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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
2, \(\Rightarrow\left\{{}\begin{matrix}\\\dfrac{5}{4}x-\dfrac{7}{2}=0\\\dfrac{5}{8}x+\dfrac{3}{5}=0\\\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{14}{5}\\\\x=\dfrac{-24}{25}\\\end{matrix}\right.\)
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 \(x\in\left\{-4;-3;-2;-1\right\}\)
a: \(\left|x\right|=3+\dfrac{1}{5}=\dfrac{16}{5}\)
mà x<0
nên x=-16/5
b: \(\left|x\right|=-2.1\)
nên \(x\in\varnothing\)
c: \(\left|x-3.5\right|=5\)
=>x-3,5=5 hoặc x-3,5=-5
=>x=8,5 hoặc x=-1,5
d: \(\left|x+\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)
=>|x+3/4|=1/2
=>x+3/4=1/2 hoặc x+3/4=-1/2
=>x=-1/4 hoặc x=-5/4
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
1, a/ \(\left|x\right|=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{-1}{2}\end{matrix}\right.\)
Vậy .............
b/ \(\left|x\right|=3,12\Leftrightarrow\left[{}\begin{matrix}x=3,12\\x=-3,12\end{matrix}\right.\)
Vậy ...........
c/ \(\left|x\right|=0\Leftrightarrow x=0\)
Vậy ..........
d/ \(\left|x\right|=2\dfrac{1}{7}\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\dfrac{1}{7}\\x=-2\dfrac{1}{7}\end{matrix}\right.\)
Vậy ..............
2, a/ \(\left|x\right|=2,1\Leftrightarrow\left[{}\begin{matrix}x=2,1\\x=-2,1\end{matrix}\right.\)
Vậy ...........
b/ \(\left|x\right|=\dfrac{17}{9}\) ; \(x< 0\)
\(\Leftrightarrow x=-\dfrac{17}{9}\)
Vậy ..........
c/ \(\left|x\right|=1\dfrac{2}{5}\Leftrightarrow\left[{}\begin{matrix}x=1\dfrac{2}{5}\\x=-1\dfrac{2}{5}\end{matrix}\right.\)
Vậy ...........
d/ \(\left|x\right|=0,35\) ; \(x>0\Leftrightarrow x=0,35\)
3, a/ \(\left|x-1,7\right|=2,3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1,7=2,3\\x-1,7=-2,3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-0,6\end{matrix}\right.\)
Vậy ...........
b/ \(\left|x+\dfrac{3}{4}\right|-\dfrac{1}{3}=0\)
\(\Leftrightarrow\left|x+\dfrac{3}{4}\right|=\dfrac{1}{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{3}{4}=\dfrac{1}{3}\\x+\dfrac{3}{4}=-\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5}{12}\\x=-\dfrac{13}{12}\end{matrix}\right.\)
Vậy ...........
a) \(x+\dfrac{3}{10}=\dfrac{-2}{5}\)
\(x=\dfrac{-2}{5}-\dfrac{3}{10}\)
\(x=\dfrac{-7}{10}\)
b) \(x+\dfrac{5}{6}=\dfrac{2}{5}-\left(-\dfrac{2}{3}\right)\)
\(x+\dfrac{5}{6}=\dfrac{2}{5}+\dfrac{2}{3}\)
\(x+\dfrac{5}{6}=\dfrac{16}{15}\)
\(x=\dfrac{16}{15}-\dfrac{5}{6}\)
\(x=\dfrac{7}{30}\)
c) \(1\dfrac{2}{5}x+\dfrac{3}{7}=-\dfrac{4}{5}\)
\(\dfrac{7}{5}x+\dfrac{3}{7}=-\dfrac{4}{5}\)
\(\dfrac{7}{5}x=-\dfrac{4}{5}-\dfrac{3}{7}\)
\(\dfrac{7}{5}x=\dfrac{-43}{35}\)
\(\Rightarrow x=\dfrac{-43}{49}\)
d) \(\left[x+\dfrac{3}{4}\right]-\dfrac{1}{3}=0\)
\(\left[x+\dfrac{3}{4}\right]=0+\dfrac{1}{3}\)
\(\left[x+\dfrac{3}{4}\right]=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}-\dfrac{3}{4}\)
\(x=\dfrac{-5}{12}\)
e) \(\left[x+\dfrac{4}{5}\right]-\left(-3,75\right)=-\left(-2,15\right)\)
\(\left[x+\dfrac{4}{5}\right]+3,75=2,15\)
\(x+\dfrac{4}{5}=2,15-3,75\)
\(x+\dfrac{4}{5}=-\dfrac{8}{5}\)
\(x=\dfrac{-8}{5}-\dfrac{4}{5}\)
\(x=\dfrac{-12}{5}\)
f) \(\left(x-2\right)^2=1\)
\(\Rightarrow x=1\)
Sức chịu đựng có giới hạn -.-
- Mình tiếp tục cho Nguyễn Phương Trâm nhé.
g, \(\left(2x-1\right)^3=-27\)
\(\Rightarrow\left(2x-1\right)^3=\left(-3\right)^3\)
\(\Rightarrow2x-1=-3\)
\(\Rightarrow2x=-2\)
=> \(x=-1\)
- Vậy x = -1
h,\(\dfrac{x-1}{-15}=-\dfrac{60}{x-1}\)
\(\Rightarrow\left(x-1\right)^2=-60.\left(-15\right)\)
\(\Rightarrow\left(x-1\right)^2=900 \)
\(\Rightarrow\left(x-1\right)^2=30^2\Rightarrow x-1=30\)
=> x = 31
i,\(x:\left(\dfrac{-1}{2}\right)^3=\dfrac{-1}{2}\)
=> \(x:\left(-\dfrac{1}{8}\right)=-\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{1}{16}\)
- Vậy x=\(\dfrac{1}{16}\)
j, \(\left(\dfrac{3}{4}\right)^5.x=\left(\dfrac{3}{4}\right)^7\)
\(\Rightarrow \left(\dfrac{3}{4}\right).x=\left(\dfrac{3}{4}\right)^2\)
\(\Rightarrow x=\left(\dfrac{3}{4}\right)^2:\dfrac{3}{4}\)
\(\Rightarrow x=\dfrac{3}{4}\)
- Vạy x = \(\dfrac{3}{4}\)
k, \(8^x:2^x=4\Rightarrow\left(8:2\right)^x=4\)
=>\(4^x=4\)
=> x = 1
- Vậy x = 1
a. \(\dfrac{-39}{7}:x=26\)
x = \(\dfrac{-39}{7}:26\)
x = \(\dfrac{-3}{14}\)
b. \(x:\dfrac{13}{5}=\dfrac{7}{4}\)
x = \(\dfrac{7}{4}.\dfrac{13}{5}\)
x = \(\dfrac{91}{20}\)
c. x = \(\dfrac{-3}{5}-\dfrac{1}{2}\)
x = \(\dfrac{-11}{10}\)
d. \(x-\dfrac{3}{4}=\dfrac{9}{4}\)
x = \(\dfrac{9}{4}+\dfrac{3}{4}\)
x = 3
e. \(\dfrac{7}{8}:x=\dfrac{14}{3}\)
x = \(\dfrac{7}{8}:\dfrac{14}{3}\)
x = \(\dfrac{3}{16}\)
f. \(x:\dfrac{8}{3}=\dfrac{13}{3}\)
x = \(\dfrac{13}{3}.\dfrac{8}{3}\)
x = \(\dfrac{104}{9}\)
g. x = \(\dfrac{4}{10}-\dfrac{2}{5}\)
x = 0
chúc bạn học tốt 
a: \(\left(2x+3\right)\left(3x-5\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-5\ge0\\2x+3\le0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>=\dfrac{5}{3}\\x< =-\dfrac{3}{2}\end{matrix}\right.\)
b: \(\dfrac{x}{3-x}>-1\)
\(\Leftrightarrow\dfrac{x}{3-x}+1>0\)
\(\Leftrightarrow\dfrac{x+3-x}{3-x}>0\)
=>3-x>0
hay x<3
c: \(\dfrac{x-1}{x+5}\ge\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{x-1}{x+5}-\dfrac{3}{2}\ge0\)
\(\Leftrightarrow\dfrac{2x-2-3x-15}{2\left(x+5\right)}>=0\)
\(\Leftrightarrow\dfrac{x+17}{2\left(x+5\right)}< =0\)
=>-17<=x<-5
d: \(\dfrac{7}{4x^2-1}\ge0\)
=>4x2-1>0
=>(2x-1)(2x+1)>0
=>x>1/2 hoặc x<-1/2
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!!!
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.....................