`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)`

____

`@` CT:

`(A - B)^3=A^3-3A^2B+3AB^2- B^3`

\(\left(2x-1\right)\left(3x+5\right)+\left(-6x^3+5x\right):x\)

\(=6x^2+10x-3x-5-6x^3+5x:x\)

\(=-6x^3+6x^2+12x-5:x\)

\(=-6x^2+6x+12-\dfrac{5}{x}=-6\left(x^2-x-12\right)-\dfrac{5}{x}\)

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\(x\) 

A = 7x

\(x^3-3x^2=0\)

\(\Leftrightarrow x^2\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

\(x^3-3x^2=0\)

\(\Leftrightarrow x^2\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2=0\\x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

Vậy pt có tập nghiệm S = { 0;3 }

khỉ nghĩ như này..

x³-3x²=0

(=)x² (x-3)=0

(=)x²=0,hoac x-3=0

(=)x=3

\(1.\left(x^3-1\right)\left(x^2+1\right)=0\)

\(< =>\left\{{}\begin{matrix}x^3-1=0\\x^2+1=0\end{matrix}\right.\)

\(< =>\left\{{}\begin{matrix}x^3=1\\x^2=-1\left(kxđ\right)\end{matrix}\right.\)

<=>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 ...

Trong Th này bn nên dùng dấu ''hoặc''

a,\(\left(x^3-1\right)\left(x^2+1\right)=0\)

\(\left[{}\begin{matrix}x^3-1=0\\x^2+1=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x^3=1\\x^2=-1\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=1\\x=\pm1\end{matrix}\right.\)

b, \(\left(2x+6\right)\left(3x^2-12\right)=0\)

\(\left[{}\begin{matrix}2x+6=0\\3x^2-12=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}2x=-6\\3x^2=12\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=-3\\x^2=4\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=-3\\x=\pm2\end{matrix}\right.\)

Bằng nhau

Bạn ghi rõ ràng đề.            

33 phần 47 < 3774 phần 5217 

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 \(x=\dfrac{-1}{3}\)

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!

\(\left(3x-2\right)^3\cdot4=256\)   

\(\left(3x-2\right)^3=256:4\)   

\(\left(3x-2\right)^3=64\)    

\(\left(3x-2\right)^3=4^3\)   

\(\Rightarrow3x-2=4\)   

\(3x=4+2\)   

\(3x=6\)   

\(x=6:3\)   

\(x=2\)