x mũ3-3x mũ2=0
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1.\left(x^3-1\right)\left(x^2+1\right)=0\)
\(< =>\left\{{}\begin{matrix}x^3-1=0\\x^2+1=0\end{matrix}\right.\)
\(< =>\left\{{}\begin{matrix}x^3=1\\x^2=-1\left(kxđ\right)\end{matrix}\right.\)
<=>x=1
vậy ...
\(2.\left(2x+6\right)\left(3x^2-12\right)=0\)
\(< =>\left\{{}\begin{matrix}2x+6=0\\3x^2-12=0\end{matrix}\right.\)
\(< =>\left\{{}\begin{matrix}2x=-6\\3x^2=12\end{matrix}\right.\)
\(< =>\left\{{}\begin{matrix}x=-3\\x^2=4\end{matrix}\right.\)
\(< =>\left\{{}\begin{matrix}x=-3\\x=2\\x=-2\end{matrix}\right.\)
vậy ...
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(5m^2-8m^2-9m^2\right)\left(-n^3+4n^3\right)=-12m^2.3n^3=-36m^2n^3\)
Để A\(\ge0\) thì \(m^2n^3\le0\)
\(\Leftrightarrow\left\{{}\begin{matrix}m\in Q\\n\le0\end{matrix}\right.\)
A=(5m2−8m2−9m2)(−n3+4n3)=−12m2.3n3=−36n5A=(5m2−8m2−9m2)(−n3+4n3)=−12m2.3n3=−36n5
Để A≥0≥0 thì n5≤0⇔n≤0
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Giải:
a) \(x\left(x-2\right)-\left(x+3\right).x+7+9x=6\)
\(\Leftrightarrow x^2-2x-\left(x^2+3x\right)+7+9x=6\)
\(\Leftrightarrow x^2-2x-x^2-3x+7+9x=6\)
\(\Leftrightarrow4x=-1\)
\(\Leftrightarrow x=-\dfrac{1}{4}\)
Vậy ...
b) \(\left(3x-5\right)\left(7-5x\right)-\left(5x+2\right)\left(2-3x\right)=4\)
\(\Leftrightarrow21x-35-15x^2+25x-\left(10x+2-15x^2+6x\right)=4\)
\(\Leftrightarrow21x-35-15x^2+25x-10x-2+15x^2-6x=4\)
\(\Leftrightarrow30x-37=4\)
\(\Leftrightarrow30x=41\)
\(\Leftrightarrow x=\dfrac{41}{30}\)
Vậy ...
c) \(\left(x+2\right)\left(x^2-2x+4\right)-\left(x^3+3\right)=14x\) (Sửa đề)
\(\Leftrightarrow x^3+8-x^3-3=14x\)
\(\Leftrightarrow5=14x\)
\(\Leftrightarrow x=\dfrac{5}{14}\)
Vậy ...
d) \(\left(x^2-x+1\right)\left(x+1\right)-x^3-3x=2\)
\(\Leftrightarrow x^3+1-x^3-3x=2\)
\(\Leftrightarrow1-3x=2\)
\(\Leftrightarrow-3x=1\)
\(\Leftrightarrow x=-\dfrac{1}{3}\)
Vậy ...
a) \(x\left(x-2\right)-\left(x+3\right)x+7+9x=6\)
=> \(x^2-2x-x-3x+7+9x=6\)
=> \(x^2-2x-x^2-3x+7+9x=6\)
=> \(\left(x^2-x^2\right)+\left(-2x-3x+9x\right)=6-7\)
=> \(4x=-1\)
Vậy \(x=\dfrac{-1}{4}\)
b) \(\left(3x-5\right)\left(7-5x\right)-\left(5x+2\right)\left(2-3x\right)=4\)
=>\(21x-15x^2-35+25x-10x+15x^2-4+6x=4\)
=> \(\left(21x+25x-10x+6x\right)\)\(+\left(-15x^2+15x^2\right)\)\(=4+35+4\)
=> \(42x=43\)
Vậy \(x=\dfrac{43}{42}\)
c) \(\left(x+2\right)\left(x^2-2x+4\right)-\left(x^3+3\right)=14\)
=> \(x^3-2x^2+4x+2x^2-4x+8-x^3-3\)\(=14x\)
=>\(\left(x^3-x^3\right)+\left(-2x^2+2x^x\right)+\left(4x-4x\right)+\left(8-3\right)\)\(=14x\)
=> \(5=14x\)
Vậy \(x=\dfrac{5}{14}\)
d) \(\left(x^2-x+1\right)\left(x+1\right)-x^3-3x=2\)
=> \(x^3+x^2+x+x^2-x+1-x^3-3x=2\)
=>\(\left(x^3-x^3\right)+\left(-x^2+x^2\right)+\left(x-x-3x\right)=2-1\)
=> \(-3x=1\)
Vậy \(x=\dfrac{-1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=\left(-2\right)^0+\left(-2\right)^1+\left(-2\right)^2+\left(-2\right)^3...+2^{2014}+2^{2015}\)
\(2S=\left(-2\right)^1+\left(-2\right)^2+\left(-2\right)^3+\left(-2\right)^4+...+\left(-2\right)^{2015}+\left(-2\right)^{^{ }2016}\)
\(2S-S=\left[\left(-2\right)^1+\left(-2\right)^2+\left(-2\right)^3+\left(-2\right)^4+...+\left(-2\right)^{2015}+\left(-2\right)^{2016}\right]\)\(-\left[\left(-2\right)^0+\left(-2\right)^1+\left(-2\right)^2+\left(-2\right)^3+...+\left(-2\right)^{2014}+\left(-2\right)^{2015}\right]\)
\(S=\left(-2\right)^{2016}-\left(-2\right)^0=\left(-2\right)^{2016}-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
125 : \(x\) = 22 - (-1)
125 : \(x\) = 4 + 1
125 : \(x\) = 5
\(x\) = 125 : 5
\(x\) = 25
18 - (\(x\) + 14) : 3 = 27
(\(x\) + 14) : 3 = 18 - 27
(\(x\) + 14) : 3 = - 9
\(x\) + 14 = - 9.3
\(x\) + 14 = -27
\(x\) = -27 - 14
\(x\) = - 41
\(x^3-3x^2=0\)
\(\Leftrightarrow x^2\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
khỉ nghĩ như này..
x3-3x2=0
(=)x2 (x-3)=0
(=)x2=0,hoac x-3=0
(=)x=3