Cho \(x+y=27\) và \(xy=24\)
Tính: \(x^3+y^3\)
K
Khách
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22 tháng 7 2023
`a, (x-y)^2 = (x+y)^2 - 4xy = 12^2 - 35 . 4 = 144 - 140 = 4`.
`b, (x+y)^2 = (x-y)^2 + 4xy = 8^2 + 20.4 = 64 + 80 = 144`
`c, x^3 + y^3 = (x+y)^3 - 3xy(x+y) = 5^3 - 3 . 6 . 5 = 125 - 90 = 35`
`d, x^3 - y^3 = (x-y)^3 - 3xy(x-y) = 3^3 - 3 .40 . 3 = 27 - 360 = -333`.
1 tháng 10 2021
\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)\)
\(=24^3-3\cdot24\cdot18\)
\(=13824-1296\)
=12528
N
3
3 tháng 6 2021
Ta có :
\(x^3\) + \(y^3\) - xy = \(-\dfrac{1}{27}\)
⇔ \(x^3\) + \(y^3\) - xy + \(\dfrac{1}{27}\) = 0
⇔ \(x^3\) + \(y^3\) + \(\dfrac{1^3}{3^3}\) - 3xy.\(\dfrac{1}{3}\) = 0
⇔ (x + y + \(\dfrac{1}{3}\))(\(x^2\) + \(y^2\) + \(\dfrac{1}{9}\) - xy - \(\dfrac{1}{3}x-\dfrac{1}{3}y\)) = 0
TH1 :
x + y + \(\dfrac{1}{3}\) = 0
⇔ x + y = - \(\dfrac{1}{3}\) (loại vì x>0 ; y>0)
TH2 :
\(x^2+y^2+\dfrac{1}{9}-xy-\dfrac{1}{3}x-\dfrac{1}{3}y=0\)\(\dfrac{1}{3}x-\dfrac{1}{3}y\)
⇔ (\(x-\dfrac{1}{3}\))\(^2\) + (\(y-\dfrac{1}{3}\))\(^2\) + (x - y)\(^2\) = 0
⇒ \(x-\dfrac{1}{3}\) = 0
\(y-\dfrac{1}{3}\) = 0
\(x-y\) = 0
⇔ x = y = \(\dfrac{1}{3}\)
Thay x = y = \(\dfrac{1}{3}\) vào \(\dfrac{x}{y^2}\) ta được :
\(\dfrac{1}{3}\) : \(\dfrac{1}{9}\)
= \(\dfrac{1}{3}\) . 9
= 3
\(\dfrac{1}{3}\)\(x^2+y^2+\dfrac{1}{9}-xy-\dfrac{1}{3}x-\dfrac{1}{3}y=0\)
G
24 tháng 12 2019
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CG
1
NM
Nguyễn Minh Quang
Giáo viên
12 tháng 8 2021
a. ta có : \(x^2+y^2=\left(x+y\right)^2-2xy=1^2-2\times\left(-6\right)=13\)
\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=1^3-3\times\left(-6\right)\times1=19\)
\(x^5+y^5=\left(x+y\right)\left[x^4-x^3y+x^2y^2-xy^3+y^4\right]\)
\(=\left(x+y\right)\left[\left(x^2+y^2\right)^2-x^2y^2-xy\left(x^2+y^2\right)\right]=1.\left(13^2-\left(-6\right)^2-\left(-6\right).13\right)=211\)
b.\(x^2+y^2=\left(x-y\right)^2+2xy=1+2\times6=13\)
\(x^3-y^3=\left(x-y\right)^3+3xy\left(x-y\right)=1^3+6.3.1=19\)
\(x^5-y^5=\left(x-y\right)\left[\left(x^4+x^3y+x^2y^2+xy^3+y^4\right)\right]\)
\(=\left(x-y\right)\left[\left(x^2+y^2\right)^2-x^2y^2+xy\left(x^2+y^2\right)\right]=1.\left(13^2-6^2+6.13\right)=211\)
TB
0
8 tháng 8 2017
(x+y)^2 =a^2
x^2 +2xy +y^2 =a^2
x^2+y^2 =a^2-2xy =a^2 -2b
x^3 +y^3 = (x+y)(x^2 -xy +y^2)
=a(a^2-2b-b)
=a(a^2-3b)
=a^3- 3ab
(x^2 +y^2)^2=(a^2-2b)^2 ( cái này tính cho x^4 + y^4)
tương tự như câu đầu tiên
x^5+ y^5 (cái đó mình không biết)
TB
0
30 tháng 5 2023
a: F=9/25x^2y^4*20/27x^3y=4/15x^5y^5
Bậc: 10
b: y=-x/3 và x+y=2
=>x+y=2 và -1/3x-y=0
=>x=3 và y=-1
Khi x=3 và y=-1 thì F=4/15*(-3)^5=-324/5
Có x^3+y^3=(x+y)^3-3xy(x+y)
=> x^3+y^3=27^3-3.24.27
=> x^3+y^3=17739.
Vậy x^3+y^3==17739
Ta có: x3 + y3 = (x + y)(x2 - xy + y2) = (x + y)(x2 + 2xy + y2) - 3xy(x + y)
= (x + y)(x + y)2 - 3xy(x + y)
= (x + y)3 - 3xy(x + y)
= 273 - 3.24.27
= 17739