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1. \(\Leftrightarrow\left(2x-1\right)\left(3x+1\right)< 0\)
\(\Rightarrow-\frac{1}{3}< x< \frac{1}{2}\)
2. \(\Leftrightarrow\left(x-2\right)\left(3-2x\right)>0\)
\(\Rightarrow\frac{3}{2}< x< 2\)
3. \(\Leftrightarrow\left(5x-3\right)^2>0\)
\(\Rightarrow x\ne\frac{3}{5}\)
4. \(\Leftrightarrow-3\left(x-\frac{1}{6}\right)-\frac{59}{12}< 0\)
\(\Rightarrow x\in R\)
5. \(\Leftrightarrow2\left(x-1\right)^2+5\ge0\)
\(\Rightarrow x\in R\)
6. \(\Leftrightarrow\left(x+2\right)\left(8x+7\right)\le0\)
\(\Rightarrow-2\le x\le-\frac{7}{8}\)
7.
\(\Leftrightarrow\left(x-1\right)^2+2>0\)
\(\Rightarrow x\in R\)
8. \(\Leftrightarrow\left(3x-2\right)\left(2x+1\right)\ge0\)
\(\Rightarrow\left[{}\begin{matrix}x\le-\frac{1}{2}\\x\ge\frac{2}{3}\end{matrix}\right.\)
9. \(\Leftrightarrow\frac{1}{3}\left(x+3\right)\left(x+6\right)< 0\)
\(\Rightarrow-6< x< -3\)
10. \(\Leftrightarrow x^2-6x+9>0\)
\(\Leftrightarrow\left(x-3\right)^2>0\)
\(\Rightarrow x\ne3\)
c: \(\Leftrightarrow\left\{{}\begin{matrix}4x+3>=0\\\left(x+2-4x-3\right)\left(x+2+4x+3\right)< 0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{3}{4}\\\left(-3x-1\right)\left(5x+5\right)< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{3}{4}\\\left(3x+1\right)\left(x+1\right)>0\end{matrix}\right.\)
\(\Leftrightarrow x>-\dfrac{1}{3}\)
d: \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}3x-2< 0\\2x+1>=0\end{matrix}\right.\\\left\{{}\begin{matrix}3x-2>=0\\\left(2x+1-3x+2\right)\left(2x+1+3x-2\right)>=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{2}{3}\\x>-\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left(-x+3\right)\left(5x-1\right)>=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-\dfrac{1}{2}< x< \dfrac{2}{3}\\\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left(x-3\right)\left(5x-1\right)< =0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{-1}{2}< x< \dfrac{2}{3}\\\dfrac{2}{3}< =x< =3\end{matrix}\right.\)
a) x3+4x2+x-6=0
<=> x3+3x2+x2+3x-2x-6=0
<=> x2(x+3)+x(x+3)-2(x+3)=0
<=> (x+3)(x2+x-2)=0
<=> \(\left[\begin{matrix}x+3=0\\x^2+x-2=0\end{matrix}\right.\)<=> \(\left[\begin{matrix}x=-3\\\left(x+\frac{1}{2}\right)^2=\frac{9}{4}\end{matrix}\right.\)
<=> \(\left[\begin{matrix}x=-3\\x=1\\x=-2\end{matrix}\right.\)
Vậy ...
b) x3-3x2+4=0
<=> x3-2x2-x2+4=0
<=> x2(x-2)-(x-2)(x+2)=0
<=> (x-2)(x2-x-2)=0
<=> \(\left[\begin{matrix}x-2=0\\x^2-x-2=0\end{matrix}\right.\)\(\Leftrightarrow\left[\begin{matrix}x=2\\\left(x-\frac{1}{2}\right)^2=\frac{9}{4}\end{matrix}\right.\)
<=> \(\left[\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
Vậy ...
c) x4+2x3+2x2-2x-3=0
<=> x4+x3+x3+x2+x2+x-3x-3=0
<=> x3(x+1)+x2(x+1)+x(x+1)-3(x+1)=0
<=> (x+1)(x3+x2+x-3)=0
<=> (x+1)(x3-x2+2x2-2x+3x-3)=0
<=> (x+1)[x2(x-1)+2x(x-1)+3(x-1)]=0
<=> (x+1)(x-1)(x2+2x+3)=0
Mà x2+2x+3=x2+2x+1+2=(x+1)2+2>0
<=> (x+1)(x-1)=0
<=>\(\left[\begin{matrix}x+1=0\\x-1=0\end{matrix}\right.\)<=> \(\left[\begin{matrix}x=-1\\x=1\end{matrix}\right.\)
Vậy ...
a/ - Với \(x>\frac{1}{4}\) PT vô nghiêm
- Với \(x\le\frac{1}{4}\)
\(\Leftrightarrow\left(x^2-1\right)^2=\left(1-4x\right)^2\)
\(\Leftrightarrow\left(x^2+4x-2\right)\left(x^2-4x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2+4x-2=0\\x^2-4x=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-2+\sqrt{6}\left(l\right)\\x=-2-\sqrt{6}\\x=4\left(l\right)\\x=0\end{matrix}\right.\)
2.
- Với \(x\ge-\frac{1}{4}\Leftrightarrow4x+1=x^2+2x-4\)
\(\Leftrightarrow x^2-2x-5=0\Rightarrow\left[{}\begin{matrix}x=1+\sqrt{6}\\x=1-\sqrt{6}\left(l\right)\end{matrix}\right.\)
- Với \(x< -\frac{1}{4}\)
\(\Leftrightarrow-4x-1=x^2+2x-4\)
\(\Leftrightarrow x^2+6x-3=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-3+2\sqrt{3}\left(l\right)\\x=-3-2\sqrt{3}\end{matrix}\right.\)
3.
- Với \(x\ge\frac{5}{3}\)
\(\Leftrightarrow3x-5=2x^2+x-3\)
\(\Leftrightarrow2x^2-2x+2=0\left(vn\right)\)
- Với \(x< \frac{5}{3}\)
\(\Leftrightarrow5-3x=2x^2+x-3\)
\(\Leftrightarrow2x^2+4x-8=0\Rightarrow\left[{}\begin{matrix}x=-1+\sqrt{5}\\x=-1-\sqrt{5}\end{matrix}\right.\)
4. Do hai vế của pt đều không âm, bình phương 2 vế:
\(\Leftrightarrow\left(x^2-2x+8\right)^2=\left(x^2-1\right)^2\)
\(\Leftrightarrow\left(x^2-2x+8\right)^2-\left(x^2-1\right)^2=0\)
\(\Leftrightarrow\left(2x^2-2x+7\right)\left(-2x+9\right)=0\)
\(\Leftrightarrow-2x+9=0\Rightarrow x=\frac{9}{2}\)
a) x3 + 4x2 - 29x + 24 = 0
<=> x3 - x2 + 5x2 - 5x - 24x + 24 = 0
<=> x2(x - 1) + 5x(x - 1) - 24(x - 1) = 0
<=> (x - 1)(x2 + 5x - 24) = 0
\(\Leftrightarrow\left[\begin{matrix}x-1=0\\x^2+5x-24=0\end{matrix}\right.\)
+) x - 1 = 0 <=> x = 1
+) x2 + 5x - 24 = 0
\(\Delta=5^2+4.1.24=121\Rightarrow\sqrt{\Delta}=11\)
Phương trình có 2 nghiệm phân biệt: \(x_1=\frac{-5+11}{2}=3;x_2=\frac{-5-11}{2}=-8\)
Vậy ...
Bài 4:
$3x^4+10x^3-3x^2-10x+3=0$
Ta đi phân tích $3x^4+10x^3-3x^2-10x+3$ thành nhân tử
Đặt $3x^4+10x^3-3x^2-10x+3=(x^2+ax+b)(3x^2+cx+d)$ với $a,b,c,d$ là các số nguyên
$\Leftrightarrow 3x^4+10x^3-3x^2-10x+3=3x^4+x^3(c+3a)+x^2(d+ac+3b)+x(ad+bc)+bd$
Đồng nhất hệ số:
\(\Rightarrow \left\{\begin{matrix} c+3a=10\\ d+ac+3b=-3\\ ad+bc=-10\\ bd=3\end{matrix}\right.\). Từ $bd=3$. Giả sử $b=-1$
$\Rightarrow d=-3$. Thay vào hệ có được $ac=3; c+3a=10\Rightarrow a=3; c=1$
Vậy $3x^4+10x^3-3x^2-10x+3=(x^2+3x-1)(3x^2+x-3)$
$\Leftrightarrow (x^2+3x-1)(3x^2+x-3)=0$
\(\Rightarrow \left[\begin{matrix} x^2+3x-1=0\\ 3x^2+x-3=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{-3\pm \sqrt{13}}{2}\\ x=\frac{-1\pm \sqrt{37}}{6}\end{matrix}\right.\)
Bài 3:
$x^4+4x^3+x^2-4x+1=0$
$\Leftrightarrow (x^4+4x^3+4x^2)-3x^2-4x+1=0$
$\Leftrightarrow (x^2+2x)^2-2(x^2+2x)-x^2+1=0$
$\Leftrightarrow (x^2+2x)^2-2(x^2+2x)+1-x^2=0$
$\Leftrightarrow (x^2+2x-1)^2-x^2=0$
$\Leftrightarrow (x^2+x-1)(x^2+3x-1)=0$
\(\Rightarrow \left[\begin{matrix} x^2+x-1=0\\ x^2+3x-1=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{-1\pm \sqrt{5}}{2}\\ x=\frac{-3\pm \sqrt{!3}}{2}\end{matrix}\right.\)
Vậy.......
