15%+1,1:(\(\dfrac{2}{5}\)-1\(\dfrac{1}{2}\))-(-\(\dfrac{1}{3}^2\))
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HH
4
TC
11 tháng 5 2022
\(=\dfrac{3}{4}+\dfrac{11}{10}:\left(\dfrac{2}{5}-\dfrac{3}{2}\right)-\dfrac{1}{9}\)
\(=\dfrac{3}{4}+\dfrac{11}{10}:-\dfrac{11}{10}-\dfrac{1}{9}=\dfrac{3}{4}+\left(-1\right)-\dfrac{1}{9}=\dfrac{27}{36}-\dfrac{36}{36}-\dfrac{4}{36}=-\dfrac{13}{36}\)
R
27 tháng 1 2022
a) \(=\dfrac{157}{8}.\dfrac{12}{7}-\dfrac{61}{4}.\dfrac{12}{7}=\dfrac{12}{7}\left(\dfrac{157}{8}-\dfrac{61}{4}\right)=\dfrac{12}{7}.\dfrac{35}{8}=\dfrac{15}{2}\)
b) \(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{2}{15}\div\dfrac{1}{5}+\dfrac{3}{5}.\dfrac{1}{3}=\dfrac{1}{3}\left(\dfrac{2}{5}+\dfrac{3}{5}\right)-\dfrac{2}{15}.5=\dfrac{1}{3}.1-\dfrac{2}{3}=\dfrac{1}{3}-\dfrac{2}{3}=-\dfrac{1}{3}\)
c) \(=-\dfrac{80}{9}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
11 tháng 7 2023
\(\dfrac{3}{16}\) - (\(x\) - \(\dfrac{5}{4}\)) - ( \(\dfrac{3}{4}\) - \(\dfrac{7}{8}\) - 1) = 2\(\dfrac{1}{2}\)
\(\dfrac{3}{16}\) - \(x\) + \(\dfrac{5}{4}\) - \(\dfrac{3}{4}\) + \(\dfrac{7}{8}\) + 1 = \(\dfrac{5}{2}\)
\(\dfrac{3}{16}\) - \(x\) + ( \(\dfrac{5}{4}\) - \(\dfrac{3}{4}\)) + (\(\dfrac{7}{8}\) + 1) = \(\dfrac{5}{2}\)
\(\dfrac{3}{16}\) - \(x\) + \(\dfrac{1}{2}\) + \(\dfrac{15}{8}\) = \(\dfrac{5}{2}\)
( \(\dfrac{3}{16}\) + \(\dfrac{1}{2}\) + \(\dfrac{15}{8}\)) - \(x\) = \(\dfrac{5}{2}\)
\(\dfrac{41}{16}\) - \(x\) = \(\dfrac{5}{2}\)
\(x\) = \(\dfrac{41}{16}\) - \(\dfrac{5}{2}\)
\(x\) = \(\dfrac{1}{16}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
11 tháng 7 2023
2, \(\dfrac{1}{2}\).( \(\dfrac{1}{6}\) - \(\dfrac{9}{10}\)) = \(\dfrac{1}{5}\) - \(x\) + ( \(\dfrac{1}{15}\) - \(\dfrac{-1}{5}\))
\(\dfrac{1}{2}\).(-\(\dfrac{11}{15}\)) = \(\dfrac{1}{5}\) - \(x\) + \(\dfrac{1}{15}\) + \(\dfrac{1}{5}\)
- \(\dfrac{11}{30}\) = ( \(\dfrac{1}{5}\)+ \(\dfrac{1}{5}\)+ \(\dfrac{1}{15}\)) - \(x\)
- \(\dfrac{11}{30}\) = \(\dfrac{7}{15}\) - \(x\)
\(x\) = \(\dfrac{7}{15}\) + \(\dfrac{11}{30}\)
\(x\) = \(\dfrac{5}{6}\)
11 tháng 7 2023
`@` `\text {Ans}`
`\downarrow`
`1,`
`3/16 - (x - 5/4) - (3/4 + (-7)/8 - 1) = 2 1/2`
`=> 3/16 - x + 5/4 - (-1/8 - 1) = 2 1/2`
`=> 3/16 - x + 5/4 - (-9/8) = 2 1/2`
`=> 3/16 - x + 19/8 = 2 1/2`
`=> 3/16 - x = 2 1/2 - 19/8`
`=> 3/16 - x =1/8`
`=> x = 3/16 - 1/8`
`=> x = 1/16`
Vậy, `x = 1/16`
`2,`
`1/2* (1/6 - 9/10) = 1/5 - x + (1/15 - (-1)/5)`
`=> 1/2 * (-11/15) = 1/5 - x + 4/15`
`=> -11/30 = x + 1/5 - 4/15`
`=> x + (-1/15) = -11/30`
`=> x = -11/30 + 1/15`
`=> x = -3/10`
Vậy, `x = -3/10.`
LM
Lê Minh Vũ
CTVHS
31 tháng 7 2023
Bài 7: Tính
a) \(4\dfrac{2}{5}\times8\dfrac{3}{4}-2\dfrac{3}{4}\)
\(=\dfrac{22}{5}\times\dfrac{35}{4}-\dfrac{11}{4}\)
\(=\dfrac{77}{2}-\dfrac{11}{4}\)
\(=\dfrac{143}{4}\)
b) \(2\dfrac{2}{3}+1\dfrac{2}{5}-\dfrac{2}{15}\)
\(=\dfrac{8}{3}+\dfrac{7}{5}-\dfrac{2}{15}\)
\(=\dfrac{47}{15}-\dfrac{2}{15}\)
\(=\dfrac{45}{15}=3\)
c) \(3\dfrac{1}{3}-2\dfrac{2}{3}+1\dfrac{5}{6}\)
\(=\dfrac{10}{3}-\dfrac{8}{3}+\dfrac{11}{6}\)
\(=\dfrac{2}{3}+\dfrac{11}{6}\)
\(=\dfrac{15}{6}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
23 tháng 9 2023
a, -4\(\dfrac{3}{5}\).2\(\dfrac{4}{3}\) < \(x\) < -2\(\dfrac{3}{5}\): 1\(\dfrac{6}{15}\)
- \(\dfrac{23}{5}\).\(\dfrac{10}{3}\) < \(x\) < - \(\dfrac{13}{5}\): \(\dfrac{21}{15}\)
- \(\dfrac{46}{3}\) < \(x\) < - \(\dfrac{13}{7}\)
\(x\) \(\in\) {-15; -14;-13;..; -2}
23 tháng 9 2023
a) Ta có \(-4\dfrac{3}{5}\cdot2\dfrac{4}{3}=-\dfrac{23}{5}\cdot\dfrac{10}{3}=-\dfrac{46}{3}\) và \(-2\dfrac{3}{5}\div1\dfrac{6}{15}=-\dfrac{13}{5}\div\dfrac{7}{5}=-\dfrac{13}{7}\)
Do đó \(-\dfrac{46}{3}< x< -\dfrac{13}{7}\)
Lại có \(-\dfrac{46}{3}\le-15\) và \(-\dfrac{13}{7}\ge-2\)
Suy ra \(-15\le x\le-2\), x ϵ Z
b) Ta có \(-4\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)=-\dfrac{13}{3}\cdot\dfrac{1}{3}=-\dfrac{13}{9}\) và \(-\dfrac{2}{3}\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{4}\right)=-\dfrac{2}{3}\cdot\dfrac{-11}{12}=\dfrac{11}{18}\)
Do đó \(-\dfrac{13}{9}< x< \dfrac{11}{18}\)
Lại có \(-\dfrac{13}{9}\le-1\) và \(\dfrac{11}{18}\ge0\)
Suy ra \(-1\le x\le0\), x ϵ Z
H9
HT.Phong (9A5)
CTVHS
16 tháng 7 2023
\(\dfrac{2}{5}\times15\dfrac{1}{3}-\dfrac{2}{5}\times10\dfrac{1}{3}\)
\(=\dfrac{2}{5}\times\left(15\dfrac{1}{3}-10\dfrac{1}{3}\right)\)
\(=\dfrac{2}{5}\times5\)
\(=2\)
____________________
\(12\dfrac{5}{11}-\left(3\dfrac{1}{4}+2\dfrac{5}{11}\right)\)
\(=12\dfrac{5}{11}-3\dfrac{1}{4}-2\dfrac{5}{11}\)
\(=\left(12\dfrac{5}{11}-2\dfrac{5}{11}\right)-3\dfrac{1}{4}\)
\(=10-3\dfrac{1}{4}\)
\(=\dfrac{27}{4}\)
_______________
\(\dfrac{34}{31}-\dfrac{19}{28}-\dfrac{3}{31}\)
\(=\left(\dfrac{34}{31}-\dfrac{3}{31}\right)-\dfrac{19}{28}\)
\(=\dfrac{31}{31}-\dfrac{19}{28}\)
\(=1-\dfrac{19}{28}\)
\(=\dfrac{9}{28}\)
\(15\%+1,1:\left(\dfrac{2}{5}-1\dfrac{1}{2}\right)-\left(-\dfrac{1}{3}\right)^2\)
\(=\dfrac{3}{2}-\dfrac{1}{9}+1,1:\left(0,4-1,5\right)\)
\(=\dfrac{27-2}{18}+1,1:\left(-1,1\right)\)
\(=\dfrac{25}{18}-1=\dfrac{7}{18}\)