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mình hỏi vs 3y^2 là 3xy^2 phải không hay chỉ là 3y^2
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 : \(x^3+y^3=a\left(b-\frac{a^2-b}{2}\right)=ab-\frac{a^3-ab}{2}\)
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(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)
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Ta có: \(P=\dfrac{1}{x^3}-\dfrac{1}{y^3}\)
\(=\left(\dfrac{1}{x}-\dfrac{1}{y}\right)^3-3\cdot\dfrac{1}{x}\cdot\dfrac{1}{y}\cdot\left(\dfrac{1}{x}-\dfrac{1}{y}\right)\)
\(=2^3-3\cdot3\cdot2\)
\(=-10\)
sai rồi kìa anh
Ta có:\(P=\dfrac{1}{x^3}-\dfrac{1}{y^3}=\left(\dfrac{1}{x}-\dfrac{1}{y}\right)\left(\dfrac{1}{x^2}+\dfrac{1}{xy}+\dfrac{1}{y^2}\right)\)
\(=\left(\dfrac{1}{x}-\dfrac{1}{y}\right)\left[\left(\dfrac{1}{x}-\dfrac{1}{y}\right)^2+3.\dfrac{1}{xy}\right]=2.\left(2^2+3.3\right)=2.13=26\)
Thử thay x=1/3;y=1 vào là thấy đúng,các cặp nghiệm khác cũng tm