K
Khách
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KH
28 tháng 2 2018
a) \(x=\dfrac{1}{4}+\dfrac{2}{13}\)
\(x=\dfrac{13}{52}+\dfrac{8}{52}\)
⇒ \(x=\dfrac{21}{52}\)
b) \(\dfrac{x}{3}=\dfrac{2}{3}+\dfrac{-1}{7}\)
\(\dfrac{x}{3}=\dfrac{14}{21}+\dfrac{-3}{21}\)
\(\dfrac{x}{3}=\dfrac{11}{21}\)
⇒ \(x=\dfrac{11.3}{21}=\dfrac{33}{21}\)
⇒ \(x=\dfrac{11}{7}\)
c) \(\dfrac{-8}{3}+\dfrac{1}{3}< x< \dfrac{-2}{7}+\dfrac{-5}{7}\)
\(\dfrac{-17}{7}< x< -1\)
⇒ \(-17< x< -7\)
⇒ \(x\in\left\{-16;-15,....;-6\right\}\)
d) \(\dfrac{1}{6}+\dfrac{2}{5}\)
\(=\dfrac{5}{30}+\dfrac{12}{30}\)
\(=\dfrac{17}{30}\)
e) \(\dfrac{3}{5}+\dfrac{-7}{4}\)
\(=\dfrac{12}{20}+\dfrac{-35}{20}\)
\(=\dfrac{-23}{20}\)
f) \(\dfrac{4}{13}+\dfrac{-12}{30}\)
\(=\dfrac{4}{13}+\dfrac{-2}{5}\)
\(=\dfrac{20}{65}+\dfrac{-26}{65}\)
\(=\dfrac{-6}{65}\)
g) \(\dfrac{-3}{29}+\dfrac{16}{58}\)
\(=\dfrac{-6}{58}+\dfrac{16}{58}\)
\(=\dfrac{10}{58}\)
h) \(\dfrac{8}{40}+\dfrac{-36}{45}\)
\(=\dfrac{1}{5}+\dfrac{-4}{5}\)
\(=\dfrac{-3}{5}\)
j) \(\dfrac{-8}{18}+\dfrac{15}{27}\)
\(=\dfrac{-2}{9}+\dfrac{5}{9}\)
\(=\dfrac{3}{9}\)
\(=\dfrac{1}{3}\)
2
8 tháng 5 2022
`[-5]/6+8/3+[-29]/6 <= x <= [-1]/2+2+5/2`
`[-5]/6+16/6+[-29]/6 <= [6x]/6 <= [-3]/6+12/6+15/6`
`-5+16-29 <= 6x <= -3+12+15`
`-18 <= 6x <= 24`
`-18:6 <= 6x:6 <= 24:6`
`-3 <= x <= 4`
Vậy `-3 <= x <= 4`
NM
Nguyễn Minh Quang
Giáo viên
8 tháng 3 2021
ta có
\(-\frac{5}{6}+\frac{8}{3}+\frac{29}{-6}=\frac{-5+16-29}{6}=-\frac{18}{6}=-3\)
\(-\frac{1}{2}+2+\frac{5}{2}=2+2=4\)
vì vậy \(-3< x< 4\Rightarrow x\in\left\{-2,-1,0,1,2,3\right\}\)
NT
0
NL
1
11 tháng 5 2020
\(\frac{-5}{6}+\frac{8}{3}+\frac{29}{-6}< x\le\frac{-1}{2}+2+\frac{5}{12}\)
\(-3< x\le\frac{23}{12}\)
\(-3< x< 2\)
\(x\in\left\{-2;-1;0;1\right\}\)
21 tháng 2 2021
1: Ta có: 7x+6(3-x)=27-20+73
\(\Leftrightarrow7x+18-6x=80\)
\(\Leftrightarrow x=80-18=62\)
Vậy: x=62
2: Ta có: \(6x-5\left(x-7\right)=\left(27-514\right)-486-73\)
\(\Leftrightarrow6x-5x+35=27-514-486-73\)
\(\Leftrightarrow x+35=-1046\)
\(\Leftrightarrow x=-1081\)
Vậy: x=-1081
Ta có: \(\dfrac{-5}{6}+\dfrac{8}{3}+\dfrac{29}{-6}\le x\le\dfrac{1}{2}+2+\dfrac{5}{2}\)
\(\Leftrightarrow\dfrac{-5}{6}+\dfrac{16}{6}+\dfrac{-29}{6}\le x\le3+2\)
\(\Leftrightarrow-3\le x\le5\)
\(\Leftrightarrow x\in\left\{-3;-2;-1;0;1;2;3;4;5\right\}\)