giải phương trình 6x^4 - 5x^3 - 38x^2 -5x +6
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\(\text{a) }\left(x^2+x\right)^2+4\left(x^2+x\right)=12\\ \Leftrightarrow\text{Đặt }x^2+x=y\\ \Leftrightarrow y^2+4y=12\\ \Leftrightarrow y^2+6y-2y-12=0\\ \Leftrightarrow\left(y^2+6y\right)-\left(2y+12\right)=0\\ \Leftrightarrow y\left(y+6\right)-2\left(y+6\right)=0\\ \Leftrightarrow\left(y+6\right)\left(y-2\right)=0\\ \Leftrightarrow\left(x^2+x+6\right)\left(x^2+x-2\right)=0\\ \Leftrightarrow\left(x^2+x+\dfrac{1}{4}+\dfrac{23}{4}\right)\left(x^2+2x-x-2\right)=0\\ \Leftrightarrow\left[\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{23}{4}\right]\left[\left(x^2+2x\right)-\left(x+2\right)\right]=0\\ \Leftrightarrow\left[\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}\right]\left[x\left(x+2\right)-\left(x+2\right)\right]=0\\ \Leftrightarrow\left(x-1\right)\left(x+2\right)=0\left(Vì\text{ }\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}\ne0\right)\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\\ \text{Vậy }S=\left\{1;-2\right\}\\ \)
\(\text{b) }6x^4-5x^3-38x^2-5x+6=0\\ \Leftrightarrow x^2\left(6x^2-5x-38-\dfrac{5}{x}+\dfrac{6}{x^2}\right)=0\\ \Leftrightarrow x^2\left[\left(6x^2+12+\dfrac{6}{x^2}\right)-\left(5x+\dfrac{5}{x}\right)-50\right]=0\\ \Leftrightarrow x^2\left[6\left(x^2+2+\dfrac{1}{x^2}\right)-5\left(x+\dfrac{1}{x}\right)-50\right]=0\\ \Leftrightarrow x^2\left[6\left(x+\dfrac{1}{x}\right)^2-5\left(x+\dfrac{1}{x}\right)-50\right]=0\\ \text{Đặt }x+\dfrac{1}{x}=y\\ \Leftrightarrow x^2\left(6y^2-5y-50\right)=0\\ \Leftrightarrow x^2\left(6y^2-20y+15y-50\right)=0\\ \Leftrightarrow x^2\left[\left(6y^2-20y\right)+\left(15y-50\right)\right]=0\\ \Leftrightarrow x^2\left[2y\left(3y-10\right)+5\left(3y-10\right)\right]=0\\ \Leftrightarrow x^2\left(2y+5\right)\left(3y-10\right)=0\\ \Leftrightarrow x^2\left(2x+\dfrac{2}{x}+5\right)\left(3x+\dfrac{3}{x}-10\right)=0\\ \Leftrightarrow\left(2x^2+2+5x\right)\left(3x^2+3-10x\right)=0\\ \Leftrightarrow\left(2x^2+4x+x+2\right)\left(3x^2-9x-x+3\right)=0\\ \Leftrightarrow\left[\left(2x^2+4x\right)+\left(x+2\right)\right]\left[\left(3x^2-9x\right)-\left(x-3\right)\right]=0\\ \Leftrightarrow\left[2x\left(x+2\right)+\left(x+2\right)\right]\left[3x\left(x-3\right)-\left(x-3\right)\right]=0\\ \Leftrightarrow\left(2x+1\right)\left(x+2\right)\left(3x-1\right)\left(x-3\right)=0\\ \)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\x+2=0\\3x-1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-1\\x=-2\\3x=1\\x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-2\\x=\dfrac{1}{3}\\x=3\end{matrix}\right.\\ \text{Vậy }S=\left\{-\dfrac{1}{2};-2;\dfrac{1}{3};3\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(\left(x^2+5x\right)^2-2\left(x^2+5x\right)-24=0\)
\(\Leftrightarrow\left(x^2+5x\right)\left(x^2+5x-2\right)-24=0\)
Đặt t = x2 + 5x - 1
Khi đó : (x2 + 5x) = t + 1 ; (x2 + 5x - 2) = t - 1
Ta có : C = (x2 + 5x - 2)2 (x2 + 5x - 2) - 24 = 0
=> (x2 + 5x - 2)3 = 24
MK chỉ giả được đến đây thôi
![](https://rs.olm.vn/images/avt/0.png?1311)
\(6x^4+5x^3-38x^2+5x+6=0\\ \Leftrightarrow6x^4+20x^3+6x^2-15x^3-50x^2-15x+6x^2+20x+6=0\\ \Leftrightarrow2x^2\left(3x^2+10x+3\right)-5x\left(3x^2+10x+3\right)+2\left(3x^2+10x+3\right)=0\\ \Leftrightarrow\left(3x^2+10x+3\right)\left(2x^2-5x+2\right)=0\\ \Leftrightarrow\left(3x^2+x+9x+3\right)\left(2x^2-x-4x+2\right)=0\\ \Leftrightarrow\left[x\left(3x+1\right)+3\left(3x+1\right)\right]\left[x\left(2x-1\right)-2\left(2x-1\right)\right]=0\\ \Leftrightarrow\left(3x+1\right)\left(x+3\right)\left(2x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x+1=0\\x+3=0\\2x-1=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{3}\\x=-3\\x=\dfrac{1}{2}\\x=2\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
pt bậc 4 => có 4 nghiệm.
bấm máy tính tìm nghiệm đẹp (-2 và 3). Chia sơ đồ hoocne.
2 nghiệm đẹp (-2 và 3) được rồi, còn 2 nghiệm còn lại thì giải pt bậc 2 là ra.
kq: x=-2, x=3, x=1/3 , x=-1/2
Ta có \(6x^4-5x^3-38x^2-5x+6=0\Leftrightarrow6x^4+12x^3-17x^3-34x^2-4x^2-8x+3x+6=0\Leftrightarrow6x^3\left(x+2\right)-17x^2\left(x+2\right)-4x\left(x+2\right)+3\left(x+2\right)=0\Leftrightarrow\left(x+2\right)\left(6x^3-17x^2-4x+3\right)=0\Leftrightarrow\left(x+2\right)\left(6x^3-18x^2+x^2-3x-x+3\right)=0\Leftrightarrow\left(x+2\right)\left[6x^2\left(x-3\right)+x\left(x-3\right)-\left(x-3\right)\right]=0\Leftrightarrow\left(x+2\right)\left(x-3\right)\left(6x^2+x-1\right)=0\Leftrightarrow\left(x+2\right)\left(x-3\right)\left(6x^2-2x+3x-1\right)=0\Leftrightarrow\left(x+2\right)\left(x-3\right)\left[2x\left(3x-1\right)+\left(3x-1\right)\right]=0\Leftrightarrow\left(x+2\right)\left(x-3\right)\left(3x-1\right)\left(2x+1\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}x+2=0\\x-3=0\\3x-1=0\\2x+1=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=-2\\x=3\\x=\dfrac{1}{3}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Vậy S={\(-\dfrac{1}{2};-2;\dfrac{1}{3};3\)}
![](https://rs.olm.vn/images/avt/0.png?1311)
Giúp tôi giải toán - Hỏi đáp, thảo luận về toán học - Học toán với OnlineMath tích mình nha
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\(6x^4+5x^3-38x^2+5x+6=0\)
\(6x^4-12x^3+17x^3-34x^2-4x^2+8x-3x+6=0\)
\(6x^3\left(x-2\right)+17x^2\left(x-2\right)-4x\left(x-2\right)-3\left(x-2\right)=0\)
\(\left(x-2\right)\left(6x^3+17x^2-4x-3\right)=0\)
\(\left(x-2\right)\left[6x^3-3x^2+20x^2-10x+6x-3\right]=0\)
\(\left(x-2\right)\left[6x^2\left(x-\dfrac{1}{2}\right)+20x\left(x-\dfrac{1}{2}\right)+6\left(x-\dfrac{1}{2}\right)\right]=0\)
\(\left(x-2\right)\left(x-\dfrac{1}{2}\right)\left(6x^2+20x+6\right)=0\)
=> \(\left[{}\begin{matrix}x-2=0\\x-\dfrac{1}{2}=0\\6x^2+20x+6=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=2\\x=\dfrac{1}{2}\\\left(3x+1\right)\left(x+3\right)=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=2\\x=\dfrac{1}{2}\\x=-3\\x=-\dfrac{1}{3}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(6x^4+5x^3-38x^2+5x+6=0\)
\(\Leftrightarrow6x^4+20x^3+6x^2-15x^3-50x^2-15x+6x^2+20x+6=0\)
\(\Leftrightarrow2x^2\left(3x^2+10x+3\right)-5x\left(3x^2+10x+3\right)+2\left(3x^2+10x+3\right)=0\)
\(\Leftrightarrow\left(3x^2+10x+3\right)\left(2x^2-5x+2\right)=0\)
\(\Leftrightarrow\left(3x^2+x+9x+3\right)\left(2x^2-x-4x+2\right)=0\)
\(\Leftrightarrow\left[x\left(3x+1\right)+3\left(3x+1\right)\right]\left[x\left(2x-1\right)-2\left(2x-1\right)\right]=0\)
\(\Leftrightarrow\left(3x+1\right)\left(x+3\right)\left(2x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\)\(3x+1=0\)
hoặc \(x+3=0\)
hoặc \(2x-1=0\)
hoặc \(x-2=0\)
\(\Leftrightarrow\)\(x=-\frac{1}{3}\)
hoặc \(x=-3\)
hoặc \(x=\frac{1}{2}\)
hoặc \(x=2\)
Vậy tập nghiệm của phương trình là \(S=\left\{-\frac{1}{3};-3;\frac{1}{2};2\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Nhận thấy \(x=0\) ko phải nghiệm
Với \(x\ne0\) chia 2 vế của pt cho \(x^2\) ta được:
\(6\left(x^2+\dfrac{1}{x^2}\right)-5\left(x+\dfrac{1}{x}\right)-38=0\)
Đặt \(x+\dfrac{1}{x}=t\Rightarrow x^2+\dfrac{1}{x^2}=t^2-2\)
\(\Rightarrow6\left(t^2-2\right)-5t-38=0\)
\(\Leftrightarrow6t^2-5t-50=0\Rightarrow\left[{}\begin{matrix}t=\dfrac{10}{3}\\t=-\dfrac{5}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}=\dfrac{10}{3}\\x+\dfrac{1}{x}=-\dfrac{5}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}3x^2-10x+3=0\\2x^2+5x+2=0\end{matrix}\right.\)
\(\Rightarrow x=\left\{-2;-\dfrac{1}{2};\dfrac{1}{3};3\right\}\)