TC
2
K
Khách
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19 tháng 7 2019
Đề a,b bạn ghi mik ko hiểu
c)Ta có : \(x+y=a=>x^2+y^2+2xy=a^2\)
Mà \(x^2+y^2=b\)nên\(b+2xy=a^2=>xy=\frac{a^2-b}{2}\)
\(x^3+y^3=\left(x+y\right)\left(x^2+y^2-xy\right)\)
Thay \(x+y=a\) ; \(x^2+y^2=b\)và \(xy=\frac{a^2-b}{2}\)ta có : \(x^3+y^3=a\left(b-\frac{a^2-b}{2}\right)=ab-\frac{a^3-ab}{2}\)
30 tháng 12 2020
bbgfhfygfdsdty64562gdfhgvfhgfhhhhh
\hvhhhggybhbghhguyg
9 tháng 7 2019
mình hỏi vs 3y^2 là 3xy^2 phải không hay chỉ là 3y^2
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)
M
0
G
24 tháng 12 2019
chịu but Merry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry ChristmasMerry Christmas
Ta có: \(P=\dfrac{1}{x^3}-\dfrac{1}{y^3}\)
\(=\left(\dfrac{1}{x}-\dfrac{1}{y}\right)^3-3\cdot\dfrac{1}{x}\cdot\dfrac{1}{y}\cdot\left(\dfrac{1}{x}-\dfrac{1}{y}\right)\)
\(=2^3-3\cdot3\cdot2\)
\(=-10\)
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