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\(a,\\ \left(x+\dfrac{1}{2}\right)^3=\dfrac{8}{125}=\dfrac{2^3}{5^3}\\ \left(x+\dfrac{1}{2}\right)^3=\left(\dfrac{2}{5}\right)^3\\ \Rightarrow\left(x+\dfrac{1}{2}\right)=\dfrac{2}{5}\\ x=\dfrac{2}{5}-\dfrac{1}{2}\\ x=-\dfrac{1}{10}\)
\(b,3\left|x\right|-27=\dfrac{1}{5}\\ 3\left|x\right|=\dfrac{1}{5}+27\\ 3\left|x\right|=\dfrac{136}{5}\\ \left|x\right|=\dfrac{136}{5}:3\\ \left|x\right|=\dfrac{136}{15}\\ Vậy:x=\dfrac{136}{15}.or.x=-\dfrac{136}{15}\)
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\(-\dfrac{7}{2}-\left(\dfrac{5}{3}\right)^2:\dfrac{10}{27}\\ =-\dfrac{7}{2}-\dfrac{25}{9}:\dfrac{10}{27}\\ =-\dfrac{7}{2}-\dfrac{25}{9}.\dfrac{27}{10}\\ =-\dfrac{7}{2}-\dfrac{675}{90}\\ =-\dfrac{7}{2}-\dfrac{75}{9}\\ =-\dfrac{63}{18}-\dfrac{150}{18}\\ =-\dfrac{213}{18}\)
`a) -3/5 xx 7/9 + (-3/5) xx 2/9 + 3/5`
`=-3/5 xx (7/9+2/9) + 3/5`
`=-3/5 xx 1 + 3/5`
`=-3/5+3/5`
`=0`
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta thấy: \(\left|\dfrac{2}{5}-x\right|\ge0\forall x\)
\(\Rightarrow Q=\dfrac{9}{2}+\left|\dfrac{2}{5}-x\right|\ge\dfrac{9}{2}\forall x\)
Dấu \("="\) xảy ra khi: \(\left|\dfrac{2}{5}-x\right|=0\Leftrightarrow\dfrac{2}{5}-x=0\Leftrightarrow x=\dfrac{2}{5}\)
Vậy \(Min_Q=\dfrac{9}{2}\) khi \(x=\dfrac{2}{5}\).
\(---\)
b) Ta thấy: \(\left|x+\dfrac{2}{3}\right|\ge0\forall x\)
\(\Rightarrow M=\left|x+\dfrac{2}{3}\right|-\dfrac{3}{5}\ge-\dfrac{3}{5}\forall x\)
Dấu \("="\) xảy ra khi: \(\left|x+\dfrac{2}{3}\right|=0\Leftrightarrow x+\dfrac{2}{3}=0\Leftrightarrow x=-\dfrac{2}{3}\)
Vậy \(Min_M=-\dfrac{3}{5}\) khi \(x=-\dfrac{2}{3}\).
\(---\)
c) Ta thấy: \(\left|\dfrac{7}{4}-x\right|\ge0\forall x\)
\(\Rightarrow-\left|\dfrac{7}{4}-x\right|\le0\forall x\)
\(\Rightarrow N=-\left|\dfrac{7}{4}-x\right|-8\le-8\forall x\)
Dấu \("="\) xảy ra khi: \(\left|\dfrac{7}{4}-x\right|=0\Leftrightarrow\dfrac{7}{4}-x=0\Leftrightarrow x=\dfrac{7}{4}\)
Vậy \(Max_N=-8\) khi \(x=\dfrac{7}{4}\).
a) Ta có: \(\left|\dfrac{2}{5}-x\right|\ge0\forall x\)
\(\Rightarrow Q=\dfrac{9}{2}+\left|\dfrac{2}{5}-x\right|\ge\dfrac{9}{2}\forall x\)
Dấu "=" xảy ra khi:
\(\dfrac{2}{5}-x=0\)
\(\Rightarrow x=\dfrac{2}{5}\)
Vậy: ...
b) Ta có: \(\left|x+\dfrac{2}{3}\right|\ge0\forall x\)
\(\Rightarrow M=\left|x+\dfrac{2}{3}\right|-\dfrac{3}{5}\ge-\dfrac{3}{5}\)
Dấu "=" xảy ra:
\(x+\dfrac{2}{3}=0\)
\(\Rightarrow x=-\dfrac{2}{3}\)
Vậy: ...
c) Ta có: \(-\left|\dfrac{7}{4}-x\right|\le0\forall x\)
\(\Rightarrow N=-\left|\dfrac{7}{4}-x\right|-8\le-8\)
Dấu "=" xảy ra:
\(\dfrac{7}{4}-x=0\)
\(\Rightarrow x=\dfrac{7}{4}\)
Vậy: ...
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\Leftrightarrow-\dfrac{1}{2}x=\dfrac{1}{4}\Leftrightarrow x=-\dfrac{1}{2}\\ b,\Leftrightarrow\dfrac{1}{6}:x=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\Leftrightarrow x=\dfrac{1}{6}:\dfrac{5}{6}=\dfrac{1}{5}\\ c,\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=3\\x+\dfrac{1}{5}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{14}{5}\\x=-\dfrac{16}{5}\end{matrix}\right.\)
\(d,\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\dfrac{22}{9}-\dfrac{7}{3}=\dfrac{1}{9}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{3}\\x+\dfrac{1}{2}=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{6}\\x=-\dfrac{5}{6}\end{matrix}\right.\\ e,\Leftrightarrow2\left|x\right|=2-\dfrac{1}{2}=\dfrac{3}{2}\\ \Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{3}{2}\\2x=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
\(f,\Leftrightarrow\left|x+\dfrac{1}{2}\right|=1+\dfrac{1}{6}=\dfrac{7}{6}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{7}{6}\\x+\dfrac{1}{2}=-\dfrac{7}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)
e: ta có: \(2\left|x\right|+\dfrac{1}{2}=2\)
\(\Leftrightarrow2\left|x\right|=\dfrac{3}{2}\)
\(\Leftrightarrow\left|x\right|=\dfrac{3}{4}\)
hay \(x\in\left\{\dfrac{3}{4};-\dfrac{3}{4}\right\}\)
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a) 32.x+2=1342176728
32.x=134217728-2
32.x=134217726
x=134217726:32
x=4194303,938
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
\(\begin{array}{l}x.\frac{{14}}{{27}} = \frac{{ - 7}}{9}\\x = \frac{{ - 7}}{9}:\frac{{14}}{{27}}\\x = \frac{{ - 7}}{9}.\frac{{27}}{{14}}\\x = \frac{{ - 3}}{2}\end{array}\)
Vậy \(x = \frac{{ - 3}}{2}\).
b)
\(\begin{array}{l}\left( {\frac{{ - 5}}{9}} \right):x = \frac{2}{3}\\x = \left( {\frac{{ - 5}}{9}} \right):\frac{2}{3}\\x = \left( {\frac{{ - 5}}{9}} \right).\frac{3}{2}\\x = \frac{{ - 5}}{6}\end{array}\)
Vậy \(x = \frac{{ - 5}}{6}\).
c)
\(\begin{array}{l}\frac{2}{5}:x = \frac{1}{{16}}:0,125\\\frac{2}{5}:x = \frac{1}{{16}}:\frac{1}{8}\\\frac{2}{5}:x = \frac{1}{{16}}.8\\\frac{2}{5}:x = \frac{1}{2}\\x = \frac{2}{5}:\frac{1}{2}\\x = \frac{2}{5}.2\\x = \frac{4}{5}\end{array}\)
Vậy \(x = \frac{4}{5}\)
d)
\(\begin{array}{l} - \frac{5}{{12}}x = \frac{2}{3} - \frac{1}{2}\\ - \frac{5}{{12}}x = \frac{4}{6} - \frac{3}{6}\\ - \frac{5}{{12}}x = \frac{1}{6}\\x = \frac{1}{6}:\left( { - \frac{5}{{12}}} \right)\\x = \frac{1}{6}.\frac{{ - 12}}{5}\\x = \frac{{ - 2}}{5}\end{array}\)
Vậy \(x = \frac{{ - 2}}{5}\).
Chú ý: Khi trình bày lời giải bài tìm x, sau khi tính xong, ta phải kết luận.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(\frac{2}{5}\right)^6:\left(\frac{2}{5}\right)^4=\left(\frac{2}{5}\right)^2=\frac{4}{25}\)
\(\left(\frac{3}{16}\right)^2:\left(\frac{9}{8}\right)^2=\frac{1}{12}\)
\(\left(\frac{2}{7}-\frac{1}{2}\right)^2=\frac{9}{196}\)
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1: x=3/4-1/2=3/4-2/4=1/4
2: x-1/5=2/11
=>x=2/11+1/5=21/55
3: x-5/6=16/42-8/56
=>x-5/6=8/21-4/28=5/21
=>x=5/21+5/6=15/14
4: x/5=5/6-19/30
=>x/5=25/30-19/30=6/30=1/5
=>x=1
5: =>|x|=1/3+1/4=7/12
=>x=7/12 hoặc x=-7/12
6: x=-1/2+3/4
=>x=3/4-1/2=1/4
11: x-(-6/12)=9/48
=>x+1/2=3/16
=>x=3/16-1/2=-5/16
1)x= 1/4
2)x= 2/11+ 1/5
x= 21/55
3)x - 5/6 = 5/21
x = 5/21+5/6
x = 15/14
4)x/5 = 5/6 + -19/30
x:5 = 1/5
x = 1/5.5
x = 1
5) |x| - 1/4 = 6/18
|x| = 6/18 - 1/4
|x| =7/12
⇒x= 7/12 hoặc -7/12
6)x = -1/2 +3/4
x= 1/4
7) x/15 = 3/5 + -2/3
x:15 = -1/15
x = -1/15. 15
x = -1
8)11/8 + 13/6 = 85/x
85/24 = 85/x
⇒ x = 24
9) x - 7/8 = 13/12
x = 13/12 + 7/8
x = 47/24
10)x - -6/15 = 4/27
x = 4/27 + (-6/15)
x = -34/135
11) -(-6/12)+x = 9/48
x= 9/48 - 6/12
x = -5/16
12) x - 4/6 = 5/25 + -7/15
x -4/6 = -4/15
x = -4/15 + 4/6
x = 2/5
a) \(-\frac{2}{3}x=\left(-\frac{1}{2}\right)^2\)
\(\Leftrightarrow-\frac{2}{3}x=\frac{1}{4}\)
\(\Rightarrow x=-\frac{3}{8}\)
b) \(x\div\left(-\frac{2}{5}\right)^3=-\frac{2}{5}\)
\(\Leftrightarrow x=\left(\frac{2}{5}\right)^4\)
\(\Rightarrow x=\frac{16}{625}\)
c) \(2^x=32\)
\(\Leftrightarrow2^x=2^5\)
\(\Rightarrow x=5\)
d) \(\frac{16}{2^x}=2\)
\(\Leftrightarrow2^x=8=2^3\)
\(\Rightarrow x=3\)
e) \(\left(x-2\right)^2=9\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=3\\x-2=-3\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}\)
a, \(-\frac{2}{3}x=\left(-\frac{1}{2}\right)^2\Leftrightarrow-\frac{2}{3}x=\frac{1}{4}\Leftrightarrow x=-\frac{3}{8}\)
b, \(\frac{x}{-2,5^3}=-\frac{2}{5}\Leftrightarrow5x=\frac{125}{4}\Leftrightarrow x=\frac{25}{4}\)
c, \(2^x=32\Leftrightarrow2^x=2^5\Leftrightarrow x=5\)
d, mk chưa hiểu đề lăm
e, \(\left(x-2\right)^2=9\Leftrightarrow\orbr{\begin{cases}x-2=3\\x-2=-3\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}}\)