Tìm x biết: 31/9|x|-5/2=8/3
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17 tháng 2 2022
\(\Leftrightarrow-\dfrac{16}{279}< \dfrac{x}{9}< =\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{x}{9}=0\)
hay x=0
6 tháng 8 2016
3/2+5/4+9/8/+17/16+33/32-6+x-1/x+1=31/32-2/2015
=(1+1/2)+(1+1/4)+(1+1/8)+(1+1/16)+(1+1/32-6+x-1/x+1=31/32-2/2015
=(1/2+1/4+1/8+1/16+1/32)+(1+1+1+1+1)-6+x-1/x+1=31/32-2/2015
=31/32+5-6+x-1/x+1=31/32-2/2015
=5-6+x-1/x+1=31/32-2/2015-31/32
=-1+x-1/x+1=-2/2015
=x-1/x+1=-2/2015- -1
=x-1/x+1=2013/2015
=>x=2014
PT
1
19 tháng 4 2020
\(\frac{31}{9}\left|x\right|-\frac{5}{2}=\frac{8}{3}\)
=> \(\frac{31}{9}\left|x\right|=\frac{8}{3}+\frac{5}{2}\)
=> \(\frac{31}{9}\left|x\right|=\frac{16}{6}+\frac{15}{6}\)
=> \(\frac{31}{9}\left|x\right|=\frac{31}{6}\)
=> \(\left|x\right|=\frac{31}{6}:\frac{31}{9}=\frac{31}{6}\cdot\frac{9}{31}=\frac{9}{6}=\frac{3}{2}\)
=> x = 3/2 hoặc x = -3/2
DL
2 tháng 7 2018
a) \(\left(2x+3\right)^3=\left(2x+3\right)^8\)
TH1 \(2x+3=1\)
\(2x=1-3=-2\)
\(x=-1\)
TH2 \(2x+3=0\)
\(2x=-3\Rightarrow x=-\frac{3}{2}\)
b) ? sai đề
c) \(\left|5-3\right|=\left|11+2x\right|\Rightarrow\left|2\right|=\left|11+2x\right|\)
\(\hept{\begin{cases}11+2x=-2\\11+2x=2\end{cases}\Rightarrow}\hept{\begin{cases}2x=13\\2x=9\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{13}{2}\\x=\frac{9}{2}\end{cases}}\)
d) \(\left(x-5\right)^4=\left(x-5\right)^6\Rightarrow\hept{\begin{cases}x-5=0\\x-5=1\end{cases}}\Rightarrow\hept{\begin{cases}x=5\\x=6\end{cases}}\)
T
7 tháng 5 2020
5/17 + -4/9 + -20/31 + -11/31 < x < -3/7 + 7/15 + 4/-7 + 8/15 + 2/3
-175/153<x\(\le\)2/3
-175/153<x\(\le\)102/153
suy ra -175<x\(\le\)153
vậy x bằng (-174/153,-173/153,-172/153,.........,101/153,102/153
14 tháng 6 2023
a: =7/8:(2/9-18+1/36)-5/12
=-7/142-5/12=-397/852
b: =3/7(4/9+5/9:6/12)=2/3
c: =5^8(16/31-47/31)+1/3=-5^8+1/3
d: =7/2(3/8+5/8:4/15)=609/64
Đề:
Giài:
\(\frac{31}{9}\left|x\right|=\frac{8}{3}+\frac{5}{2}\)
\(\frac{31}{9}\left|x\right|=\frac{31}{6}\)
\(\left|x\right|=\frac{31}{6}:\frac{31}{9}\)
\(\left|x\right|=\frac{3}{2}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{-3}{2}\end{cases}}\)
Vậy x = \(\frac{3}{2}\)hoặc x = \(\frac{-3}{2}\)
\(\frac{31}{9}\left|x\right|-\frac{5}{2}=\frac{8}{3}\)
\(\frac{31}{9}\left|x\right|=\frac{8}{3}+\frac{5}{2}\)
\(\frac{31}{9}\left|x\right|=\frac{8\cdot2}{6}+\frac{5\cdot3}{6}\)
\(\frac{31}{9}\left|x\right|=\frac{16}{6}+\frac{15}{6}\)
\(\frac{31}{9}\left|x\right|=\frac{31}{6}\)
\(\left|x\right|=\frac{31}{6}:\frac{31}{9}\)
\(\left|x\right|=\frac{31}{6}\cdot\frac{9}{31}\)
\(\left|x\right|=\frac{1}{2}\cdot\frac{3}{1}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{3}{2}\\x=-\frac{3}{2}\end{cases}}\)
Vậy \(x\in\left\{\frac{3}{2};-\frac{3}{2}\right\}\)