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`#3107.101107`
a,
\(\text{A = }\left\{x\in R\text{ | }\left(2x-x^2\right)\left(3x-2\right)=0\right\}\)
`<=> (2x - x^2)(3x - 2) = 0`
`<=>`\(\left[{}\begin{matrix}2x-x^2=0\\3x-2=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x\left(2-x\right)=0\\3x=2\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=0\\2-x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=0\\x=2\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy, `A = {0; 2; 2/3}`
b,
\(\text{B = }\left\{x\in R\text{ | }2x^3-3x^2-5x=0\right\}\)
`<=> 2x^3 - 3x^2 - 5x = 0`
`<=> x(2x^2 - 3x - 5) = 0`
`<=>`\(\left[{}\begin{matrix}x=0\\2x^2-3x-5=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=0\\2x^2-2x+5x-5=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=0\\\left(2x^2-2x\right)+\left(5x-5\right)=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=0\\2x\left(x-1\right)+5\left(x-1\right)=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=0\\\left(2x+5\right)\left(x-1\right)=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=0\\2x+5=0\\x-1=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=0\\x=-\dfrac{5}{2}\\x=1\end{matrix}\right.\)
Vậy, `B = {-5/2; 0; 1}.`
c,
\(\text{C = }\left\{x\in Z\text{ | }2x^2-75x-77=0\right\}\)
`<=> 2x^2 - 75x - 77 = 0`
`<=> 2x^2 - 2x + 77x - 77 = 0`
`<=> (2x^2 - 2x) + (77x - 77) = 0`
`<=> 2x(x - 1) + 77(x - 1) = 0`
`<=> (2x + 77)(x - 1) = 0`
`<=>`\(\left[{}\begin{matrix}2x+77=0\\x-1=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}2x=-77\\x=1\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=-\dfrac{77}{2}\\x=1\end{matrix}\right.\)
Vậy, `C = {-77/2; 1}`
d,
\(\text{D = }\left\{x\in R\text{ | }\left(x^2-x-2\right)\left(x^2-9\right)=0\right\}\)
`<=> (x^2 - x - 2)(x^2 - 9) = 0`
`<=>`\(\left[{}\begin{matrix}x^2-x-2=0\\x^2-9=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x^2+x-2x-2=0\\x^2=9\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}\left(x^2+x\right)-\left(2x+2\right)=0\\x^2=\left(\pm3\right)^2\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x\left(x+1\right)-2\left(x+1\right)=0\\x=\pm3\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}\left(x-2\right)\left(x+1\right)=0\\x=\pm3\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x-2=0\\x+1=0\\x=\pm3\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=2\\x=-1\\x=\pm3\end{matrix}\right.\)
Vậy, `D = {-1; -3; 2; 3}.`
Để pt có 2 nghiệm dương (ko yêu cầu pb?) \(\left\{{}\begin{matrix}a\ne0\\\Delta\ge0\\x_1+x_2=-\frac{b}{a}>0\\x_1x_2=\frac{c}{a}>0\end{matrix}\right.\)
a/ \(\left\{{}\begin{matrix}\Delta=\left(2m-1\right)^2+4m-4\ge0\\x_1+x_2=2m+1>0\\x_1x_2=-m+1>0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4m^2-3\ge0\\m>-\frac{1}{2}\\m< 1\end{matrix}\right.\) \(\Rightarrow\frac{\sqrt{3}}{2}\le m< 1\)
b/ \(\left\{{}\begin{matrix}\Delta=\left(m+2\right)^2-4\left(-2m+1\right)\ge0\\-m-2>0\\-2m+1>0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m^2+12m\ge0\\m< -2\\m< \frac{1}{2}\end{matrix}\right.\) \(\Rightarrow m\le-12\)
e/
\(\left\{{}\begin{matrix}\Delta=\left(m+1\right)^2-4m\ge0\\x_1+x_2=m+1>0\\x_1x_2=m>0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(m-1\right)^2\ge0\\m>-1\\m>0\end{matrix}\right.\) \(\Rightarrow m>0\)
f/
\(\left\{{}\begin{matrix}m-2\ne0\\\Delta'=\left(2m-3\right)^2-\left(m-2\right)\left(5m-6\right)\ge0\\x_1+x_2=\frac{2\left(3-2m\right)}{m-2}>0\\x_1x_2=\frac{5m-6}{m-2}>0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m\ne2\\-m^2+4m-3\ge0\\\frac{3-2m}{m-2}>0\\\frac{5m-6}{m-2}>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}m\ne2\\1\le m\le3\\\frac{3}{2}< m< 2\\\left[{}\begin{matrix}m< \frac{6}{5}\\m>2\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\) Không tồn tại m thỏa mãn
Để (1) có 2 nghiệm thỏa mãn \(x_1< 2< x_2\)
\(\Leftrightarrow f\left(2\right)< 0\Leftrightarrow2^2-2.2-m< 0\)
\(\Leftrightarrow-m< 0\Rightarrow m>0\)
Bài 1:
a: \(\Leftrightarrow x^2-5x+6< =0\)
=>(x-2)(x-3)<=0
=>2<=x<=3
b: \(\Leftrightarrow\left(x-6\right)^2< =0\)
=>x=6
c: \(\Leftrightarrow x^2-2x+1>=0\)
\(\Leftrightarrow\left(x-1\right)^2>=0\)
hay \(x\in R\)
$\textbf{Bài 1d}$
$x^2-2|x-3|+3x\ge0.$
$\bullet\ x\ge3:$
$x^2-2(x-3)+3x\ge0$
$\Leftrightarrow x^2+x+6\ge0.$
Bất phương trình luôn đúng.
$\bullet\ x<3:$
$x^2-2(3-x)+3x\ge0$
$\Leftrightarrow x^2+5x-6\ge0$
$\Leftrightarrow(x+6)(x-1)\ge0.$
$\Leftrightarrow x\le-6$ hoặc $x\ge1.$
Kết hợp điều kiện $x<3$.
Vậy $x\le-6$ hoặc $1\le x.$
$\textbf{Bài 1e}$
$x-|x+3|-10\le0.$
$\bullet\ x\ge-3:$
$x-(x+3)-10\le0$
$-13\le0.$
Luôn đúng.
$\bullet\ x<-3:$
$x-(-x-3)-10\le0$
$2x-7\le0.$
$x\le\dfrac72.$
Điều này luôn đúng vì $x<-3.$
Vậy bất phương trình đúng với mọi $x.$
Bài 2a
$A=(-x^2+x-1)(6x^2-5x+1).$
$-x^2+x-1=-(x^2-x+1)<0,\ \forall x.$
$6x^2-5x+1=(3x-1)(2x-1).$
Suy ra $A>0$ khi $\dfrac16<x<\dfrac12.$
$A=0$ khi $x=\dfrac16,\ \dfrac12.$
$A<0$ khi $x<\dfrac16$ hoặc $x>\dfrac12.$
$\textbf{Bài 2b}$
$B=\dfrac{x^2-x-2}{-x^2+3x+4}.$
$=\dfrac{(x-2)(x+1)}{-(x-4)(x+1)}.$
$=-\dfrac{x-2}{x-4},\qquad x\ne-1,\ 4.$
Suy ra $B>0$ khi $2<x<4.$
$B=0$ khi $x=2.$
$B<0$ khi $x<2$ hoặc $x>4.$
Loại $x=-1,\ 4.$
Bài 2c
$C=x^2-5x+2.$
$\Delta=25-8=17.$
$x_1=\dfrac{5-\sqrt{17}}2,\qquad x_2=\dfrac{5+\sqrt{17}}2.$
Suy ra $C>0$ khi $x<\dfrac{5-\sqrt{17}}2$
hoặc $x>\dfrac{5+\sqrt{17}}2.$
$C=0$ tại $x=\dfrac{5-\sqrt{17}}2,\ \dfrac{5+\sqrt{17}}2.$
$C<0$ khi $\dfrac{5-\sqrt{17}}2<x<\dfrac{5+\sqrt{17}}2.$