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a) \(x^4-4x^3+12x-9=0\)
\(\Leftrightarrow x^4-x^3-3x^3+3x^2-3x^2+3x+9x-9=0\)
\(\Leftrightarrow x^3\left(x-1\right)-3x^2\left(x-1\right)-3x\left(x-1\right)+9\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3-3x^2-3x+9\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x-3\right)-3\left(x-3\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-3\right)\left(x-3\right)=0\)
\(\Leftrightarrow x-1=0\)hoặc \(x^2-3=0\)hoặc \(x-3=0\)
\(\Leftrightarrow x=1\)hoặc \(x=\pm\sqrt{3}\)hoặc \(x=3\)
Vậy tập nghiệm của phương trình là : \(S=\left\{1;\pm\sqrt{3};3\right\}\)
b) \(x^5-5x^3+4x=0\)
\(\Leftrightarrow x^5-x^3-4x^3+4x=0\)
\(\Leftrightarrow x^3\left(x^2-1\right)-4x\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x^3-4x\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow x\left(x^2-4\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow x=0\)hoặc \(x=\pm2\)hoặc \(x=\pm1\)
Vậy tập nghiệm của phương trình là : \(S=\left\{0;\pm2;\pm1\right\}\)
c) \(x^4-4x^3+3x^2+4x-4=0\)
\(\Leftrightarrow x^4-x^3-3x^3+3x^2+4x-4=0\)
\(\Leftrightarrow x^3\left(x-1\right)-3x^2\left(x-1\right)+4\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3-3x^2+4=0\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x^3-2x^2-x^2+4=0\right)\)
\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x-2\right)-\left(x-2\right)\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x^2+x+2\right)=0\)
\(\Leftrightarrow x-1=0\)
hoặc \(x^2+x+2=\left(x+\frac{1}{2}^2\right)+\frac{7}{4}=0\left(ktm\right)\)
hoặc \(x-2=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\)
Vậy tập nghiệm của phương trình là \(S=\left\{1;2\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^4+4x^3+5x^2-4x+4=0\)
\(\Leftrightarrow x^4+4x^3+4x^2+x^2-4x+4=0\)
\(\Leftrightarrow x^2\left(x+2\right)^2+\left(x-2\right)^2=0\)
Vì \(x^2\left(x+2\right)^2\ge0\forall x;\left(x-2^2\right)\ge0\forall x\)
\(\Rightarrow x^2\left(x+2\right)^2+\left(x-2\right)^2\ge0\)
Mà \(x^2\left(x+2\right)^2+\left(x-2\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}x\left(x+2\right)=0\\x-2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=0;x=-2\\x=2\end{cases}}\)
Mà ko cùng một lúc tồn tại 2 giá trị của x
\(\Rightarrow\)Phương trình vô nghiệm
Vậy ...
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(^{x^3}\) - 7x+6=0
\(\Leftrightarrow\) \(^{x^3}\) - x-6x+6=0
\(\Leftrightarrow\) \(\left(x^3-x\right)\) - \(\left(6x-6\right)\) =0
\(\Leftrightarrow\) x\(\left(x^2-1\right)\) - 6\(\left(x-1\right)\) =0
\(\Leftrightarrow\) x\(\left(x+1\right)\)\(\left(x-1\right)\) - 6\(\left(x-1\right)\) =0
\(\Leftrightarrow\) \(\left(x-1\right)\) \(\left[x-6\left(x+1\right)\right]\) =0
\(\Leftrightarrow\) \(\left(x-1\right)\) \(\left(6-5x\right)\) =0
\(\Leftrightarrow\) \(\left[\begin{matrix}x-1=0\\6-5x=0\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left[\begin{matrix}x=1\\5x=-6\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left[\begin{matrix}x=1\\x=-\frac{6}{5}\end{matrix}\right.\)
Những câu sau dùng phương pháp phân tích đa thức thành nhân tử nhé!
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
\(x^4-4x^3+12x-9=0\)
\(\Rightarrow x^4-4x^3+3x^2-3x^2+12x-9=0\)
\(\Rightarrow x^2\left(x^2-4x+3\right)-3\left(x^2-4x+3\right)=0\)
\(\Rightarrow\left(x^2-4x+3\right)\left(x^2-3\right)=0\)
\(\Rightarrow\left(x^2-3x-x+3\right)\left(x^2-3\right)=0\)
\(\Rightarrow\left[x\left(x-3\right)-\left(x-3\right)\right]\left(x^2-3\right)=0\)
\(\Rightarrow\left(x-3\right)\left(x-1\right)\left(x^2-3\right)=0\)
\(\Rightarrow\left[\begin{matrix}x-3=0\\x-1=0\\x^2-3=0\end{matrix}\right.\)\(\Rightarrow\left[\begin{matrix}x=3\\x=1\\x=\pm\sqrt{3}\end{matrix}\right.\)
Bài 2:
\(x^4-4x^3+3x^2+4x-4=0\)
\(\Rightarrow x^4-4x^3+4x^2-x^2+4x-4=0\)
\(\Rightarrow x^2\left(x^2-4x+4\right)-\left(x^2-4x+4\right)=0\)
\(\Rightarrow\left(x^2-4x+4\right)\left(x^2-1\right)=0\)
\(\Rightarrow\left(x-2\right)^2\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\left[\begin{matrix}x-2=0\\x-1=0\\x+1=0\end{matrix}\right.\)\(\Rightarrow\left[\begin{matrix}x=2\\x=1\\x=-1\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
3\(x^2\) - 4\(x\) - 4 = 0
3(\(x^2\) - 2. \(\dfrac{2}{3}\)\(x\) + \(\dfrac{4}{9}\)) - \(\dfrac{16}{3}\) = 0
3.(\(x-\dfrac{2}{3}\))2 = \(\dfrac{16}{3}\)
(\(x-\dfrac{2}{3}\))2 = \(\dfrac{16}{9}\)
\(\left[{}\begin{matrix}x-\dfrac{2}{3}=\dfrac{4}{3}\\x-\dfrac{2}{3}=-\dfrac{4}{3}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{4}{3}+\dfrac{2}{3}\\x=-\dfrac{4}{3}+\dfrac{2}{3}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=2\\x=-\dfrac{2}{3}\end{matrix}\right.\)
S = { -\(\dfrac{2}{3}\); 2}
3x² - 4x - 4 = 0
⇔ 3x² - 6x + 2x - 4 = 0
⇔ (3x² - 6x) + (2x - 4) = 0
⇔ 3x(x - 2) + 2(x - 2) = 0
⇔ (x - 2)(3x + 2) = 0
⇔ x - 2 = 0 hoặc 3x + 2 = 0
*) x - 2 = 0
⇔ x = 2
*) 3x + 2 = 0
⇔ 3x = -2
⇔ x = -2/3
Vậy S = {-2/3; 2}
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^4-4x^3+3x^2+4x-4=0\)
\(\Leftrightarrow\) \(x^4-4x^3+4x^2-x^2+4x-4=0\)
\(\Leftrightarrow\) \(x^2\left(x^2-4x+4\right)-\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\) \(x^2\left(x-2\right)^2-\left(x-2\right)^2=0\)
\(\Leftrightarrow\) \(\left(x-2\right)^2\left(x^2-1\right)=0\)
\(\Leftrightarrow\) \(^{\left(x-2\right)^2=0}_{x^2-1=0}\) \(\Leftrightarrow\) \(^{x-2=0}_{x^2=1}\) \(\Leftrightarrow\) \(^{x=2}_{x=^+_-1}\)
Vậy, \(S=\left\{-1;1;2\right\}\)
\(3x^2+4x-4=0\)
\(\Leftrightarrow3x^2+6x-2x-4=0\)
\(\Leftrightarrow\left(3x^2+6x\right)-\left(2x+4\right)=0\)
\(\Leftrightarrow3x\left(x+2\right)-2\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\3x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=\frac{2}{3}\end{cases}}}\)