(-3x mũ3+6xy-3x) (1/3xy mũ3)
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`x^3 - 3x^2y + x + 3xy^2 - y - y^3`
`=(x)^3 - 3*(x)^2*y + 3*x*y^2 - (y)^3 + (x - y)`
`= (x - y)^3 + (x - y)`
`= (x - y)[(x - y)^2 + 1]`
`= (x - y)(x - y - 1)(x - y + 1)`
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`@` CT:
`(A - B)^3=A^3-3A^2B+3AB^2- B^3`
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\(\left(2x-1\right)\left(3x+5\right)+\left(-6x^3+5x\right):x\)
\(=6x^2+10x-3x-5-6x^3+5x:x\)
\(=-6x^3+6x^2+12x-5:x\)
\(=-6x^2+6x+12-\dfrac{5}{x}=-6\left(x^2-x-12\right)-\dfrac{5}{x}\)
A = (2\(x\) - 1)(3\(x\) + 5) + (-6\(x\)3 + 5\(x\)): \(x\)
A = 6\(x^2\) + 10\(x\) - 3\(x\) - 5 + \(x\)(- 6\(x^2\) + 5): \(x\)
A = 6\(x^2\) + 7\(x\) - 5 - 6\(x^2\) + 5
A = (6\(x^2\) - 6\(x^2\)) + 7\(x\) - (5 - 5)
A = 7\(x\)
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\(x^3-3x^2=0\)
\(\Leftrightarrow x^2\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
khỉ nghĩ như này..
x3-3x2=0
(=)x2 (x-3)=0
(=)x2=0,hoac x-3=0
(=)x=3
![](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 ...
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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 \(x=\dfrac{-1}{3}\)
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a) \(\left(x-\frac{1}{2}\right)^3=27\)
=> \(\left(x-\frac{1}{2}\right)^3=3^3\)
=> \(x-\frac{1}{2}=3\)
=> \(x=3+\frac{1}{2}\)
=> \(x=\frac{7}{2}\)
Vậy \(x=\frac{7}{2}.\)
b) \(\left(2x-1\right)^3=-27\)
=> \(\left(2x-1\right)^3=\left(-3\right)^3\)
=> \(2x-1=-3\)
=> \(2x=\left(-3\right)+1\)
=> \(2x=-2\)
=> \(x=\left(-2\right):2\)
=> \(x=-1\)
Vậy \(x=-1.\)
Chúc bạn học tốt!
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\(\left(3x-2\right)^3\cdot4=256\)
\(\left(3x-2\right)^3=256:4\)
\(\left(3x-2\right)^3=64\)
\(\left(3x-2\right)^3=4^3\)
\(\Rightarrow3x-2=4\)
\(3x=4+2\)
\(3x=6\)
\(x=6:3\)
\(x=2\)
(-3x3 + 6xy - 3x)(1/3 xy3)
= -3x3 . 1/3 xy3 + 6xy . 1/3 xy3 - 3x . 1/3 xy3
= -x4y3 + 2x2y4 - x2y3
`(-3x^3+6xy -3x)(1/3 xy^3)`
`= -3x^3 . 1/3 xy^3 + 6xy . 1/3 xy^3 - 3x . 1/3xy^3`
`= -x^4y^3 + 2x^2y^4 - x^2y^3`