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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. \(\Leftrightarrow x^2-6x+9>0\)
\(\Leftrightarrow\left(x-3\right)^2>0\)
\(\Rightarrow x\ne3\)
Mình giải mẫu pt đầu thôi nhé, những pt sau ttự.
1,\(x^4-\frac{1}{2}x^3-x^2-\frac{1}{2}x+1=0\)
Ta thấy x=0 ko là nghiệm.
Chia cả 2 vế cho x2 >0:
pt\(\Leftrightarrow x^2-\frac{1}{2}x-1-\frac{1}{2x}+\frac{1}{x^2}=0\)
Đặt \(t=x-\frac{1}{x}\left(t\in R\right)\)
\(\Rightarrow x^2+\frac{1}{x^2}=t^2+2\)
pt\(\Leftrightarrow t^2-\frac{1}{2}t+1=0\)(vô n0)
Vậy pt vô n0.
#Walker
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/ Đặt \(\left|x\right|=t\ge0\Rightarrow t^2-t-2=0\Rightarrow\left[{}\begin{matrix}t=-1\left(l\right)\\t=2\end{matrix}\right.\)
\(\Rightarrow\left|x\right|=2\Rightarrow x=\pm2\)
b/ \(\Leftrightarrow\left(x+1\right)^2+\left|x+1\right|-6=0\)
Đặt \(\left|x+1\right|=t\ge0\Rightarrow t^2+t-6=0\Rightarrow\left[{}\begin{matrix}t=-3\left(l\right)\\t=2\end{matrix}\right.\)
\(\Rightarrow\left|x+1\right|=2\Rightarrow\left[{}\begin{matrix}x+1=2\\x+1=-2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)
c/ \(\Leftrightarrow\left(x+1\right)^2-5\left|x+1\right|+4=0\)
Đặt \(\left|x+1\right|=t\ge0\Rightarrow t^2-5t+4=0\Rightarrow\left[{}\begin{matrix}t=1\\t=4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left|x+1\right|=1\\\left|x+1\right|=4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x+1=1\\x+1=-1\\x+1=4\\x+1=-4\end{matrix}\right.\)
d. \(\Leftrightarrow\left(x-1\right)^2+5\left|x-1\right|+4=0\)
Đặt \(\left|x+1\right|=t\ge0\Rightarrow t^2+5t+4=0\Rightarrow\left[{}\begin{matrix}t=-1\left(l\right)\\t=-4\left(l\right)\end{matrix}\right.\)
Vậy pt vô nghiệm
e. \(\Leftrightarrow\left(x-2\right)^2+2\left|x-2\right|-3=0\)
Đặt \(\left|x-2\right|=t\ge0\)
\(\Rightarrow t^2+2t-3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-3\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\left|x-2\right|=1\Leftrightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)
f. \(\Leftrightarrow\left(2x-5\right)^2+4\left|2x-5\right|-12=0\)
Đặt \(\left|2x-5\right|=t\ge0\)
\(\Rightarrow t^2+4t-12=0\Rightarrow\left[{}\begin{matrix}t=2\\t=-6\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\left|2x-5\right|=2\Rightarrow\left[{}\begin{matrix}2x-5=2\\2x-5=-2\end{matrix}\right.\)
b: Trường hợp 1: x<-3
Pt sẽ là \(x^2+6x-x-3+10=0\)
\(\Leftrightarrow x^2+5x+7=0\)
\(\Delta=5^2-4\cdot1\cdot7=-3< 0\)
Do đó: Phương trình vô nghiệm
Trường hợp 2: x>=-3
Pt sẽ là \(x^2+6x+3+x+3+10=0\)
\(\Leftrightarrow x^2+7x+16=0\)
\(\Delta=7^2-4\cdot1\cdot16=49-64=-15< 0\)
Do đó: Phương trình vô nghiệm