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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
Bạn không nói gì, mình sẽ rút gọn nhé.
\(2+\dfrac{1}{2+\dfrac{1}{1+\dfrac{68}{288}}}\\ =2+\dfrac{1}{2+\dfrac{1}{1+\dfrac{17}{72}}}\\ =2+\dfrac{1}{2+\dfrac{1}{\dfrac{89}{72}}}\\ =2+\dfrac{1}{2+\dfrac{72}{89}}\\ =2+\dfrac{1}{\dfrac{250}{89}}\\ =2+\dfrac{89}{250}=\dfrac{589}{250}\)
Vậy thôi, chúc bạn học tốt nhé.
a, \(\left(x-3\right)\left(2x+5\right)>0\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-3>0\\2x+5>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-3< 0\\2x+5< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>3\\x>-\dfrac{5}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x< 3\\x< -\dfrac{5}{2}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>3\\x< -\dfrac{5}{2}\end{matrix}\right.\)
b,\(\left(1-4x\right)\left(x-2\right)< 0\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}1-4x>0\\x-2< 0\end{matrix}\right.\\\left\{{}\begin{matrix}1-4x< 0\\x-2>0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{1}{4}\\x< 2\end{matrix}\right.\\\left\{{}\begin{matrix}x>\dfrac{1}{4}\\x>2\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< 2\\x>2\end{matrix}\right.\)
c, \(\dfrac{-3}{x+2}< 0\Leftrightarrow x+2>0\Leftrightarrow x>-2\)
1/3 . (2x - 1/3) - 25/27 = 0
1/3 . (2x - 1/3) = 25/27
2x - 1/3 = 25/27 : 1/3
2x - 1/3 = 25/9
2x = 25/9 + 1/3
2x = 28/9
x = 28/9 : 2
x = 14/9
\(\left(x-1\right)\left(x+5\right)>0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1>0\Rightarrow x>1\\x+5>0\Rightarrow x>-5\end{matrix}\right.\\\left\{{}\begin{matrix}x-1< 0\Rightarrow x< 1\\x+5< 0\Rightarrow x< -5\end{matrix}\right.\end{matrix}\right.\)
\(\left(x-1\right)\left(x+5\right)< 0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1>0\Rightarrow x>1\\x+5< 0\Rightarrow x< -5\end{matrix}\right.\\\left\{{}\begin{matrix}x-1< 0\Rightarrow x< 1\\x+5>0\Rightarrow x>-5\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow-5< x< 1\)
câu dễ tự làm
\(\Rightarrow x>-5;x< -5\)
mình làm lại câu b) nha
b) |x-3|=-4
th1: x-3=-4
x=3+(-4)
x=-1
th2: x-3=4
x=3+4
x=7
b) \(\left|x-3\right|=-4\)
t/h1:\(x-3=-4\)
\(x=3-\left(-4\right)\)
\(x=7\)
t/h2:\(x-3=4\)
\(x=3-4\)
\(x=-1\)
Ta có:\(\left(2x-\dfrac{1}{2}\right)\left(\dfrac{5}{4}-3x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-\dfrac{1}{2}=0\\\dfrac{5}{4}-3x=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=\dfrac{1}{2}\\3x=\dfrac{5}{4}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=\dfrac{5}{12}\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=\dfrac{5}{12}\end{matrix}\right.\)
a, \(4.3^{2x}-2.9^x-54=0\)
\(\Rightarrow3^{2x}\left(4-2\right)=54\)
\(\Rightarrow3^{2x}=27=3^3\)
Vì \(3\ne\pm1;3\ne0\) nên \(2x=3\Rightarrow x=\dfrac{3}{2}\)
b, \(\dfrac{1}{2}.2^x+4.2^x-288=0\)
\(\Rightarrow2^x\left(\dfrac{1}{2}+4\right)=288\)
\(\Rightarrow2^x=64=2^6\)
Vì \(2\ne\pm1;2\ne0\) nên \(x=6\)