| \(\dfrac{\chi}{4}\) +0,75|=\(\dfrac{5}{8}+2\cdot\dfrac{3}{18}\)
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a: \(=\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(\dfrac{-5}{13}-\dfrac{8}{13}\right)+\left(\dfrac{-18}{35}-\dfrac{17}{35}\right)\)
=1-1-1
=-1
b: \(=\dfrac{-3}{8}\left(\dfrac{1}{6}+\dfrac{5}{6}\right)+\dfrac{-5}{8}=\dfrac{-3}{8}-\dfrac{5}{8}=-1\)
c: \(=\dfrac{4}{4}\cdot\dfrac{5}{15}\cdot\dfrac{11}{11}=\dfrac{1}{3}\)
a)\(=\left(-\dfrac{5}{13}+\dfrac{-8}{13}\right)+\left(-\dfrac{18}{35}-\dfrac{17}{35}\right)+\left(\dfrac{3}{14}+\dfrac{14}{17}\right)=-1-1+1=-1\)
b)\(=\dfrac{-3}{8}.\left(\dfrac{1}{6}+\dfrac{5}{6}\right)-\dfrac{10}{16}=-\dfrac{3}{8}.1-\dfrac{10}{16}=-\dfrac{6}{16}-\dfrac{10}{16}=-\dfrac{16}{16}=-1\)
c)\(\dfrac{-4.5.11}{11.5.3.-4}=\dfrac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\sqrt{\dfrac{4}{25}+\dfrac{4}{5}-\dfrac{9}{5}\cdot\dfrac{1}{9}+\dfrac{3}{4}}=\sqrt{\dfrac{4}{25}+\dfrac{4}{5}-\dfrac{1}{5}+\dfrac{3}{4}}\)
\(=\sqrt{\dfrac{4}{25}+\dfrac{3}{5}+\dfrac{3}{4}}\)
\(=\sqrt{\dfrac{16+60+75}{100}}=\dfrac{\sqrt{151}}{10}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(=\dfrac{13\left(3-18\right)}{40\left(15-2\right)}=\dfrac{13}{15-2}\cdot\dfrac{-15}{40}=\dfrac{-3}{8}\)
b: \(=\dfrac{18\left(34-124\right)}{36\left(-17-13\right)}=\dfrac{1}{2}\cdot\dfrac{-90}{-30}=\dfrac{3}{2}\)
c: \(=\dfrac{3\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}{4\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}+\dfrac{\dfrac{-1}{4}\cdot\dfrac{-2}{3}-\dfrac{3}{4}:\dfrac{1}{6}}{\dfrac{3}{2}\cdot\left(\dfrac{-2}{3}-\dfrac{3}{4}\cdot\dfrac{-2}{3}\right)}\)
\(=\dfrac{3}{4}+\dfrac{\dfrac{2}{12}-\dfrac{9}{2}}{\dfrac{3}{2}\cdot\dfrac{-1}{6}}=\dfrac{3}{4}+\dfrac{-13}{3}:\dfrac{-3}{12}=\dfrac{3}{4}+\dfrac{13}{3}\cdot\dfrac{12}{3}\)
\(=\dfrac{3}{4}+\dfrac{156}{9}=\dfrac{217}{12}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a.\)
\(-\dfrac{2}{3}\cdot\dfrac{?}{4}=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{?}{4}=\dfrac{1}{2}:-\dfrac{2}{3}=\dfrac{1}{2}\cdot-\dfrac{3}{2}=-\dfrac{3}{4}\)
\(\Leftrightarrow?=-3\)
\(b.\)
\(\dfrac{?}{3}\cdot\dfrac{5}{8}=-\dfrac{5}{12}\)
\(\Leftrightarrow\dfrac{?}{3}=\dfrac{-5}{12}:\dfrac{5}{8}=\dfrac{-5}{12}\cdot\dfrac{8}{5}=-\dfrac{2}{3}\)
\(\Leftrightarrow?=-2\)
\(c.\)
\(\dfrac{5}{6}\cdot\dfrac{3}{?}=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{3}{?}=\dfrac{1}{4}:\dfrac{5}{6}=\dfrac{1}{4}\cdot\dfrac{6}{5}=\dfrac{3}{10}\)
\(\Leftrightarrow?=10\)
Mk gọi ? = x nha
a) \(\dfrac{-2}{3}.\dfrac{x}{4}=\dfrac{1}{2}\)
\(\dfrac{x}{4}=\dfrac{1}{2}:\dfrac{-2}{3}\)
\(\dfrac{x}{4}=\dfrac{-3}{4}\)
⇒x=-3
b)\(\dfrac{x}{3}.\dfrac{5}{8}=\dfrac{-5}{12}\)
\(\dfrac{x}{3}=\dfrac{-5}{12}:\dfrac{5}{8}\)
\(\dfrac{x}{3}=\dfrac{-2}{3}\)
⇒x=-2
c)\(\dfrac{5}{6}.\dfrac{3}{x}=\dfrac{1}{4}\)
\(\dfrac{3}{x}=\dfrac{1}{4}:\dfrac{5}{6}\)
\(\dfrac{3}{x}=\dfrac{3}{10}\)
⇒x=10
![](https://rs.olm.vn/images/avt/0.png?1311)
`a)1/2 . [-3]/4 . [-5]/8 . [-8]/9=[1. (-3).(-5).(-8)]/[2.4.8.3.3]=[-5]/[2.4.3]=[-5]/24`
`b)(2/[1.3]+2/[3.5]+2/[5.7]).([10.13]/3-[2^2]/3-[5^3]/3)`
`=(1-1/3+1/3-1/5+1/5-1/7).[10.13-2^2-5^3]/3`
`=(1-1/7).[130-4-125]/3`
`=6/7 . 1/3 = 2/7`
____________________________________________________
`8/9+1/9 . 2/9+1/9 . 7/9`
`=8/9+1/9.(2/9+7/9)`
`=8/9+1/9 . 9/9`
`=8/9+1/9=9/9=1`
a) \(\dfrac{1}{2}\cdot\dfrac{-3}{4}\cdot\dfrac{-5}{8}\cdot\dfrac{-8}{9}\)
\(=\dfrac{1\cdot\left(-3\right)\cdot\left(-5\right)\cdot\left(-8\right)}{2\cdot4\cdot8\cdot9}\)
\(=-\dfrac{5}{24}\)
b) \(\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}\right)\cdot\left(\dfrac{10\cdot13}{3}-\dfrac{2^2}{3}-\dfrac{5^3}{3}\right)\)
\(=\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}\right)\cdot\left(\dfrac{130}{3}-\dfrac{4}{3}-\dfrac{125}{3}\right)\)
\(=\left(1-\dfrac{1}{7}\right)\cdot\dfrac{1}{3}\)
\(=\dfrac{6}{7}\cdot\dfrac{1}{3}\)
\(=\dfrac{2}{7}\)
\(\dfrac{8}{9}+\dfrac{1}{9}\cdot\dfrac{2}{9}+\dfrac{1}{9}\cdot\dfrac{7}{9}\)
\(=\dfrac{8}{9}+\dfrac{2}{81}+\dfrac{7}{81}\)
\(=\dfrac{72}{81}+\dfrac{2}{81}+\dfrac{7}{81}\)
\(=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
\(a,=\frac{2}{3}-\frac{16}{3}=\frac{-14}{3}\)
\(b,=\left(\frac{3}{7}+\frac{4}{7}\right)+\left(-\frac{6}{19}+\frac{-13}{19}\right)=1-1=0\)
\(c,=\frac{3}{5}.\left(\frac{8}{9}-\frac{7}{9}+\frac{26}{9}\right)=\frac{3}{5}.3=\frac{9}{5}\)
a,\(\dfrac{1}{2}\).\(\dfrac{4}{3}\)-\(\dfrac{20}{3}\).\(\dfrac{4}{5}\)=\(\dfrac{2}{3}\)-\(\dfrac{16}{3}\)=-\(\dfrac{14}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(6-\dfrac{14}{5}\right).\dfrac{25}{8}-\dfrac{18}{5}.\dfrac{1}{4}\)
\(=\left(\dfrac{30}{5}-\dfrac{14}{5}\right).\dfrac{25}{8}-\dfrac{18}{5}.\dfrac{1}{4}\)
\(=\dfrac{16}{5}.\dfrac{25}{8}-\dfrac{18}{20}\)
\(=\dfrac{16.25}{5.8}-\dfrac{18}{20}\)
\(=\dfrac{2.5}{1}-\dfrac{9}{10}\)
\(=10-\dfrac{9}{10}\)
\(=\dfrac{100}{10}-\dfrac{9}{10}\)
\(=\dfrac{91}{10}=9\dfrac{1}{10}\)
\(\left(6-\dfrac{14}{5}\right)\) . \(\dfrac{25}{8}\) - \(\dfrac{18}{5}\) . \(\dfrac{1}{4}\)
=\(\left(\dfrac{30}{5}-\dfrac{14}{5}\right)\) . \(\dfrac{25}{8}\) - \(\dfrac{18}{5}\) .\(\dfrac{1}{4}\)
= \(\dfrac{16}{5}\) . \(\dfrac{25}{8}\) - \(\dfrac{9}{10}\)
= \(\dfrac{16.25}{5.8}\) - \(\dfrac{9}{10}\)
= \(\dfrac{2.5}{1}\) - \(\dfrac{9}{10}\)
= 10 -\(\dfrac{9}{10}\)
= \(\dfrac{100}{10}\) - \(\dfrac{9}{10}\)
= \(\dfrac{91}{10}\) = 9,1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\dfrac{7}{8}\left(\dfrac{2}{9}-\dfrac{1}{18}+\dfrac{1}{36}-\dfrac{5}{12}\right)=\dfrac{7}{8}\cdot\dfrac{8-2+1-15}{36}\)
\(=\dfrac{7}{8}\cdot\dfrac{-8}{36}=\dfrac{-7}{36}\)
`|x/4+0,75|=5/8+2. 3/18`
`|x/4+3/4|=5/8+ 1/3`
`|x/4+3/4|=23/24`
`[(x/4+3/4=23/24),(x/4+3/4=-23\24):}`
`[(x/4=5/4),(x/4=-59/24):}`
`[(x=5),(x=-59/6):}`
`|x/4 + 0,75| = 5/8 + 2 . 3/18`
=`> |x/4 +3/4| = 5/8 + 1/3`
=>` |x/4 + 3/4| = 23/24`
Xét `x/4 + 3/4 = 23/24`
`=> x/4 = 5/24`
`=> x = 5/6`
Xét `x/4 + 3/4 = -23/24`
`=> x/4 = -41/24`
`=> x = -41/6`
Vậy `x = 5/6; -41/6`
(Chúc bạn học tốt)