`#3107.101107`

`D = x^3 - y^3 - 3xy` biết `x - y - 1 = 0`

Ta có:

`x - y - 1 = 0`

`=> x - y = 1`

`D = x^3 - y^3 - 3xy`

`= (x - y)(x^2 + xy + y^2) - 3xy`

`= 1 * (x^2 + xy + y^2) - 3xy`

`= x^2+ xy + y^2 - 3xy`

`= x^2 - 2xy + y^2`

`= x^2 - 2*x*y + y^2`

`= (x - y)^2`

`= 1^2 = 1`

Vậy, với `x - y = 1` thì `D = 1`

________

`E = x^3 + y^3` với `x + y = 5; x^2 + y^2 = 17`

`x + y = 5`

`=> (x + y)^2 = 25`

`=> x^2 + 2xy + y^2 = 25`

`=> 2xy = 25 - (x^2 + y^2)`

`=> 2xy = 25 - 17`

`=> 2xy = 8`

`=> xy = 4`

Ta có:

`E = x^3 + y^3`

`= (x + y)(x^2 - xy + y^2)`

`= 5 * [ (x^2 + y^2) - xy]`

`= 5 * (17 - 4)`

`= 5 * 13`

`= 65`

Vậy, với `x + y = 5; x^2 + y^2 = 17` thì `E = 65`

________

`F = x^3 - y^3` với `x - y = 4; x^2 + y^2 = 26`

Ta có:

`x - y = 4`

`=> (x - y)^2 = 16`

`=> x^2 - 2xy + y^2 = 16`

`=> (x^2 + y^2) - 2xy = 16`

`=> 2xy = (x^2 + y^2) - 16`

`=> 2xy = 26 - 16`

`=> 2xy = 10`

`=> xy = 5`

Ta có:

`F = x^3 - y^3`

`= (x - y)(x^2 + xy + y^2)`

`= 4 * [ (x^2 + y^2) + xy]`

`= 4 * (26 + 5)`

`= 4*31`

`= 124`

Vậy, với `x - y = 4; x^2 + y^2 = 26` thì `F = 124.`

Biến đổi tương đương nhé bạn.

a: Ta có: \(\left(x+y\right)^2\)

\(=x^2+2xy+y^2\)

\(\Leftrightarrow x^2+y^2=\dfrac{\left(x+y\right)^2}{2xy}\ge\dfrac{\left(x+y\right)^2}{2}\forall x,y>0\)

\(a,x+y=1\Leftrightarrow\left(x+y\right)^3=1\Leftrightarrow x^3+y^3+3xy\left(x+y\right)=1\\ \Leftrightarrow x^3+y^3+3xy\cdot1=1\Leftrightarrow x^3+y^3+3xy=1\)

\(b,x^3-y^3-3xy\\ =x^3-3x^2y+3xy^2-y^3-3xy+3x^2y-3xy^2\\ =\left(x-y\right)^3-3xy\left(x-y-1\right)\\ =1^3-3xy\left(1-1\right)=1-0=1\)

\(c,x^3+y^3+3xy\left(x^2+y^2\right)+6x^2y^2\left(x+y\right)\\ =\left(x+y\right)\left(x^2-xy+y^2\right)+3xy\left[\left(x+y\right)^2-2xy\right]+6x^2y^2\\ =x^2-xy+y^2+3xy-6x^2y^2+6x^2y^2\\ =x^2+2xy+y^2=\left(x+y\right)^2=1\)

có `x-y=4`

`<=>x^2 -2xy+y^2 =16`

`<=>12-2xy+16`

`<=>-2xy=4`

`<=>xy=-2`

`x^3 -y^3`

`=(x-y)(x^2 +xy+y^2)`

`=4(12-2)`

`=4*10`

`=40`

Có: \(x-y=4\)
\(\Rightarrow\left(x-y\right)^2=16\)

\(\Rightarrow x^2-2xy+y^2=16\)

\(\Rightarrow x^2+y^2-2xy=16\)

\(\Rightarrow12-2xy=16\) \(\Leftrightarrow2xy=-4\Leftrightarrow xy=-2\)

Lại có: \(A=x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)\)

\(=4.\left(x^2+y^2+xy\right)\) (do \(x-y=4\))

\(=4.\left(12-2\right)\) (do \(x^2+y^2=12;xy=-2\))

\(=4.10=40\)

                                                                            Vậy \(A=40\).

10: \(x\left(x-y\right)+x^2-y^2\)

\(=x\left(x-y\right)+\left(x-y\right)\left(x+y\right)\)

\(=\left(x-y\right)\left(x+x+y\right)\)

\(=\left(x-y\right)\left(2x+y\right)\)

11: \(x^2-y^2+10x-10y\)

\(=\left(x^2-y^2\right)+\left(10x-10y\right)\)
\(=\left(x-y\right)\left(x+y\right)+10\left(x-y\right)\)

\(=\left(x-y\right)\left(x+y+10\right)\)

12: \(x^2-y^2+20x+20y\)

\(=\left(x^2-y^2\right)+\left(20x+20y\right)\)

\(=\left(x-y\right)\left(x+y\right)+20\left(x+y\right)\)

\(=\left(x+y\right)\left(x-y+20\right)\)

13: \(4x^2-9y^2-4x-6y\)

\(=\left(4x^2-9y^2\right)-\left(4x+6y\right)\)

\(=\left(2x-3y\right)\left(2x+3y\right)-2\left(2x+3y\right)\)

\(=\left(2x+3y\right)\left(2x-3y-2\right)\)

14: \(x^3-y^3+7x^2-7y^2\)

\(=\left(x^3-y^3\right)+\left(7x^2-7y^2\right)\)

\(=\left(x-y\right)\left(x^2+xy+y^2\right)+7\cdot\left(x^2-y^2\right)\)

\(=\left(x-y\right)\left(x^2+xy+y^2\right)+7\left(x-y\right)\left(x+y\right)\)

\(=\left(x-y\right)\left(x^2+xy+y^2+7x+7y\right)\)

15: \(x^3+4x-\left(y^3+4y\right)\)

\(=x^3-y^3+4x-4y\)

\(=\left(x^3-y^3\right)+\left(4x-4y\right)\)

\(=\left(x-y\right)\left(x^2+xy+y^2\right)+4\left(x-y\right)\)

\(=\left(x-y\right)\left(x^2+xy+y^2+4\right)\)

16: \(x^3+y^3+2x+2y\)

\(=\left(x^3+y^3\right)+\left(2x+2y\right)\)

\(=\left(x+y\right)\left(x^2-xy+y^2\right)+2\left(x+y\right)\)

\(=\left(x+y\right)\left(x^2-xy+y^2+2\right)\)

17: \(x^3-y^3-2x^2y+2xy^2\)

\(=\left(x^3-y^3\right)-\left(2x^2y-2xy^2\right)\)

\(=\left(x-y\right)\left(x^2+xy+y^2\right)-2xy\left(x-y\right)\)

\(=\left(x-y\right)\left(x^2+xy+y^2-2xy\right)\)

\(=\left(x-y\right)\left(x^2-xy+y^2\right)\)

18: \(x^3-4x^2+4x-xy^2\)

\(=x\left(x^2-4x+4-y^2\right)\)

\(=x\left[\left(x^2-4x+4\right)-y^2\right]\)

\(=x\left[\left(x-2\right)^2-y^2\right]\)

\(=x\left(x-2-y\right)\left(x-2+y\right)\)

Phân tích đa thức thành nhân tử nha

 a) Ta thấy \(xy=\dfrac{\left(x+y\right)^2-\left(x^2+y^2\right)}{2}=\dfrac{3^2-5}{2}=2\)

\(\Rightarrow x^3+y^3=\left(x+y\right)\left(x^2+y^2-xy\right)\) \(=3\left(5-2\right)=9\)

 b) Ta thấy \(xy=\dfrac{-\left(x-y\right)^2+\left(x^2+y^2\right)}{2}=\dfrac{15-5^2}{2}=-5\)

\(\Rightarrow x^3-y^3=\left(x-y\right)\left(x^2+y^2+xy\right)\) \(=5\left(15-5\right)=50\)

\(x^3+y^3+3\left(x^2+y^2\right)+4\left(x+y\right)+4=0\)

\(\Leftrightarrow\left(x+y\right)^3-3xy\left(x+y\right)+3\left(x+y\right)^2-6xy+4\left(x+y\right)+4=0\)

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

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

\(\Leftrightarrow\left(x+y+2\right)\left[\left(x-y\right)^2+\left(x+1\right)^2+\left(y+1\right)^2+2\right]=0\)

\(\Leftrightarrow x+y+2=0\)

\(\Leftrightarrow x+y=-2\)

\(M=\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}=\frac{4}{-2}=-2\)

Dấu \(=\)khi \(x=y=-1\).

Đề bài yêu cầu gì vậy em.

\(x+y+4=0\Rightarrow\left\{{}\begin{matrix}y=-4-x\\x+y=-4\end{matrix}\right.\)

\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=\left(-4\right)^3-3xy.\left(-4\right)=12xy-64\)

\(\Rightarrow P=2\left(12xy-64\right)+3\left(x^2+y^2\right)+10x\)

\(=24xy+3x^2+3y^2+10x-128\)

\(=24x\left(-4-x\right)+3x^2+3\left(-4-x\right)^2+10x-128\)

\(=-18x^2-62x-80=-18\left(x+\dfrac{31}{18}\right)^2-\dfrac{479}{18}\le-\dfrac{479}{18}\)

\(P_{max}=-\dfrac{479}{18}\) khi \(\left(x;y\right)=\left(-\dfrac{31}{18};-\dfrac{41}{18}\right)\)