a/(x-3).(2x-1)>0
b/(2-3x).(-5x+1)<0
c/(x+1).(x-2).(3-x)>0
K
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K
30 tháng 6 2021
a) 3x(4x-3)-2x(5-6x)=0
\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)
\(\Leftrightarrow24x^2-19x=0\)
\(\Leftrightarrow x\left(24x-19\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{24}\end{matrix}\right.\)
Vậy x=0 hoặc x=\(\dfrac{19}{24}\)
30 tháng 7 2021
a) \(\left(2x-1\right)^2-25=0\)
⇔ \(\left(2x-1\right)^2-5^2=0\)
⇔ \(\left(2x-1-5\right)\left(2x-1+5\right)=0\)
⇒ \(2x-1-5=0\) hoặc \(2x-1+5=0\)
⇔ \(x=3\) hoặc \(x=-2\)
TC
30 tháng 7 2021
Bài 1: Tìm x
a) (2x-1) ² - 25 = 0
<=> (2x-1)2 = 25
<=> 2x-1 = 5 hay 2x-1 =-5
<=> 2x= 6 hay 2x=-4
<=> x=3 hay x= -2
Vậy S={3; -2}
b) 3x (x-1) + x - 1 = 0
<=> (x-1)(3x+1)=0
<=> x-1=0 hay 3x+1=0
<=> x=1 hay 3x=-1
<=> x=1 hay x=\(\dfrac{-1}{3}\)
Vậy S={1;\(\dfrac{-1}{3}\)}
c) 2(x+3) - x ² - 3x = 0
<=> 2(x+3)- x(x+3)=0
<=> (x+3)(2-x)=0
<=> x+3=0 hay 2-x=0
<=> x=-3 hay x=2
Vậy S={-3;2}
d) x(x - 2) + 3x - 6 = 0
<=> x(x-2)+3(x-2)=0
<=> (x-2)(x+3)=0
<=> x-2=0 hay x+3=0
<=> x=2 hay x=-3
Vậy S={2;-3}
e) 4x ² - 4x +1 = 0
<=> (2x-1)2=0
<=> 2x-1=0
<=> 2x=1
<=> x=\(\dfrac{1}{2}\)
Vậy S={\(\dfrac{1}{2}\)}
f) x +5x2 = 0
<=> x(1+5x)=0
<=>x=0 hay 1+5x=0
<=> x=0 hay 5x=-1
<=> x=0 hay x= \(\dfrac{-1}{5}\)
Vậy S={0;\(\dfrac{-1}{5}\)}
g) x ²+ 2x -3 = 0
<=> x2-x+3x-3=0
<=> x(x-1)+3(x-1)=0
<=> (x-1)(x+3)=0
<=> x-1=0 hay x+3=0
<=> x=1 hay x=-3
Vậy S={1;-3}
a)\(\left(x-3\right)\left(2x-1\right)>0.\)
\(Th1:x-3>0;2x-1>0\)
\(x-3>0\Rightarrow x>3_{\left(1\right)}\)
\(2x-1>0\Rightarrow2x>1\Rightarrow x>\frac{1}{2}_{\left(2\right)}\)
\(\left(1\right),\left(2\right)\Rightarrow x>3`\)
\(Th2:x-3< 0;2x-1< 0\)
\(x-3< 0\Rightarrow x< 3_{\left(1\right)}\)
\(2x-1< 0\Rightarrow2x< 1\Rightarrow x< \frac{1}{2}_{\left(2\right)}\)
\(\left(1\right),\left(2\right)\Rightarrow x< \frac{1}{2}\)
b) \(\left(2-3x\right)\left(-5x+1\right)< 0\)
\(Th1:2-3x>0;-5x+1< 0\)
\(2-3x>0\Rightarrow3x>2\Rightarrow x>\frac{2}{3}_{\left(1\right)}\)
\(-5x+1< 0\Rightarrow-5x< -1\Rightarrow x< \frac{1}{5}_{\left(2\right)}\)
\(_{\left(1\right),\left(2\right)\Rightarrow}\)không xảy ra trường hợp này
\(Th2:2-3x< 0;-5x+1>0\)
\(2-3x< 0\Rightarrow3x< 2\Rightarrow x< \frac{2}{3}_{\left(1\right)}\)
\(-5x+1>0\Rightarrow-5x>-1\Rightarrow x>\frac{1}{5}_{\left(2\right)}\)
\(\left(1\right),\left(2\right)\Rightarrow\frac{1}{5}< x< \frac{2}{3}\)
\(\left(x+1\right)\left(x-2\right)\left(3-x\right)>0.\)
\(Th1:x+1>0;x-2>0;3-x>0\)
\(Th2:x+1< 0;x-2< 0;3-x>0\)
\(Th3:x+1>0;x-2< 0;3-x< 0\)
\(Th4:x+1< 0;x-2>0;3-x< 0\)
Mình ghi từng trường hợp nhé ! Bạn tự xét !