrac{2}{3}\)
\(\Leftrightarrow x=-\dfrac{2}{3}-\dfrac{1}{7}=\dfrac{-14}{21}-\dfrac{3}{21}\)
hay \(x=-\dfrac{17}{21}\)
Vậy: \(x=-\dfrac{17}{21}\)
b) Ta có: \(\dfrac{-2}{3}:x=\dfrac{-5}{6}\)
\(\Leftrightarrow x=\dfrac{-2}{3}:\dfrac{-5}{6}=\dfrac{-2}{3}\cdot\dfrac{6}{-5}=\dfrac{-12}{-15}=\dfrac{4}{5}\)
Vậy: \(x=\dfrac{4}{5}\)
c) Ta có: \(\left(\dfrac{3}{5}-2x\right)\cdot\dfrac{5}{8}=1\)
\(\Leftrightarrow\left(\dfrac{3}{5}-2x\right)=1:\dfrac{5}{8}=\dfrac{8}{5}\)
\(\Leftrightarrow-2x=\dfrac{8}{5}-\dfrac{3}{5}=1\)
hay \(x=-\dfrac{1}{2}\)
Vậy: \(x=-\dfrac{1}{2}\)
d) Ta có: \(\dfrac{3}{4}+\dfrac{2}{5}x=\dfrac{29}{60}\)
\(\Leftrightarrow x\cdot\dfrac{2}{5}=\dfrac{29}{60}-\dfrac{3}{4}=\dfrac{29}{60}-\dfrac{45}{60}=\dfrac{-16}{60}=\dfrac{-4}{15}\)
hay \(x=\dfrac{-4}{15}:\dfrac{2}{5}=\dfrac{-4}{15}\cdot\dfrac{5}{2}=\dfrac{-20}{30}=-\dfrac{2}{3}\)
Vậy: \(x=-\dfrac{2}{3}\)
e) Ta có: \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}=\dfrac{8}{20}-\dfrac{15}{20}=\dfrac{-7}{20}\)
hay \(x=-\dfrac{1}{4}:\dfrac{7}{20}=\dfrac{-1}{4}\cdot\dfrac{20}{7}=\dfrac{-20}{28}=\dfrac{-5}{7}\)
Vậy: \(x=-\dfrac{5}{7}\)
f) Ta có: \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
\(\Leftrightarrow-x+\dfrac{11}{12}-\dfrac{2}{5}-\dfrac{2}{3}=0\)
\(\Leftrightarrow-x+\dfrac{55}{60}-\dfrac{24}{60}-\dfrac{40}{60}=0\)
\(\Leftrightarrow-x-\dfrac{9}{60}=0\)
\(\Leftrightarrow-x=\dfrac{9}{60}=\dfrac{3}{20}\)
hay \(x=-\dfrac{3}{20}\)
Vậy: \(x=-\dfrac{3}{20}\)
g) Ta có: \(\left|x+\dfrac{1}{3}\right|-4=\dfrac{-1}{2}\)
\(\Leftrightarrow\left|x+\dfrac{1}{3}\right|=\dfrac{-1}{2}+4=\dfrac{-1}{2}+\dfrac{8}{2}=\dfrac{7}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=\dfrac{7}{2}\\x+\dfrac{1}{3}=-\dfrac{7}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}-\dfrac{1}{3}=\dfrac{21}{6}-\dfrac{2}{6}=\dfrac{19}{6}\\x=-\dfrac{7}{2}-\dfrac{1}{3}=\dfrac{-21}{6}-\dfrac{2}{6}=\dfrac{-23}{6}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{19}{6};-\dfrac{23}{6}\right\}\)
16 tháng 6 2021
`A=(8 2/7-4 2/7)-3 4/9`
`=8+2/7-4-2/7-3-4/9`
`=4-3-4/9`
`=1-4/9=5/9`
`B=(10 2/9-6 2/9)+2 3/5`
`=10+2/9-6-2/9+2+3/5`
`=4+2+3/5`
`=6+3/5=33/5`
Bài 2:
`a)5 1/2*3 1/4`
`=11/2*13/4`
`=143/8`
`b)6 1/3:4 2/9`
`=19/3:38/9`
`=19/3*9/38=3/2`
`c)4 3/7*2`
`=31/7*2`
`=62/7`
16 tháng 6 2021
Bài 1:
\(A=\left(8\dfrac{2}{7}-4\dfrac{2}{7}\right)-3\dfrac{4}{9}\)
\(A=\left(\dfrac{58}{7}-\dfrac{30}{7}\right)-\dfrac{31}{9}\)
\(A=4-\dfrac{31}{9}\)
\(A=\dfrac{5}{9}\)
\(B=\left(10\dfrac{2}{9}-6\dfrac{2}{9}\right)+2\dfrac{3}{5}\)
\(B=\left(\dfrac{92}{9}-\dfrac{56}{9}\right)+\dfrac{13}{5}\)
\(B=4+\dfrac{13}{5}\)
\(B=\dfrac{33}{5}\)
MH
9 tháng 5 2022
Bài 1:
+) \(\dfrac{7}{8}\times y=\dfrac{3}{2}+\dfrac{6}{4}=3\)
\(y=3:\dfrac{7}{8}=\dfrac{24}{7}\)
+) \(\dfrac{1}{y}\times\left(\dfrac{2}{5}+\dfrac{1}{5}\right)=\dfrac{10}{3}\)
\(\dfrac{1}{y}=\dfrac{10}{3}:\dfrac{3}{5}=\dfrac{50}{9}\)
\(y=\dfrac{9}{50}\)
C
27 tháng 4 2022
1)
a)\(\dfrac{20}{30}+\dfrac{9}{30}=\dfrac{29}{30}\)
b)\(\dfrac{16}{24}-\dfrac{15}{24}=\dfrac{1}{24}\)
c)\(\dfrac{12}{63}=\dfrac{4}{21}\)
d) \(\dfrac{5}{6}x\dfrac{3}{4}=\dfrac{15}{24}=\dfrac{5}{8}\)
2)
a)\(x=\dfrac{5}{4}-\dfrac{2}{3}\)
\(x=\dfrac{7}{12}\)
b) \(x=4x\dfrac{3}{5}\)
\(x=\dfrac{12}{5}\)
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
3 tháng 8 2023
\(a,2\dfrac{2}{5}:y\times1\dfrac{3}{4}=\dfrac{7}{8}\\ \dfrac{12}{5}:y\times\dfrac{7}{4}=\dfrac{7}{8}\\ \dfrac{12}{5}:y=\dfrac{7}{8}:\dfrac{7}{4}\\ \dfrac{12}{5}:y=\dfrac{1}{2}\\ y=\dfrac{12}{5}:\dfrac{1}{2}=\dfrac{24}{5}\\ b,3\dfrac{2}{5}:y:1\dfrac{1}{4}=2\dfrac{3}{5}\\ \dfrac{17}{5}:y:\dfrac{5}{4}=\dfrac{13}{5}\\ y:\dfrac{5}{4}=\dfrac{17}{5}:\dfrac{13}{5}\\ y:\dfrac{5}{4}=\dfrac{17}{13}\\ y=\dfrac{17}{13}\times\dfrac{5}{4}=\dfrac{85}{52}\)
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
3 tháng 8 2023
\(c,\dfrac{12}{5}-2\dfrac{2}{5}\times y=1\dfrac{1}{4}\\ \dfrac{12}{5}-\dfrac{12}{5}\times y=\dfrac{5}{4}\\ \dfrac{12}{5}\times y=\dfrac{12}{5}-\dfrac{5}{4}\\ \dfrac{12}{5}\times y=\dfrac{23}{20}\\ y=\dfrac{23}{20}:\dfrac{12}{5}\\ y=\dfrac{23}{48}\)
3 tháng 8 2023
a) \(...=\dfrac{19}{8}:\dfrac{15}{4}x\dfrac{8}{3}=\dfrac{19}{8}x\dfrac{4}{15}x\dfrac{8}{3}=\dfrac{76}{45}\)
b) \(...=\dfrac{3}{2}:\dfrac{7}{3}:\dfrac{17}{6}=\dfrac{3}{2}x\dfrac{3}{7}x\dfrac{6}{17}=\dfrac{27}{119}\)
c) \(...=\dfrac{14}{3}-\dfrac{7}{4}:\dfrac{12}{5}=\dfrac{14}{3}-\dfrac{7}{4}x\dfrac{5}{12}=\dfrac{14}{3}-\dfrac{35}{48}=\dfrac{14x16}{48}-\dfrac{35}{48}=\dfrac{224}{48}-\dfrac{35}{48}=\dfrac{189}{48}=\dfrac{63}{16}\)
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
3 tháng 8 2023
\(a,2\dfrac{3}{8}:3\dfrac{3}{4}\times2\dfrac{2}{3}\\ =\dfrac{2\times8+3}{8}:\dfrac{3\times4+3}{4}\times\dfrac{2\times3+2}{3}\\ =\dfrac{19}{8}:\dfrac{15}{4}\times\dfrac{8}{3}\\ =\dfrac{19\times4\times8}{8\times15\times3}=\dfrac{76}{45}\)
\(b,1\dfrac{1}{2}:\dfrac{7}{3}:2\dfrac{5}{6}\\ =\dfrac{3}{2}:\dfrac{7}{3}:\dfrac{2\times6+5}{6}\\ =\dfrac{3}{2}\times\dfrac{3}{7}\times\dfrac{6}{17}\\ =\dfrac{54}{238}=\dfrac{27}{119}\)
\(c,4\dfrac{2}{3}-1\dfrac{3}{4}:2\dfrac{2}{5}\\ =\dfrac{4\times3+2}{3}-\dfrac{1\times4+3}{4}:\dfrac{2\times5+2}{5}\\ =\dfrac{14}{3}-\dfrac{7}{4}:\dfrac{12}{5}\\ =\dfrac{14}{3}-\dfrac{7}{4}.\dfrac{5}{12}\\ =\dfrac{14}{3}-\dfrac{35}{48}\\ =\dfrac{14\times16-35}{48}=\dfrac{189}{48}=\dfrac{63}{16}\)
31 tháng 7 2023
\(4\dfrac{2}{3}+3\dfrac{2}{7}=\dfrac{14}{3}+\dfrac{23}{7}=\dfrac{14x7+23x3}{21}=\dfrac{167}{21}\)
\(8\dfrac{5}{9}:5\dfrac{1}{2}=\dfrac{77}{9}:\dfrac{11}{2}=\dfrac{77}{9}x\dfrac{2}{11}=\dfrac{14}{9}\)
\(6\dfrac{5}{7}:2\dfrac{1}{6}=\dfrac{47}{7}:\dfrac{13}{6}=\dfrac{47}{7}x\dfrac{6}{13}=\dfrac{282}{91}\)
\(1\dfrac{3}{4}x2\dfrac{5}{6}=\dfrac{7}{4}x\dfrac{17}{6}=\dfrac{119}{24}\)
\(5\dfrac{3}{4}-2=\dfrac{23}{4}-2=\dfrac{23}{4}-\dfrac{8}{4}=\dfrac{15}{4}\)
a) \(4\dfrac{1}{2}+\dfrac{1}{2}:5\dfrac{1}{2}=\dfrac{9}{2}+\dfrac{1}{2}:\dfrac{11}{2}=\dfrac{9}{2}+\dfrac{1}{2}x\dfrac{2}{11}=\dfrac{9}{2}+\dfrac{1}{11}=\dfrac{99}{22}+\dfrac{2}{22}=\dfrac{101}{22}\)
b) \(\dfrac{7}{6}:\left(\dfrac{1}{2}x\dfrac{3}{4}\right)-\dfrac{5}{8}=\dfrac{7}{6}:\dfrac{3}{8}-\dfrac{5}{8}=\dfrac{7}{6}x\dfrac{8}{3}-\dfrac{5}{8}=\dfrac{28}{9}-\dfrac{5}{8}=\dfrac{224}{72}-\dfrac{45}{72}=\dfrac{179}{72}\)