K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
TD
0
CX
0
ML
3 tháng 4 2017
\(6x^4-5x^3-38x^2-5x+6=0\)
\(\Leftrightarrow6x^4-12x^3+17x^3-34^2-4x^2+8x-3x+6=0\)
\(\Leftrightarrow6x^3\left(x-2\right)+17x^2\left(x-2\right)-4x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(6x^3+18x^2-4x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(6x^3+18x^2-x^2-3x-x-3=0\right)\)
\(\Leftrightarrow\left(x-2\right)\left[6x^2\left(x+3\right)-x\left(x+3\right)-\left(x+3\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)\left(6x^2-x-1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)\left(6x^2-3x+2x-1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)\left[6x\left(x-\frac{1}{2}\right)+2\left(x-\frac{1}{2}\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)\left(x-\frac{1}{2}\right)\left(6x+2\right)=0\)
15 tháng 10 2015
Giúp tôi giải toán - Hỏi đáp, thảo luận về toán học - Học toán với OnlineMath tích mình nha
28 tháng 12 2017
a, \(x^4-6x^3+11x^2-6x+1=0\)
\(\Rightarrow\left(x^2-3x+1\right)^2=0\)
\(\Rightarrow x^2-3x+1=0\)
\(\Rightarrow x=\frac{\pm\sqrt{5}+3}{2}\)
Chúc bạn học tốt
28 tháng 12 2017
\(x^4-\left(6x^2-2x^2\right)+\left(9x^2-6x+1\right)=0\)
\(x^4-2x^2\left(3x-1\right)+\left(3x-1\right)^2=0\)
\(\left(x^2-3x+1\right)^2=0\)
tự làm
B) \(\left(6x^4-18x^3\right)+\left(13x^{^3}-39x^2\right)+\left(x-3x\right)-\left(2x-6\right)=0\)
\(6x^3\left(x-3\right)+13x^2\left(x-3\right)+x\left(x-3\right)-2\left(x-3\right)=0\)
\(\left(x-3\right)\left(6x^3+13x^2-2\right)=0\)
\(\left(x-3\right)\left(6x^3+12x^2+x^2+2x-x-2\right)\)
\(\left(x-3\right)\left\{6x^2\left(x+2\right)+x\left(x+2\right)-\left(x+2\right)\right\}\)
\(\left(x-3\right)\left(x+2\right)\left(6x^2-x-1\right)\)
\(\left(x-3\right)\left(x+2\right)\left(6x^2-3x+2x-1\right)\)
\(\left(x-3\right)\left(x+2\right)\left(3x\left(2x-1\right)+\left(2x-1\right)\right)\)
\(\left(x-3\right)\left(x+2\right)\left(2x-1\right)\left(3x+1\right)=0\)
câu C nghĩ đã
MS
7 tháng 1 2018
\(6x^4+5x^3-38x^2+5x+6=0\)
\(6x^4-12x^3+17x^3-34x^2-4x^2+8x-3x+6=0\)
\(6x^3\left(x-2\right)+17x^2\left(x-2\right)-4x\left(x-2\right)-3\left(x-2\right)=0\)
\(\left(x-2\right)\left(6x^3+17x^2-4x-3\right)=0\)
\(\left(x-2\right)\left[6x^3-3x^2+20x^2-10x+6x-3\right]=0\)
\(\left(x-2\right)\left[6x^2\left(x-\dfrac{1}{2}\right)+20x\left(x-\dfrac{1}{2}\right)+6\left(x-\dfrac{1}{2}\right)\right]=0\)
\(\left(x-2\right)\left(x-\dfrac{1}{2}\right)\left(6x^2+20x+6\right)=0\)
=> \(\left[{}\begin{matrix}x-2=0\\x-\dfrac{1}{2}=0\\6x^2+20x+6=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=2\\x=\dfrac{1}{2}\\\left(3x+1\right)\left(x+3\right)=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=2\\x=\dfrac{1}{2}\\x=-3\\x=-\dfrac{1}{3}\end{matrix}\right.\)
AN
27 tháng 12 2017
a, \(x^4-6x^3+11x^2-6x+1=0\)
=> \(x^4-6x^3+9x^2+2x^2-6x+1=0\)
=> \(x^2+3x+1=0\)
=> \(\Delta\) =\(b^2-4c\)
=\(3^2.4=5\)
Nên \(\sqrt{\Delta}=5\)
x= \(\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{-3+\sqrt{5}}{2}\)
hoặc x= \(\dfrac{b+\sqrt{\Delta}}{2a}=\dfrac{3+\sqrt{5}}{2}\)
LC
1
4 tháng 2 2018
1 ) \(\left(x^2+x\right)^2+4\left(x^2+x\right)=12\)
Đặt \(t=x^2+x\), ta được :
\(t^2+4t-12=0\)
\(\Leftrightarrow t^2-2t+6t-12=0\)
\(\Leftrightarrow\left(t-2\right)\left(t+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=2\\t=-6\end{matrix}\right.\)
+ ) Khi \(t=2,\) thì :
\(x^2+x=2\)
\(\Leftrightarrow x^2+x-2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
+ ) Khi \(t=-6,\) thì :
\(x^2+x=-6\)
\(\Leftrightarrow x^2+x+6=0\)
\(\Leftrightarrow x^2+2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{23}{4}=0\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}=0\) ( vô lí )
Vậy .........
2 ) \(6x^4-5x^3-38x^2-5x+6=0\)
\(\Leftrightarrow6x^4-18x^3+13x^3-39x^2+x^2-3x-2x+6=0\)
\(\Leftrightarrow6x^3\left(x-3\right)+13x^2\left(x-3\right)+x\left(x-3\right)-2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(6x^3+13x^2+x-2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(6x^3+12x^2+x^2+2x-x-2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left[6x^2\left(x+2\right)+x\left(x+2\right)-\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)\left(6x^2+x-1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)\left(6x^2+3x-2x-1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)\left[3x\left(2x+1\right)-\left(2x+1\right)\right]=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)\left(3x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\\x=\dfrac{1}{3}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
\(6x^4+5x^3-38x^2+5x+6=0\\ \Leftrightarrow6x^4+20x^3+6x^2-15x^3-50x^2-15x+6x^2+20x+6=0\\ \Leftrightarrow2x^2\left(3x^2+10x+3\right)-5x\left(3x^2+10x+3\right)+2\left(3x^2+10x+3\right)=0\\ \Leftrightarrow\left(3x^2+10x+3\right)\left(2x^2-5x+2\right)=0\\ \Leftrightarrow\left(3x^2+x+9x+3\right)\left(2x^2-x-4x+2\right)=0\\ \Leftrightarrow\left[x\left(3x+1\right)+3\left(3x+1\right)\right]\left[x\left(2x-1\right)-2\left(2x-1\right)\right]=0\\ \Leftrightarrow\left(3x+1\right)\left(x+3\right)\left(2x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x+1=0\\x+3=0\\2x-1=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{3}\\x=-3\\x=\dfrac{1}{2}\\x=2\end{matrix}\right.\)