3x^3-4x^2+x
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
31 tháng 10 2021
\(a,\Rightarrow4x^2-20x-4x^2+3x+4x-3=5\\ \Rightarrow-13x=8\Rightarrow x=-\dfrac{8}{13}\\ b,\Rightarrow3x^2-10x+8-3x^2+27x=-3\\ \Rightarrow17x=-11\Rightarrow x=-\dfrac{11}{17}\\ c,\Rightarrow\left(x+3\right)\left(2-x\right)=0\Rightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\\ d,\Rightarrow2x\left(4x^2-25\right)=0\\ \Rightarrow2x\left(2x-5\right)\left(2x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{5}\\x=-\dfrac{2}{5}\end{matrix}\right.\\ e,Sửa:\left(4x-3\right)^2-3x\left(3-4x\right)=0\\ \Rightarrow\left(4x-3\right)^2+3x\left(4x-3\right)=0\\ \Rightarrow\left(4x-3\right)\left(7x-3\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{7}\end{matrix}\right.\)
31 tháng 10 2021
a.
4x(x-5) - (x-1)(4x-3)-5=0
4x^2-20x-4x^2+3x+4x+3=0
(4x^2-4x^2)+(-20x+3x+4x)+3=0
13x+3 = 0
13x=-3
x=-3/13
b,
(3x-4)(x-2)-3x(x-9)+3=0
3x^2-6x-4x+8 - 3x^2+27x+3=0
(3x^2-3x^2)+(-6x-4x+27x)+(8+3)=0
17x+11=0
17x=-11
x=-11/17
c, 2(x+3)-x^2-3x=0
2(x+3) - x(x+3)=0
(x+3)(2-x)=0
TH1: x+3 = 0; x=-3
TH2: 2-x=0;x=2
29 tháng 11 2023
a: \(x^3-4x^2-x+4=0\)
=>\(\left(x^3-4x^2\right)-\left(x-4\right)=0\)
=>\(x^2\left(x-4\right)-\left(x-4\right)=0\)
=>\(\left(x-4\right)\left(x^2-1\right)=0\)
=>\(\left[{}\begin{matrix}x-4=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x^2=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;1;-1\right\}\)
b: Sửa đề: \(x^3+3x^2+3x+1=0\)
=>\(x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=0\)
=>\(\left(x+1\right)^3=0\)
=>x+1=0
=>x=-1
c: \(x^3+3x^2-4x-12=0\)
=>\(\left(x^3+3x^2\right)-\left(4x+12\right)=0\)
=>\(x^2\cdot\left(x+3\right)-4\left(x+3\right)=0\)
=>\(\left(x+3\right)\left(x^2-4\right)=0\)
=>\(\left(x+3\right)\left(x-2\right)\left(x+2\right)=0\)
=>\(\left[{}\begin{matrix}x+3=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\\x=-2\end{matrix}\right.\)
d: \(\left(x-2\right)^2-4x+8=0\)
=>\(\left(x-2\right)^2-\left(4x-8\right)=0\)
=>\(\left(x-2\right)^2-4\left(x-2\right)=0\)
=>\(\left(x-2\right)\left(x-2-4\right)=0\)
=>(x-2)(x-6)=0
=>\(\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)
21 tháng 5 2022
a: \(P\left(x\right)=2x^3+x^2+x+2\)
\(Q\left(x\right)=x^3+x^2+x+1\)
b: \(P\left(-1\right)=2\cdot\left(-1\right)+1-1+2=0\)
\(Q\left(-1\right)=-1+1-1+1=0\)
Do đó: x=-1 là nghiệm chung của P(x), Q(x)
C
21 tháng 5 2022
\(P\left(x\right)=2x^3-2x+x^2+3x+2\)
\(P\left(x\right)=2x^3+x^2+x+2\)
\(Q\left(x\right)=4x^3-3x^2-3x+4x-3x^3+4x^2+1\)
\(Q\left(x\right)=x^3+x^2+x+1\)
__________________________________________________
\(P\left(-1\right)=2.\left(-1\right)^3+\left(-1\right)^2+\left(-1\right)+2\)
\(P\left(-1\right)=0\)
\(Q\left(-1\right)=\left(-1\right)^3+\left(-1\right)^2+\left(-1\right)+1\)
\(Q\left(-1\right)=0\)
Vậy x = -1 là nghiệm của P(x),Q(x)
TA
1
13 tháng 10 2021
c: Ta có: \(x^3+3x^2+3x-7=0\)
\(\Leftrightarrow x+1=2\)
hay x=1
b: Ta có: \(x\left(x-3\right)-4x+12=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
13 tháng 5 2018
a) \(P_{\left(x\right)}=2x^3-2x+x^2+3x+2\)
\(P_{\left(x\right)}=2x^3+x^2+x+2\)
\(Q_{\left(x\right)}=4x^3-3x^2-3x+4x-3x^3+4x^2+1\)
\(Q_{\left(x\right)}=x^3+x^2+x+1\)
b) \(P_{\left(x\right)}+Q_{\left(x\right)}=\left(2x^3+x^2+x+2\right)+\left(x^3+x^2++x+1\right)\)
\(=3x^3+2x^2+2x+3\)
\(3x^3-4x^2+x\)
\(=x\left(3x^2-4x+1\right)\)
\(=x\left[3x^2-3x-x+1\right]\)
\(=x\left[3x\left(x-1\right)-\left(x-1\right)\right]\)
\(=x\left(x-1\right)\left(3x-1\right)\)