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DH
19 tháng 6 2017
a, \(\left(2x-3\right)\left(\dfrac{3}{4}x+1\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}2x-3=0\\\dfrac{3}{4}x+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x=3\\\dfrac{3}{4}x=-1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{-4}{3}\end{matrix}\right.\)
Vậy......
b, \(\left(5x-1\right)\left(2x-\dfrac{1}{3}\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}5x-1=0\\2x-\dfrac{1}{3}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x=1\\2x=\dfrac{1}{3}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{1}{6}\end{matrix}\right.\)
Vậy......
c, \(\dfrac{3}{7}+\dfrac{1}{7}:x=\dfrac{3}{14}\)
\(\Rightarrow\dfrac{1}{7}:x=\dfrac{-3}{14}\)
\(\Rightarrow x=\dfrac{1}{7}:\dfrac{-3}{14}=\dfrac{-2}{3}\)
Vậy.....
Chúc bạn học tốt!!!
19 tháng 6 2017
a,(2x-3)(\(\dfrac{3}{4}\)x+1)=0
*2x-3=0\(\rightarrow\)2x=3\(\rightarrow\)x=\(\dfrac{3}{2}\)
*\(\dfrac{3}{4}\)x+1=0\(\rightarrow\)\(\dfrac{3}{4}\)x=\(-1\)\(\rightarrow\)x=\(\dfrac{-3}{4}\)
Vậy x\(\in\){\(\dfrac{-3}{4};\dfrac{3}{2}\)}
b,(5x-1)(2x-1/3)=0
*5x-1=0\(\rightarrow\)5x=1\(\rightarrow\)x=1/5
*2x-1/3=0\(\rightarrow\)2x=1/3\(\rightarrow\)x=1/6
Vậy x\(\in\){1/5;1/6}
c,3/7+1/7:x=3/14
1/7:x=-3/14
x=-14/21
Vậy x=-14/21
N
27 tháng 7 2017
a) \(\frac{3-2x}{5}=\frac{2}{7}\)
\(\Rightarrow7.\left(3-2x\right)=2.5\)
\(\Rightarrow21-14x=10\)
\(\Rightarrow14x=11\)
\(\Rightarrow x=\frac{11}{14}\)
27 tháng 7 2017
b) ( 5x - 6 ) : 7 = \(4\frac{1}{2}+0,25\%\)
( 5x - 6 ) : 7 = \(\frac{19}{4}\)
5x - 6 = \(\frac{19}{4}\). 7
5x - 6 = \(\frac{133}{4}\)
5x = \(\frac{133}{4}\)+ 6
5x = \(\frac{157}{4}\)
x = \(\frac{157}{4}\): 5
x = \(\frac{157}{20}\)
16 tháng 7 2019
\(\left(x+1\right)\left(x+7\right)< 0\)
thì \(x+1;x+7\)khác dấu
th1\(\hept{\begin{cases}x+1< 0\\x+7>0\end{cases}\Leftrightarrow\hept{\begin{cases}x< -1\\x>-7\end{cases}\Rightarrow}-7< x< -1\left(tm\right)}\)
th2\(\hept{\begin{cases}x+1>0\\x+7< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x>-1\\x< -7\end{cases}\Rightarrow}-1< x< -7\left(vl\right)}\)
vậy với\(-7< x< -1\)thì \(\left(x+1\right)\left(x+7\right)< 0\)
16 tháng 7 2019
a) (2x - 3) = 5
<=> 2x - 3 = 5
<=> 2x = 5 + 3
<=> 2x = 8
<=> x = 4
=> x = 4
b) (5x - 3) = 1/2
<=> 5x - 3 = 1/2
<=> 5x = 1/2 + 3
<=> 5x = 7/2
<=> x = 7/10
=> x = 7/10
c) (x + 1)(x + 7) < 0
<=> x = -1; -7
<=> x < -7 <=> x = -8 <=> (-8 + 1)(-8 + 7) < 0 <=> 7 < 0 (loại)
<=> -7 < x < -1 <=> x = -6 <=> (-6 + 1)(-6 + 7) < 0 <=> -5 < 0 (nhận)
<=> x > -1 <=> x = 0 <=> (x + 1)(x + 7) < 0 <=> 7 < 0 (loại)
Vậy: -7 < x < -1
25 tháng 5 2022
a: \(\Leftrightarrow\dfrac{2}{3}x-\dfrac{5}{6}=\dfrac{-3}{2}x+\dfrac{3}{4}\)
=>13/6x=3/4+5/6
=>13/6x=9/12+10/12=19/12
hay x=19/26
b: \(\left(5x-3\right)\left(2x+5\right)=0\)
=>5x-3=0 hoặc 2x+5=0
=>x=3/5 hoặc x=-5/2
c: \(\left(\dfrac{5}{6}:x-\dfrac{5}{4}\right)^4=\dfrac{81}{16}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{6}:x-\dfrac{5}{4}=\dfrac{3}{2}\\\dfrac{5}{6}:x-\dfrac{5}{4}=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{6}:x=\dfrac{11}{4}\\\dfrac{5}{6}:x=-\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{10}{33}\\x=-\dfrac{10}{3}\end{matrix}\right.\)
d: \(\left|\dfrac{2}{5}x-\dfrac{1}{5}\right|\cdot\dfrac{5}{4}-2=\dfrac{3}{2}\)
\(\Leftrightarrow\left|\dfrac{2}{5}x-\dfrac{1}{5}\right|\cdot\dfrac{5}{4}=\dfrac{3}{2}+2=\dfrac{7}{2}\)
\(\Leftrightarrow\left|\dfrac{2}{5}x-\dfrac{1}{5}\right|=\dfrac{7}{2}:\dfrac{5}{4}=\dfrac{7}{2}\cdot\dfrac{4}{5}=\dfrac{28}{10}=\dfrac{14}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{5}x-\dfrac{1}{5}=\dfrac{14}{5}\\\dfrac{2}{5}x-\dfrac{1}{5}=-\dfrac{14}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3:\dfrac{2}{5}=\dfrac{15}{2}\\x=-\dfrac{13}{5}:\dfrac{2}{5}=\dfrac{-13}{2}\end{matrix}\right.\)
13 tháng 7 2017
a) \(-\dfrac{2}{3}\left(x-\dfrac{1}{4}\right)=\dfrac{1}{3}\left(2x-1\right)\)
\(\Rightarrow-\dfrac{2}{3}x+\dfrac{1}{6}=\dfrac{2}{3}x-\dfrac{1}{3}\)
\(\Rightarrow\dfrac{1}{6}+\dfrac{1}{3}=\dfrac{2}{3}x+\dfrac{2}{3}x\)
\(\Rightarrow\dfrac{1}{2}=\dfrac{4}{3}x\)
\(\Rightarrow x=\dfrac{1}{2}:\dfrac{4}{3}=\dfrac{3}{8}\)
Vậy \(x=\dfrac{3}{8}\).
2 tháng 5 2017
a) \(\left(2x-3\right)\left(6-2x\right)=0\)
\(\circledast\)TH1: \(2x-3=0\\ 2x=0+3\\ 2x=3\\ x=\dfrac{3}{2}\)
\(\circledast\)TH2: \(6-2x=0\\ 2x=6-0\\ 2x=6\\ x=\dfrac{6}{2}=3\)
Vậy \(x\in\left\{\dfrac{3}{2};3\right\}\).
b) \(\dfrac{1}{3}x+\dfrac{2}{5}\left(x-1\right)=0\)
\(\dfrac{1}{3}x=0-\dfrac{2}{5}\left(x-1\right)\)
\(\dfrac{1}{3}x=-\dfrac{2}{5}\left(x-1\right)\)
\(-\dfrac{2}{5}-\dfrac{1}{3}=-x\left(x-1\right)\)
\(-\dfrac{11}{15}=-x\left(x-1\right)\)
\(\Rightarrow x=1.491631652\)
Vậy \(x=1.491631652\)
c) \(\left(3x-1\right)\left(-\dfrac{1}{2}x+5\right)=0\)
\(\circledast\)TH1: \(3x-1=0\\ 3x=0+1\\ 3x=1\\ x=\dfrac{1}{3}\)
\(\circledast\)TH2: \(-\dfrac{1}{2}x+5=0\\ -\dfrac{1}{2}x=0-5\\ -\dfrac{1}{2}x=-5\\ x=-5:-\dfrac{1}{2}\\ x=10\)
Vậy \(x\in\left\{\dfrac{1}{3};10\right\}\).
d) \(\dfrac{x}{5}=\dfrac{2}{3}\\ x=\dfrac{5\cdot2}{3}\\ x=\dfrac{10}{3}\)
Vậy \(x=\dfrac{10}{3}\).
e) \(\dfrac{x}{3}-\dfrac{1}{2}=\dfrac{1}{5}\\ \)
\(\dfrac{x}{3}=\dfrac{1}{5}+\dfrac{1}{2}\)
\(\dfrac{x}{3}=\dfrac{7}{10}\)
\(x=\dfrac{3\cdot7}{10}\)
\(x=\dfrac{21}{10}\)
Vậy \(x=\dfrac{21}{10}\).
f) \(\dfrac{x}{5}-\dfrac{1}{2}=\dfrac{6}{10}\)
\(\dfrac{x}{5}=\dfrac{6}{10}+\dfrac{1}{2}\)
\(\dfrac{x}{5}=\dfrac{11}{10}\)
\(x=\dfrac{5\cdot11}{10}\)
\(x=\dfrac{55}{10}=\dfrac{11}{2}\)
Vậy \(x=\dfrac{11}{2}\).
g) \(\dfrac{x+3}{15}=\dfrac{1}{3}\\ x+3=\dfrac{15}{3}=5\\ x=5-3\\ x=2\)
Vậy \(x=2\).
h) \(\dfrac{x-12}{4}=\dfrac{1}{2}\\ x-12=\dfrac{4}{2}=2\\ x=2+12\\ x=14\)
Vậy \(x=14\).
29 tháng 7 2018
Bài 3: A=2018-|x+2019|. Vì |x+2019|\(\ge\)0 nên -|x+2019|\(\le\)0=>2018-|x+2019|\(\le\) 2. Vậy A có GTLN = 2 khi x+2019=0 hay x=-2019. B=-10-\(\left|2x-\dfrac{1}{1009}\right|\). Vì \(\left|2x-\dfrac{1}{1009}\right|\ge0\Rightarrow-\left|2x-\dfrac{1}{1009}\right|\le0\Rightarrow-10-\left|2x-\dfrac{1}{1009}\right|\le-10\). Vậy B có GTLN = -10 khi 2x-\(\dfrac{1}{1009}=0\) => \(2x=\dfrac{1}{1009}\Rightarrow x=\dfrac{1}{1009}:2=\dfrac{1}{2018}\)
29 tháng 7 2018
Bài 2: A=\(\left|5x+1\right|-\dfrac{3}{8}\). Vì \(\left|5x+1\right|\ge0\Rightarrow\left|5x+1\right|-\dfrac{3}{8}\ge\dfrac{-3}{8}\). Vậy A có GTNN = \(\dfrac{-3}{8}\) khi 5x+1= 0=> 5x= -1=> x = \(\dfrac{-1}{5}\). B=\(\left|2-\dfrac{1}{6}x\right|+0,25\) , vì \(\left|2-\dfrac{1}{6}x\right|\ge0\Rightarrow\left|2-\dfrac{1}{6}x\right|+0,25\ge0,25\) . Vậy B có GTNN = 0,25 khi \(2-\dfrac{1}{6}x=0\Rightarrow\dfrac{x}{6}=2\Rightarrow x=2.6=12\)
(5x-1)(2x-1/3)=0
=>\(\left\{{}\begin{matrix}5x-1=0=>x=\dfrac{1}{5}\\2x-\dfrac{1}{3}=0=>x=\dfrac{1}{6}\end{matrix}\right.\)
Vậy x=\(\dfrac{1}{5};\dfrac{1}{6}\)
\(\left(5x-1\right)\left(2x-\dfrac{1}{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-1=0\\2x-\dfrac{1}{3}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=1\\2x=\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=\dfrac{1}{6}\end{matrix}\right.\)
Vậy ....