![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(x+\dfrac{1}{3}=\dfrac{2}{6}\)
\(\Leftrightarrow x+\dfrac{1}{3}=\dfrac{1}{3}\)
hay x=0
Vậy: x=0
b) Ta có: \(x-\dfrac{1}{4}=\dfrac{1}{-2}\)
\(\Leftrightarrow x-\dfrac{1}{4}=\dfrac{-1}{2}\)
\(\Leftrightarrow x=\dfrac{-1}{2}+\dfrac{1}{4}=\dfrac{-2}{4}+\dfrac{1}{4}=\dfrac{-1}{4}\)
Vậy: \(x=-\dfrac{1}{4}\)
c) Ta có: \(\dfrac{-1}{6}=\dfrac{3}{2}x\)
\(\Leftrightarrow x=\dfrac{-1}{6}:\dfrac{3}{2}=\dfrac{-1}{6}\cdot\dfrac{2}{3}\)
hay \(x=\dfrac{-1}{9}\)
Vậy: \(x=\dfrac{-1}{9}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(x\) \(\times\) \(\dfrac{1}{2}\) - \(\dfrac{3}{4}\) = \(\dfrac{5}{6}\)
\(x\) \(\times\) \(\dfrac{1}{2}\) = \(\dfrac{5}{6}\) + \(\dfrac{3}{4}\)
\(x\) \(\times\) \(\dfrac{1}{2}\) = \(\dfrac{19}{12}\)
\(x\) = \(\dfrac{19}{12}\) : \(\dfrac{1}{2}\)
\(x\) = \(\dfrac{19}{6}\)
b, \(x\) : \(\dfrac{1}{2}\) - \(\dfrac{3}{4}\) = \(\dfrac{5}{6}\)
\(x\): \(\dfrac{1}{2}\) = \(\dfrac{5}{6}\) + \(\dfrac{3}{4}\)
\(x\) : \(\dfrac{1}{2}\) = \(\dfrac{19}{12}\)
\(x\) = \(\dfrac{19}{12}\) \(\times\) \(\dfrac{1}{2}\)
\(x\) = \(\dfrac{19}{24}\)
c, \(x\) \(\times\) \(\dfrac{3}{4}\) + \(x\) \(\times\) \(\dfrac{1}{4}\) = \(\dfrac{7}{8}\)
\(x\) \(\times\) ( \(\dfrac{3}{4}\) + \(\dfrac{1}{4}\)) = \(\dfrac{7}{8}\)
\(x\) \(\times\) 1 = \(\dfrac{7}{8}\)
\(x\) = \(\dfrac{7}{8}\)
d, \(x\times\) \(\dfrac{3}{4}\) - \(x\) \(\times\) \(\dfrac{1}{4}\) = \(\dfrac{7}{8}\)
\(x\) \(\times\) ( \(\dfrac{3}{4}\) - \(\dfrac{1}{4}\)) = \(\dfrac{7}{8}\)
\(x\) \(\times\) \(\dfrac{1}{2}\) = \(\dfrac{7}{8}\)
\(x\) = \(\dfrac{7}{8}\) : \(\dfrac{1}{2}\)
\(x\) = \(\dfrac{7}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 6: Khôg có cau nào đúng
Câu 7: C
Câu 8: B
Câu 9: B
Câu 10: D
![](https://rs.olm.vn/images/avt/0.png?1311)
nhìn cái đề con hơi bị ''sốc'' , thế này ạ ???
Sửa đề \(4+\frac{1}{3}x\left(\frac{1}{6}-\frac{1}{2}\right)\le x\le\frac{2}{3}x\left(\frac{1}{3}-\frac{1}{2}-\frac{3}{4}\right)\)
\(4+\frac{1}{3}x\left(-\frac{1}{3}\right)\le x\le\frac{2}{3}x\left(-\frac{11}{12}\right)\)
\(4-\frac{1}{9}x\le x\le-\frac{11}{18}x\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\frac{1}{2}-\left(\frac{1}{3}+\frac{1}{4}\right)< x< \frac{1}{48}-\left(\frac{1}{16}-\frac{1}{6}\right)\)
=) \(\frac{-1}{12}< x< \frac{1}{8}\)
Vì \(\frac{-1}{12}< 0;\frac{1}{8}>0\)và \(< 1\)
mà x là số nguyên =) \(x=0\)
b) \(\frac{3}{4}-\frac{5}{6}\le\frac{x}{12}< 1-\left(\frac{2}{3}-\frac{1}{4}\right)\)
=) \(\frac{-1}{12}\le\frac{x}{12}< \frac{7}{12}\)
=) \(-1\le x< 7\)=) \(x=\left\{-1;0;1;2...;6\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1: =>x=3/5-1/5=2/5
b: =>x/3=5/8+1/8=3/4
=>x=9/4
3: =>10/3x=3+1/4+6+3/4=10
=>x=10:10/3=3
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 9,
62x73+36x33=36x73+36x27=36(73+27)=36x100=3600.
197-\([\)6x(5-1)2+20220\(]\):5=197-\([\)6x16+1\(]\):5=197-97:5=197-97/5=888/5.
Bài 10,
21-4x=13
=>4x=21-13=8
=>x=8:4=2.
30:(x-3)+1=45:43=42=16
=>30:(x-3)=16-1=15
=>x-3=30:15=2
=>x=2+3=5.
(x-1)3+5x6=38
=>(x-1)3+30=38
=>(x-1)3=38-30=8=23
=>x-1=2
=>x=3.
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a: \(x=\dfrac{2}{3}:\dfrac{3}{5}=\dfrac{2}{3}\cdot\dfrac{5}{3}=\dfrac{10}{9}\)
b: \(x=\dfrac{17}{8}:\dfrac{7}{17}=\dfrac{17}{8}\cdot\dfrac{17}{7}=\dfrac{289}{56}\)
c: \(x=-\dfrac{3}{4}:\dfrac{7}{12}=\dfrac{-3}{4}\cdot\dfrac{12}{7}=\dfrac{-63}{28}=-\dfrac{9}{4}\)
d: \(\Leftrightarrow x\cdot\dfrac{1}{6}=\dfrac{3}{8}-\dfrac{1}{4}=\dfrac{1}{4}\)
hay \(x=\dfrac{1}{4}:\dfrac{1}{6}=\dfrac{3}{2}\)
e: \(\Leftrightarrow\dfrac{1}{2}:x=-4-\dfrac{1}{3}=-\dfrac{17}{3}\)
hay \(x=-\dfrac{1}{2}:\dfrac{17}{3}=\dfrac{-3}{34}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
d,
\(|x-\frac{1}{3}|=\frac{5}{6}\Rightarrow \left[\begin{matrix} x-\frac{1}{3}=\frac{5}{6}\\ x-\frac{1}{3}=-\frac{5}{6}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{7}{6}\\ x=\frac{-1}{2}\end{matrix}\right.\)
e,
\(\frac{3}{4}-2|2x-\frac{2}{3}|=2\)
\(\Leftrightarrow 2|2x-\frac{2}{3}|=\frac{3}{4}-2=\frac{-5}{4}\)
\(\Leftrightarrow |2x-\frac{2}{3}|=-\frac{5}{8}<0\) (vô lý vì trị tuyệt đối của 1 số luôn không âm)
Vậy không tồn tại $x$ thỏa mãn đề bài.
f,
\(\frac{2x-1}{2}=\frac{5+3x}{3}\Leftrightarrow 3(2x-1)=2(5+3x)\)
\(\Leftrightarrow 6x-3=10+6x\)
\(\Leftrightarrow 13=0\) (vô lý)
Vậy không tồn tại $x$ thỏa mãn đề bài.
a,
$0-|x+1|=5$
$|x+1|=0-5=-5<0$ (vô lý do trị tuyệt đối của một số luôn không âm)
Do đó không tồn tại $x$ thỏa mãn điều kiện đề.
b,
\(2-|\frac{3}{4}-x|=\frac{7}{12}\)
\(|\frac{3}{4}-x|=2-\frac{7}{12}=\frac{17}{12}\)
\(\Rightarrow \left[\begin{matrix} \frac{3}{4}-x=\frac{17}{12}\\ \frac{3}{4}-x=\frac{-17}{12}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{-2}{3}\\ x=\frac{13}{6}\end{matrix}\right.\)
c,
\(2|\frac{1}{2}x-\frac{1}{3}|-\frac{3}{2}=\frac{1}{4}\)
\(2|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{4}\)
\(|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{8}\)
\(\Rightarrow \left[\begin{matrix} \frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\ \frac{1}{2}x-\frac{1}{3}=-\frac{7}{8}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{29}{12}\\ x=\frac{-13}{12}\end{matrix}\right.\)
x = \(\frac{-1}{18}\)