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Ta có: \(x-y=5\Rightarrow\left(x-y\right)^2=25\)
\(\Rightarrow x^2+y^2-2xy=25\)
\(\Rightarrow15-2xy=25\)
\(\Rightarrow2xy=-10\Rightarrow xy=-5\)
\(M=x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)=5\left(15-5\right)=5.10=50\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`#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.`
![](https://rs.olm.vn/images/avt/0.png?1311)
1)
Ta có: x+y=2
nên \(\left(x+y\right)^2=4\)
\(\Leftrightarrow x^2+y^2+2xy=4\)
\(\Leftrightarrow2xy=2\)
hay xy=1
Ta có: \(x^3+y^3\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)\)
\(=2^3-3\cdot1\cdot2\)
=2
2)\(x^2+y^2=\left(x+y\right)^2-2xy=8^2-2\cdot\left(-20\right)=104\)
\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=8^3-3\cdot\left(-20\right)\cdot8=512+480=992\)
\(x^2+y^2+xy=\left(x+y\right)^2-xy=8^2-\left(-20\right)=64+20=84\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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: \(x^3-4x^2+4x-xy^2\)
\(=x\left(x^2-4x+4-y^2\right)\)
\(=x\left[\left(x^2-4x+4\right)-y^2\right]\)
\(=x\left[\left(x-2\right)^2-y^2\right]\)
\(=x\left(x-2-y\right)\left(x-2+y\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có
N = x 3 + 3 x 2 y + 3 x y 2 + y 3 + x 2 + 2 x y + y 2 = ( x 3 + 3 x 2 y + 3 x y 2 + y 3 ) + ( x 2 + 2 x y + y 2 ) = ( x + y ) 3 + ( x + y ) 2 = ( x + y ) 2 ( x + y + 1 )
Từ đề bài x = 10 – y ó x + y = 10. Thay x + y = 10 vào N = ( x + y ) 2 (x + y + 1) ta được
N = 10 2 (10 + 1) = 1100
Suy ra N > 1000 khi x = 10 – y
Đáp án cần chọn là: D
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,x+y=1\Leftrightarrow\left(x+y\right)^3=1\Leftrightarrow x^3+y^3+3xy\left(x+y\right)=1\\ \Leftrightarrow x^3+y^3+3xy\cdot1=1\Leftrightarrow x^3+y^3+3xy=1\)
\(b,x^3-y^3-3xy\\ =x^3-3x^2y+3xy^2-y^3-3xy+3x^2y-3xy^2\\ =\left(x-y\right)^3-3xy\left(x-y-1\right)\\ =1^3-3xy\left(1-1\right)=1-0=1\)
\(c,x^3+y^3+3xy\left(x^2+y^2\right)+6x^2y^2\left(x+y\right)\\ =\left(x+y\right)\left(x^2-xy+y^2\right)+3xy\left[\left(x+y\right)^2-2xy\right]+6x^2y^2\\ =x^2-xy+y^2+3xy-6x^2y^2+6x^2y^2\\ =x^2+2xy+y^2=\left(x+y\right)^2=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
a.
$x^3+y^3=(x+y)^3-3xy(x+y)=9^3-3.9.18=243$
$x^4+y^4=(x^2+y^2)^2-2x^2y^2=[(x+y)^2-2xy]^2-2x^2y^2$
$=[9^2-2.18]^2-2.18^2=1377$
Nếu $x\geq y$ thì:
$x^3-y^3=(x-y)(x^2+xy+y^2)$
$=|x-y|[(x+y)^2-xy]=\sqrt{(x+y)^2-4xy}[(x+y)^2-xy]$
$=\sqrt{9^2-4.18}(9^2-18)=189$
Nếu $x< y$ thì $x^3-y^3=-189$
b.
$A=(x+y)^2-6(x+y)+y-5$
$=(-9)^2-6(-9)+y-5=130+y$
Chưa đủ cơ sở để tính biểu thức.
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Ta có: \(\left(x+y\right)^2\)
\(=x^2+2xy+y^2\)
\(\Leftrightarrow x^2+y^2=\dfrac{\left(x+y\right)^2}{2xy}\ge\dfrac{\left(x+y\right)^2}{2}\forall x,y>0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`a, x^3 + y^3 + x + y`
`= (x+y)(x^2-xy+y^2)+x+y`
`= (x+y)(x^2-xy+y^2+1)`
`b, x^3 - y^3 + x -y`
`= (x-y)(x^2+xy+y^2)+x-y`
`= (x-y)(x^2+xy+y^2+1)`
`c, (x-y)^3 + (x+y)^3`
`= (x-y+x+y)(x^2-2xy+y^2 - x^2 + y^2 + x^2 + 2xy + y^2)`
`= (2x)(x^2 + 3y^2)`
`d, x^3 - 3x^2y + 3xy^2 - y^3 + y^2 - x^2`
`= (x-y)^3 + (y-x)(x+y)`
`=(x-y)(x^2+2xy+y^2-x-y)`
a: =(x+y)(x^2-xy+y^2)+(x+y)
=(x+y)(x^2-xy+y^2+1)
b: =(x-y)(x^2+xy+y^2)+(x-y)
=(x-y)(x^2+xy+y^2+1)
c: =x^3-3x^2y+3xy^2-y^3+x^3+3x^2y+3xy^2-y^3
=2x^3+6xy^2
d: =(x-y)^3+(y-x)(y+x)
=(x-y)[(x-y)^2-(x+y)]
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
$x^3+y^3=(x+y)^3-3xy(x+y)=2^3-3xy.2=8-6xy$
$=8-3[(x+y)^2-(x^2+y^2)]=8-3(2^2-20)=56$
( x- y)^2 = 5^2
=> x^2 - 2xy + y^2 = 25
=> 15 - 2xy = 25
=> 2xy = -10
=> xy = -5
x^3 - y^3 = ( x- y)(x^2+xy+y^2) = 5.(15 - 5 ) = 5.10 = 50