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a)\(\left(2x-3\right)\left(x+1\right)< 0\)
\(\Leftrightarrow\begin{cases}2x-3>0\\x+1< 0\end{cases}\) hoặc \(\begin{cases}2x-3< 0\\x+1>0\end{cases}\)
\(\Leftrightarrow\begin{cases}x>\frac{3}{2}\\x< -1\end{cases}\) (loại) hoặc \(\begin{cases}x< \frac{3}{2}\\x>-1\end{cases}\)
\(\Leftrightarrow-1< x< \frac{3}{2}\)
b) \(\left(x-\frac{1}{2}\right)\left(x+3\right)>0\)
\(\Leftrightarrow\begin{cases}x-\frac{1}{2}>0\\x+3>0\end{cases}\) hoặc \(\begin{cases}x-\frac{1}{2}< 0\\x+3< 0\end{cases}\)
\(\Leftrightarrow\begin{cases}x>\frac{1}{2}\\x>-3\end{cases}\) hoặc \(\begin{cases}x< \frac{1}{2}\\x< -3\end{cases}\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x>\frac{1}{2}\\x< -3\end{array}\right.\)
c) Sai đề phải là \(\frac{x}{\left(x+3\right)\left(x+7\right)}\)
Có: \(\frac{3}{\left(x+3\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+3\right)\left(x+17\right)}\)
\(\Leftrightarrow\)\(\frac{1}{x+3}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+7}=\frac{x}{\left(x+3\right)\left(x+7\right)}\)
\(\Leftrightarrow\)\(\frac{1}{x+3}-\frac{1}{x+7}=\frac{x}{\left(x+3\right)\left(x+7\right)}\)
\(\Leftrightarrow\)\(\frac{4}{\left(x+3\right)\left(x+7\right)}=\frac{x}{\left(x+3\right)\left(x+7\right)}\)
\(\Leftrightarrow x=4\)
\(\frac{1}{4}+\frac{1}{3}:2x=-5\)
\(\frac{1}{3}:2x=-5-\frac{1}{4}\)
\(\frac{1}{3}:2x=\frac{-21}{4}\)
\(2x=\frac{1}{3}:\frac{-21}{4}\)
\(2x=\frac{-4}{63}\)
\(x=\frac{-4}{63}:2\)
\(x=\frac{-2}{63}\)
\(\)
\(\frac{1}{4}+\frac{1}{3}:2x=-5\)
\(\Rightarrow\frac{1}{3}:2x=-\frac{21}{4}\)
\(\Rightarrow2x=\frac{-4}{63}\)
\(\Rightarrow x=\frac{-2}{63}\)
\(\left(3x-\frac{1}{4}\right)\left(x+\frac{1}{2}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-\frac{1}{4}=0\\x+\frac{1}{2}=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{12}\\x=\frac{-1}{2}\end{cases}}}\)
\(\left(2x-5\right)\left(\frac{3}{2}x+9\right)\left(0,3x-12\right)=0\)
Th1 : \(2x-5=0\Rightarrow x=\frac{5}{2}\)
Th2 : \(\frac{3}{2}x+9=0\Rightarrow x=-6\)
Th3 : \(0,3x-12=0\Rightarrow x=\frac{12}{0,3}\)
a)\(1-2x< 1\)
\(\Leftrightarrow2x>0\)
\(\Leftrightarrow x>0\)
b)\(\left(x-2\right)^2\left(x+1\right)\left(x-4\right)< 0\)
\(\Leftrightarrow\hept{\begin{cases}x\ne2\\\left(x+1\right)\left(x-4\right)< 0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\ne2\\x+1< 0\\x-4>0\end{cases}}\)hoặc \(\hept{\begin{cases}x\ne2\\x+1>0\\x-4< 0\end{cases}}\)
mà \(x+1>x-4\forall x\)
nên \(\hept{\begin{cases}x\ne2\\x+1>0\\x-4< 0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x\ne2\\x>-1\\x< 4\end{cases}}\)
hay \(\hept{\begin{cases}x\ne2\\-1< x< 4\end{cases}}\)
c)\(x-2< 0\)
\(\Leftrightarrow x< 2\)
d)\(\frac{x^2\left(x-3\right)}{x-9}< 0\left(x\ne9\right)\)
\(\Leftrightarrow\hept{\begin{cases}x\ne0\\\frac{x-3}{x-9}< 0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\ne0\\x-3< 0\\x-9>0\end{cases}}\)hoặc \(\hept{\begin{cases}x\ne0\\x-3>0\\x-9< 0\end{cases}}\)
mà \(x-3>x-9\forall x\)
\(\Leftrightarrow\hept{\begin{cases}x\ne0\\x-3>0\\x-9< 0\end{cases}}\)\(\Leftrightarrow3< x< 9\)
e)\(\frac{5}{x}< 1\left(x\ne0\right)\)
\(\Leftrightarrow x>5\)
f)\(8x>2x\)
\(\Leftrightarrow6x>0\)
\(\Leftrightarrow x>0\)
g)\(x+a< a\)
\(\Leftrightarrow x< 0\)
h)\(x^3< x^2\)
\(\Leftrightarrow x^2\left(x-1\right)< 0\)
\(\Leftrightarrow\hept{\begin{cases}x\ne0\\x-1< 0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\ne0\\x< 1\end{cases}}\)
x+7/2010+x+6/2011=x+5/2012+x+4/2013
((x+7/2010)-1)+((x+6/2011)-1)=(x+5/2012)-1)+(x+4/2013)-1)
x+2017/2010+x+2017/2011-x+2017/2012-x+2017/2013=0
x+2017(1/2010+1/2011-1/2012-1/2013)=0
x+2017=0(vì 1/2010+1/2011-1/2012-1/2013<0)
x=-2017
vậy.......
tk mk nha bn
1) Tìm x:
a) \(\frac{11}{12}-\frac{5}{12}.\left(\frac{2}{5}+x\right)=\frac{2}{3}\)
\(\Leftrightarrow\frac{5}{12}.\left(\frac{2}{5}+x\right)=\frac{11}{12}-\frac{2}{3}=\frac{1}{4}\)
\(\Leftrightarrow\frac{2}{5}+x=\frac{1}{4}:\frac{5}{12}=\frac{3}{5}\)
\(\Leftrightarrow x=\frac{3}{5}-\frac{2}{5}=\frac{1}{5}\)
b) \(\frac{3}{4}+\frac{1}{4}:x=\frac{2}{5}\)
\(\Leftrightarrow\frac{1}{4}:x=\frac{2}{5}-\frac{3}{4}=-\frac{7}{20}\)
\(\Leftrightarrow x=-\frac{7}{20}:\frac{1}{4}=\frac{-7}{5}\)
a) \(\frac{11}{12}-\frac{5}{12}\left(\frac{2}{5}+x\right)=\frac{2}{3}\)
\(\Leftrightarrow\frac{11}{12}-\frac{5}{12}.\frac{2}{5}-\frac{5}{12}x=\frac{2}{3}\)
\(\Leftrightarrow\frac{11}{12}-\frac{1}{6}-\frac{5}{12}x=\frac{2}{3}\)
\(\Leftrightarrow\frac{-5}{12}x=\frac{2}{3}-\frac{11}{12}+\frac{1}{6}\)
\(\Leftrightarrow-\frac{5}{12}x=\frac{8}{12}-\frac{11}{12}+\frac{2}{12}=-\frac{1}{12}\)
\(\Leftrightarrow x=\frac{-1}{12}:\left(-\frac{5}{12}\right)=-\frac{1}{12}.\left(-\frac{12}{5}\right)=\frac{1}{5}\)
Vậy x = 1/5
b) \(\frac{3}{4}+\frac{1}{4}:x=\frac{2}{5}\)
\(\Leftrightarrow\frac{1}{4}:x=\frac{2}{5}-\frac{3}{4}=\frac{8}{20}-\frac{15}{20}=-\frac{7}{20}\)
\(\Leftrightarrow x=\frac{1}{4}:\left(-\frac{7}{20}\right)=\frac{1}{4}.\left(-\frac{20}{7}\right)=-\frac{5}{7}\)
Vậy x = -5/7
c) \(2x\left(x-\frac{1}{7}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-\frac{1}{7}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\frac{1}{7}\end{matrix}\right.\)
d) \(\left(x+1\right)\left(x-2\right)< 0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1< 0\\x-2>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+1>0\\x-2< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< -1\\x>2\end{matrix}\right.\\\left\{{}\begin{matrix}x>-1\\x< 2\end{matrix}\right.\end{matrix}\right.\)
Ta thấy x <-1 và x >2 vô lí
Do đó: x >-1 và x <2
Vậy -1 < x <2
e) \(\left(x-2\right)\left(x+\frac{2}{3}\right)>0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2>0\\x+\frac{2}{3}>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2< 0\\x+\frac{2}{3}< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>2\\x>-\frac{2}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}x< 2\\x< -\frac{2}{3}\end{matrix}\right.\end{matrix}\right.\)
Vậy x > 2 hoặc x < -2/3
a) \(2\frac{1}{3}+\left(x-\frac{3}{2}\right)=\left(3-\frac{3}{2}\right)x\)
\(2\frac{1}{3}+x-\frac{3}{2}=3x-\frac{3}{2}x\)
\(2\frac{1}{3}-\frac{3}{2}=3x-\frac{3}{2}x-x\)
\(\frac{5}{6}=3x-\frac{3}{2}x-x\)
\(\frac{5}{6}=\left(3-\frac{3}{2}-1\right)x\)
\(\frac{5}{6}=\frac{1}{2}x\)
\(x=\frac{5}{6}:\frac{1}{2}\)
\(x=\frac{5}{3}\)
b) |3x-4|+|3y+5|=0
ĐK : \(\hept{\begin{cases}\left|3x-4\right|\ge0\\\left|3y+5\right|\ge0\end{cases}}\Leftrightarrow\left|3x-4\right|+\left|3y+5\right|\ge0\)
Mà |3x-4|+|3y+5|=0 nên :
\(\Rightarrow\hept{\begin{cases}3x-4=0\\3y+5=0\end{cases}}\Rightarrow\hept{\begin{cases}3x=4\\3y=-5\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{4}{3}\\y=\frac{-5}{3}\end{cases}}\)
Vậy x=4/3 ; y=-5/3
c) \(\left|x+\frac{19}{5}\right|+\left|y+\frac{1890}{1975}\right|+\left|z-2004\right|=0\)
ĐK : \(\hept{\begin{cases}\left|x+\frac{19}{5}\right|\ge0\\\left|y+\frac{1890}{1975}\right|\ge0\\\left|z-2004\right|\ge0\end{cases}}\Leftrightarrow\left|x+\frac{19}{5}\right|+\left|y+\frac{1890}{1975}\right|+\left|z-2004\right|\ge0\)
Mà \(\left|x+\frac{19}{5}\right|+\left|y+\frac{1890}{1975}\right|+\left|z-2004\right|=0\) nên :
\(\Rightarrow\hept{\begin{cases}x+\frac{19}{5}=0\\y+\frac{1890}{1975}=0\\z-2004=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-\frac{19}{5}\\y=-\frac{1890}{1975}\\z=2004\end{cases}}\)
Vậy ...
X:(\(\frac{2}{9}-\frac{1}{5}\))=\(\frac{8}{16}\)
x:\(\frac{1}{45}\) =\(\frac{8}{16}\)
x: =\(\frac{8}{16}.\frac{1}{45}\)
x: =\(\frac{1}{90}\)