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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)
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)
`#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)
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)
\(a,x+y=1\Leftrightarrow\left(x+y\right)^3=1\Leftrightarrow x^3+y^3+3xy\left(x+y\right)=1\\ \Leftrightarrow x^3+y^3+3xy\cdot1=1\Leftrightarrow x^3+y^3+3xy=1\)
\(b,x^3-y^3-3xy\\ =x^3-3x^2y+3xy^2-y^3-3xy+3x^2y-3xy^2\\ =\left(x-y\right)^3-3xy\left(x-y-1\right)\\ =1^3-3xy\left(1-1\right)=1-0=1\)
\(c,x^3+y^3+3xy\left(x^2+y^2\right)+6x^2y^2\left(x+y\right)\\ =\left(x+y\right)\left(x^2-xy+y^2\right)+3xy\left[\left(x+y\right)^2-2xy\right]+6x^2y^2\\ =x^2-xy+y^2+3xy-6x^2y^2+6x^2y^2\\ =x^2+2xy+y^2=\left(x+y\right)^2=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=2\left(x-y\right)\left(x^2+xy+y^2\right)-3\left(x-y\right)^2-6xy\)
\(=4\left[\left(x-y\right)^2+3xy\right]-12-6xy\)
\(=4\left(4+3xy\right)-12-6xy=4+6xy\)
Không thể rút gọn thêm
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
$x^3+y^3=(x+y)^3-3xy(x+y)=2^3-3xy.2=8-6xy$
$=8-3[(x+y)^2-(x^2+y^2)]=8-3(2^2-20)=56$
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(A=40\).
a)
Ta có :
\(x+y=3\)
\(x^2+y^2=5\Leftrightarrow\left(x+y\right)^2-2xy=5\Leftrightarrow9-2xy=5\Leftrightarrow2xy=4\Rightarrow xy=2\)
\(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=3.\left(5-2\right)=9\)
b)
Ta có :
\(x-y=5\)
\(x^2+y^2=15\Leftrightarrow\left(x-y\right)^2+2xy=15\Leftrightarrow25+2xy=15\Rightarrow xy=-5\)
=> \(x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)=\left(5\right)\left(15+-5\right)=50\)