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12 tháng 1 2018
2, \(\Rightarrow\left\{{}\begin{matrix}\\\dfrac{5}{4}x-\dfrac{7}{2}=0\\\dfrac{5}{8}x+\dfrac{3}{5}=0\\\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{14}{5}\\\\x=\dfrac{-24}{25}\\\end{matrix}\right.\)
TN
8 tháng 6 2017
a)\(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
\(\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{2}{3}\)
\(\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{8}{12}\)
\(\dfrac{2}{5}+x=\dfrac{3}{12}\)
\(\dfrac{2}{5}+x=\dfrac{1}{4}\)
\(x=\dfrac{1}{4}-\dfrac{2}{5}\)
\(x=\dfrac{5}{20}-\dfrac{8}{20}\)
\(x=\dfrac{-3}{20}\)
b)\(2x\left(x-\dfrac{1}{7}\right)=0\)
\(\Rightarrow2x=0\) hoặc \(x-\dfrac{1}{7}=0\)
\(x=0:2\) \(x=0+\dfrac{1}{7}\)
\(x=0\) \(x=\dfrac{1}{7}\)
\(\Rightarrow x=0\) hoặc \(x=\dfrac{1}{7}\)
c)\(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)
\(\dfrac{1}{4}:x=\dfrac{8}{20}-\dfrac{15}{20}\)
\(\dfrac{1}{4}:x=\dfrac{-7}{20}\)
\(x=\dfrac{1}{4}:\dfrac{-7}{20}\)
\(x=\dfrac{1}{4}.\dfrac{-20}{7}\)
x= \(\dfrac{1.\left(-5\right)}{1.7}\)
\(x=\dfrac{-5}{7}\)
PT
2 tháng 10 2017
a) \(\left|\dfrac{5}{3}x\right|=\left|\dfrac{1}{6}\right|\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{5}{3}x=\dfrac{-1}{6}\\\dfrac{5}{3}x=\dfrac{1}{6}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1}{10}\\x=\dfrac{1}{10}\end{matrix}\right.\)
b) \(\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|-\dfrac{3}{4}=\left|-\dfrac{3}{4}\right|\)
\(\Rightarrow\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|-\dfrac{3}{4}=\dfrac{3}{4}\)
\(\Rightarrow\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|=\dfrac{3}{2}\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}x-\dfrac{3}{4}=\dfrac{-3}{2}\\\dfrac{3}{4}x-\dfrac{3}{4}=\dfrac{3}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}x=-\dfrac{3}{4}\\\dfrac{3}{4}x=\dfrac{9}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)
c) \(\left|x+\dfrac{3}{5}\right|=\left|x-\dfrac{7}{3}\right|\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{3}{5}=x-\dfrac{7}{3}\\x+\dfrac{3}{5}=-x+\dfrac{7}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x\in\varnothing\\x=\dfrac{13}{15}\end{matrix}\right.\)
19 tháng 11 2022
a: =>1/6x=-49/60
=>x=-49/60:1/6=-49/60*6=-49/10
b: =>3/2x-1/5=3/2 hoặc 3/2x-1/5=-3/2
=>x=17/15 hoặc x=-13/15
c: =>1,25-4/5x=-5
=>4/5x=1,25+5=6,25
=>x=125/16
d: =>2^x*17=544
=>2^x=32
=>x=5
i: =>1/3x-4=4/5 hoặc 1/3x-4=-4/5
=>1/3x=4,8 hoặc 1/3x=-0,8+4=3,2
=>x=14,4 hoặc x=9,6
j: =>(2x-1)(2x+1)=0
=>x=1/2 hoặc x=-1/2
AT
20 tháng 6 2017
a/dễ --> tự lm
b/ \(\left(x-\dfrac{4}{7}\right)\left(1\dfrac{3}{5}+2x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{5}=0\\1\dfrac{3}{5}+2x=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\2x=\dfrac{8}{5}\Rightarrow x=\dfrac{4}{5}\end{matrix}\right.\)
Vậy...............
c/ \(\left(x-\dfrac{4}{7}\right):\left(x+\dfrac{1}{2}\right)>0\)
TH1: \(\left\{{}\begin{matrix}x-\dfrac{4}{7}>0\\x+\dfrac{1}{2}>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x>\dfrac{4}{7}\\x>-\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow x>\dfrac{4}{7}\)
TH2: \(\left\{{}\begin{matrix}x-\dfrac{4}{7}< 0\\x+\dfrac{1}{2}< 0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x< \dfrac{4}{7}\\x< -\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow x< -\dfrac{1}{2}\)
Vậy \(x>\dfrac{4}{7}\) hoặc \(x< -\dfrac{1}{2}\) thì thỏa mãn đề
d/ \(\left(2x-3\right):\left(x+1\dfrac{3}{4}\right)< 0\)
TH1: \(\left\{{}\begin{matrix}2x-3>0\\x+1\dfrac{3}{4}< 0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x>1,5\\x< -\dfrac{7}{4}\end{matrix}\right.\)(vô lý)
TH2: \(\left\{{}\begin{matrix}2x-3< 0\\x+1\dfrac{3}{4}>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x< 1,5\\x>-\dfrac{7}{4}\end{matrix}\right.\)\(\Rightarrow-\dfrac{7}{4}< x< 1,5\)
Vậy...................
22 tháng 5 2022
b: 2x-3<0
=>2x<3
hay x<3/2
c: \(\left(2x-4\right)\left(9-3x\right)>0\)
=>(x-2)(x-3)<0
=>2<x<3
d: \(\dfrac{2}{3}x-\dfrac{3}{4}>0\)
=>2/3x>3/4
hay x>9/8
27 tháng 6 2017
a, \(\dfrac{3}{4}+x=\dfrac{8}{13}\)
\(x=\dfrac{8}{13}-\dfrac{3}{4}\)
\(x=-\dfrac{7}{52}\)
b,\(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
\(\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{2}{3}\)
\(\dfrac{2}{5}+x=\dfrac{1}{4}\)
\(x=\dfrac{1}{4}-\dfrac{2}{5}\)
\(x=-\dfrac{3}{20}\)
c, \(2x\left(x-\dfrac{1}{7}\right)=0\)
\(2x-\dfrac{1}{7}=0\)
\(x-\dfrac{1}{7}=0:2\)
\(x-\dfrac{1}{7}=0\)
\(x=0-\dfrac{1}{7}\)
\(x=\dfrac{1}{7}\)
d, \(\dfrac{3}{4}+\dfrac{1}{4}\div x=\dfrac{2}{5}\)
\(\left(\dfrac{3}{4}+\dfrac{1}{4}\right):x=\dfrac{2}{5}\)
\(1:x=\dfrac{2}{5}\)
\(x=1:\dfrac{2}{5}\)
\(x=\dfrac{5}{2}\)
27 tháng 6 2017
a) \(\dfrac{3}{4}+x=\dfrac{8}{13}\)\(\Leftrightarrow\) \(x=\dfrac{8}{13}-\dfrac{3}{4}=\dfrac{-7}{52}\) vậy \(x=\dfrac{-7}{52}\)
b) \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\) \(\Leftrightarrow\) \(\dfrac{11}{12}-\dfrac{2}{5}-x=\dfrac{2}{3}\) \(\Leftrightarrow\) \(x=\dfrac{11}{12}-\dfrac{2}{5}-\dfrac{2}{3}=\dfrac{-3}{20}\) vậy \(x=\dfrac{-3}{20}\)
c) \(2x\left(x-\dfrac{1}{7}\right)=0\) \(\Leftrightarrow\) \(2x^2-\dfrac{2}{7}x=0\)
\(\Delta\) = \(\left(\dfrac{-2}{7}\right)^2-4.2.0=\dfrac{4}{49}>0\)
\(\Rightarrow\) phương trình có 2 nghiệm phân biệt
\(x_1=\dfrac{\dfrac{2}{7}+\sqrt{\dfrac{4}{49}}}{4}=\dfrac{1}{7}\)
\(x_2=\dfrac{\dfrac{2}{7}-\sqrt{\dfrac{4}{49}}}{4}=0\)
vậy \(x=0;x=\dfrac{1}{7}\)
6 tháng 1 2018
a/ \(\dfrac{x+1}{2}=\dfrac{2x+3}{5}\)
\(\Leftrightarrow5\left(x+1\right)=2\left(2x+3\right)\)
\(\Leftrightarrow5x+5=4x+6\)
\(\Leftrightarrow5x-4x=6-5\)
\(\Leftrightarrow x=1\left(tm\right)\)
Vậy ...
b/ \(\left|x-1\right|+3\left|y+1\right|+\left|z+2\right|=0\)
Mà với \(\forall x;y;z\) ta có :
\(\left\{{}\begin{matrix}\left|x-1\right|\ge0\\3\left|y+1\right|\ge0\\\left|z+2\right|\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left|x-1\right|=0\\3\left|y+1\right|=0\\\left|z+2\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y+1=0\\z+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\\z=-2\end{matrix}\right.\)
Vậy ...
c/ \(\dfrac{x-2}{4}=\dfrac{5-3x}{4}\)
\(\Leftrightarrow x-2=5-3x\)
\(\Rightarrow x+3x=5+2\)
\(\Leftrightarrow4x=7\)
\(\Leftrightarrow x=\dfrac{7}{4}\)
Vậy ......
d/ \(\dfrac{x+2}{4}=\dfrac{4}{x+2}\)
\(\Leftrightarrow\left(x+2\right)\left(x+2\right)=16\)
\(\Leftrightarrow\left(x+2\right)^2=4^2=\left(-4\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)
Vậy ...
e/ \(\dfrac{x-1}{5}=\dfrac{-20}{x-1}\)
\(\Leftrightarrow\left(x-1\right)\left(x-1\right)=-100\)
\(\Leftrightarrow\left(x-1\right)^2=-100\)
Lại có : \(\left(x-1\right)^2\ge0\)
\(\Leftrightarrow\) k tồn tại x
DH
26 tháng 6 2017
a, \(\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{2}{5}\right)>0\)
\(\Rightarrow\left\{{}\begin{matrix}x-\dfrac{1}{3}>0\\x+\dfrac{2}{5}>0\end{matrix}\right.\) hay \(\left\{{}\begin{matrix}x-\dfrac{1}{3}< 0\\x+\dfrac{2}{5}< 0\end{matrix}\right.\)
+,Xét \(\left\{{}\begin{matrix}x-\dfrac{1}{3}>0\\x+\dfrac{2}{5}>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x>\dfrac{1}{3}\\x>-\dfrac{2}{5}\end{matrix}\right.\)
\(\Rightarrow x>\dfrac{1}{3}\)
+, Xét \(\left\{{}\begin{matrix}x-\dfrac{1}{3}< 0\\x+\dfrac{2}{5}< 0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x< \dfrac{1}{3}\\x< -\dfrac{2}{5}\end{matrix}\right.\)
\(\Rightarrow x< -\dfrac{2}{5}\)
Vậy...........
b, \(\left(x+\dfrac{3}{5}\right)\left(x+1\right)< 0\)
Vì \(x+\dfrac{3}{5}< x+1\) với mọi \(x\in R\)
\(\Rightarrow\left\{{}\begin{matrix}x+\dfrac{3}{5}< 0\\x+1>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x< -\dfrac{3}{5}\\x>-1\end{matrix}\right.\)
Vậy...........
c, \(\dfrac{3}{7}x-\dfrac{2}{5}x=\dfrac{-17}{35}\)
\(\Rightarrow\dfrac{1}{35}x=\dfrac{-17}{35}\)
\(\Rightarrow x=-17\)
d, \(\left(\dfrac{3}{4}x-\dfrac{9}{10}\right)\left(\dfrac{1}{3}+\dfrac{-3}{5}x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}x-\dfrac{9}{10}=0\\\dfrac{1}{3}+\dfrac{-3}{5}x=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}x=\dfrac{9}{10}\\-\dfrac{3}{5}x=-\dfrac{1}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{6}{5}\\x=\dfrac{5}{9}\end{matrix}\right.\)
Vậy.........
Chúc bạn học tốt!!!
AT
26 tháng 6 2017
a/ \(\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{2}{5}\right)>0\)
TH1:\(\left\{{}\begin{matrix}x-\dfrac{1}{3}>0\\x+\dfrac{2}{5}>0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x>\dfrac{1}{3}\\x>-\dfrac{2}{5}\end{matrix}\right.\)\(\Rightarrow x>\dfrac{1}{3}\)
TH2:\(\left\{{}\begin{matrix}x-\dfrac{1}{3}< 0\\x+\dfrac{2}{5}< 0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x< \dfrac{1}{3}\\x< -\dfrac{2}{5}\end{matrix}\right.\)\(\Rightarrow x< -\dfrac{2}{5}\)
Vậy \(x>\dfrac{1}{3}\) hoặc \(x< -\dfrac{2}{5}\) thì tm
b/ \(\left(x+\dfrac{3}{5}\right)\left(x+1\right)< 0\)
TH1:\(\left\{{}\begin{matrix}x+\dfrac{3}{5}< 0\\x+1>0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x< -\dfrac{3}{5}\\x>-1\end{matrix}\right.\) \(\Rightarrow-1< x< -\dfrac{3}{5}\)
TH2:\(\left\{{}\begin{matrix}x+\dfrac{3}{5}>0\\x+1< 0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x>-\dfrac{3}{5}\\x< -1\end{matrix}\right.\)(vô lý)
Vậy....................
c/ \(\dfrac{3}{7}x-\dfrac{2}{5}x=-\dfrac{17}{35}\)
\(\Rightarrow\left(\dfrac{3}{7}-\dfrac{2}{5}\right)x=-\dfrac{17}{35}\)
\(\Rightarrow\dfrac{1}{35}x=-\dfrac{17}{35}\)
\(\Rightarrow x=-\dfrac{17}{35}:\dfrac{1}{35}=-17\)
Vậy.............
d/ \(\left(\dfrac{3}{4}x-\dfrac{9}{10}\right)\left(\dfrac{1}{3}+\dfrac{-3}{5}x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}x-\dfrac{9}{10}=0\\\dfrac{1}{3}-\dfrac{3}{5}x=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}x=\dfrac{9}{10}\\\dfrac{3}{5}x=\dfrac{1}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{6}{5}\\x=\dfrac{5}{9}\end{matrix}\right.\)
Vậy.....................
26 tháng 5 2022
a: 2x(x-1/7)=0
=>x(x-1/7)=0
=>x=0 hoặc x=1/7
b: \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}=\dfrac{8}{20}-\dfrac{15}{20}=\dfrac{-7}{20}\)
nên \(x=\dfrac{-1}{4}:\dfrac{7}{20}=\dfrac{-20}{4\cdot7}=\dfrac{-5}{7}\)
c: \(\Leftrightarrow\dfrac{41}{9}:\dfrac{41}{18}-7< x< \left(3.2:3.2+\dfrac{45}{10}\cdot\dfrac{31}{45}\right):\left(-21.5\right)\)
\(\Leftrightarrow2-7< x< \dfrac{\left(1+3.1\right)}{-21.5}\)
\(\Leftrightarrow-5< x< \dfrac{-41}{215}\)
mà x là số nguyên
nên \(x\in\left\{-4;-3;-2;-1\right\}\)
Lời giải:
$|x-\frac{4}{5}|=0+\frac{3}{5}=\frac{3}{5}$
$\Rightarrow x-\frac{4}{5}=\frac{3}{5}$ hoặc $x-\frac{4}{5}=\frac{-3}{5}$
$\Rightarrow x=\frac{7}{5}$ hoặc $x=\frac{1}{5}$
\(\left|x-\dfrac{4}{5}\right|\) \(-\dfrac{3}{5}=0\)
\(\left|x-\dfrac{4}{5}\right|\) \(=0+\dfrac{3}{5}=\dfrac{3}{5}\)
trường hợp 1 :
\(x-\dfrac{4}{5}=\dfrac{3}{5}\)
\(x=\dfrac{3}{5}+\dfrac{4}{5}\) \(=\dfrac{7}{5}\)
vậy \(x\) \(=\dfrac{7}{5}\)
trường hợp 2 :
\(x-\dfrac{4}{5}=\dfrac{-3}{5}\)
\(x=\dfrac{-3}{5}+\dfrac{4}{5}\)\(=\dfrac{1}{5}\)
vậy \(x=\dfrac{1}{5}\)