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a)
\(\left(3x+\dfrac{1}{3}\right)\left(x-\dfrac{1}{2}\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x+\dfrac{1}{3}=0\\x-\dfrac{1}{2}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{9}\\x=\dfrac{1}{2}\end{matrix}\right.\)
b)
\(\left(x-\dfrac{3}{2}\right)\left(2x+1\right)>0\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-\dfrac{3}{2}>0\\2x+1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-\dfrac{3}{2}< 0\\2x+1< 0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>\dfrac{3}{2}\\x>-\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x< \dfrac{3}{2}\\x< -\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x>\dfrac{3}{2}\\x< -\dfrac{1}{2}\end{matrix}\right.\)
a) Ta có: x(x-1)<0
\(\Leftrightarrow\)x; x-1 khác dấu
*Trường hợp 1:
\(\left\{{}\begin{matrix}x>0\\x-1< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>0\\x< 1\end{matrix}\right.\Leftrightarrow0< x< 1\)
*Trường hợp 2:
\(\left\{{}\begin{matrix}x< 0\\x-1>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 0\\x>1\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
Vậy: 0<x<1
b) Ta có: (2-x)(3x-12)>0
\(\Leftrightarrow\)2-x; 3x-12 cùng dấu
*Trường hợp 1:
\(\left\{{}\begin{matrix}2-x>0\\3x-12>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>2\\3x>12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>2\\x>4\end{matrix}\right.\Leftrightarrow x>4\)
*Trường hợp 2:
\(\left\{{}\begin{matrix}2-x< 0\\3x-12< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 2\\3x< 12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 2\\x< 4\end{matrix}\right.\Leftrightarrow x< 2\)
Vậy: 2<x<4
c) Ta có: \(\left(x+1\right)^2\cdot\left(5-2x\right)\le0\)
*Trường hợp 1:
\(\left(x+1\right)^2\cdot\left(5-2x\right)< 0\)
\(\Leftrightarrow\)(x+1)2; 5-2x khác dấu
-Trường hợp 1:
\(\left\{{}\begin{matrix}\left(x+1\right)^2< 0\\5-2x>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+1< 0\\2x< 5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 1\\x< \frac{5}{2}\end{matrix}\right.\Leftrightarrow x< 1\)
-Trường hợp 2:
\(\left\{{}\begin{matrix}\left(x+1\right)^2>0\\5-2x< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+1>0\\2x>5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>1\\x>\frac{5}{2}\end{matrix}\right.\Leftrightarrow x>\frac{5}{2}\)
Vậy: \(1< x< \frac{5}{2}\)
câu d tương tự nhé bạn
a: 1-2x<7
=>-2x<6
hay x>-3
b: (x-1)(x-2)>0
=>x-2>0 hoặc x-1<0
=>x>2 hoặc x<1
c: \(\left(x-2\right)^2\cdot\left(x+1\right)\left(x-4\right)< 0\)
=>(x+1)(x-4)<0
=>-1<x<4
a) \(\left(x-1\right)\left(2x-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\Rightarrow x=1\\2x-4=0\Rightarrow x=2\end{matrix}\right.\)
b) \(\left(x^2+5\right)\left(x-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2+5=0\Rightarrow x=-\sqrt{5}\\x-5=0\Rightarrow x=5\end{matrix}\right.\)
mà \(x\in Z\Rightarrow x=5\)
c) \(\left(x^2+5\right)\left(x^2-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2+5=0\Rightarrow x=-\sqrt{5}\\x^2-2=0\Rightarrow x=\sqrt{2}\end{matrix}\right.\)
mà \(x\in Z\Rightarrow x\in\varnothing\)
a) \(x^2-3x+3=0\)
\(\Leftrightarrow x^2-2.x.\frac{3}{2}+\left(\frac{3}{2}\right)^2+\frac{3}{4}=0\)
\(\Leftrightarrow\left(x-\frac{3}{2}\right)^2+\frac{3}{4}=0\)
\(\Leftrightarrow\left(x-\frac{3}{2}\right)^2=\frac{-3}{4}\)
Vô lí => Phương trình vô nghiệm
\(\Rightarrow x\in\varnothing\)
b) \(x-\left(x-2\right)-\left(x-2\right)=0\)
\(\Leftrightarrow x-x+2-x+2=0\)
\(\Leftrightarrow-x=-4\)
\(\Leftrightarrow x=4\)
c) \(\left(x-2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=1\end{cases}}}\)
Vậy x = {1;2}
d) \(x^2-2x-x+2=0\)
\(\Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}}\)
Vậy x = {1;2}