Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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
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. \(\Leftrightarrow x^2-6x+9>0\)
\(\Leftrightarrow\left(x-3\right)^2>0\)
\(\Rightarrow x\ne3\)
b: \(\Leftrightarrow\left(x^2+3x+2\right)\left(x^2+3x-18\right)=-36\)
\(\Leftrightarrow\left(x^2+3x\right)^2-16\left(x^2+3x\right)=0\)
\(\Leftrightarrow\left(x^2+3x\right)\left(x^2+3x-16\right)=0\)
hay \(x\in\left\{0;-3;\dfrac{-3+\sqrt{73}}{2};\dfrac{-3-\sqrt{73}}{2}\right\}\)
c: \(\Leftrightarrow6x^4-18x^3-17x^3+51x^2+11x^2-33x-2x+6=0\)
\(\Rightarrow\left(x-3\right)\left(6x^3-17x^2+11x-2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(6x^3-12x^2-5x^2+10x+x-2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-2\right)\left(6x^2-5x+1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-2\right)\left(3x-1\right)\left(2x-1\right)=0\)
hay \(x\in\left\{3;2;\dfrac{1}{3};\dfrac{1}{2}\right\}\)
d: \(\Leftrightarrow\left(x-1\right)^2\cdot\left(x^2+3x+1\right)=0\)
hay \(x\in\left\{1;\dfrac{-3+\sqrt{5}}{2};\dfrac{-3-\sqrt{5}}{2}\right\}\)
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+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 ...