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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
Tìm x biết:
b/\(\left(2x+3\right)^2-\left(5x-4\right)\left(5x+4\right)=\left(x+5\right)^2-\left(3x-1\right)\left(7x+2\right)-\left(x^2-x+1\right)\)
<=> \(4x^2 +12x+9-25x^2+16-x^2-10x-25+21x^2+6x-7x-2+x^2-x+1=0\)
<=>0x-1=0
<=>0x=1 (vô lí) (dòng này không cần ghi thêm cũng được)
=> Không có giá trị x nào thỏa mãn
c/ \((1-3x)^2-(x-2)(9x+1)=(3x-4)(3x+4)-9(x+3)^2\)
<=>\(1-6x+9x^2-9x^2-x+18x+2-9x^2+16+9x^2+54x+81=0\)
<=> 65x+100=0
<=> x=\(\dfrac{-20}{13}\)
d/\((3x+4)(3x-4)-(2x+5)^2=(x-5)^2+(2x+1)^2-(x^2-2x)+(x-1)^2\)
<=> \(9x^2-16-4x^2-20x-25-x^2+10x-25-4x^2-4x-1+x^2+2x-x^2+2x-1=0\)
<=> -10x-68=0
<=> x=\(\dfrac{-34}{5}\)
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 \(\left|2x-5\right|=t\ge0\)
\(\Rightarrow t^2+4t-12=0\Rightarrow\left[{}\begin{matrix}t=2\\t=-6\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\left|2x-5\right|=2\Rightarrow\left[{}\begin{matrix}2x-5=2\\2x-5=-2\end{matrix}\right.\)
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. \(\Leftrightarrow x^2-6x+9>0\)
\(\Leftrightarrow\left(x-3\right)^2>0\)
\(\Rightarrow x\ne3\)
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:
\(\Leftrightarrow\left(x^2-2x+8\right)^2=\left(x^2-1\right)^2\)
\(\Leftrightarrow\left(x^2-2x+8\right)^2-\left(x^2-1\right)^2=0\)
\(\Leftrightarrow\left(2x^2-2x+7\right)\left(-2x+9\right)=0\)
\(\Leftrightarrow-2x+9=0\Rightarrow x=\frac{9}{2}\)