Cho x+y=a,x^2+y^2=b.Tính x^3+y^3 theo a và b
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:\(\left(x+y\right)^3-3x^2y-3xy^2=x^3+y^3\)
Tại x+y=a và x.y=b ta đc:
\(x^3+y^3=a^3-3xy\left(x+y\right)\)
\(x^3+y^3=a^3-3ab\)
Ta có: \(x^3\)+\(y^3\)=(x+y)(x2+y2+xy)=(x+y)[(x+y)2-xy]= a(a2-b)=\(a^3\)-ab. Chúc bạn học tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
1) Cho x+y=2 và x^2+y^2=10. Tính x^3+y^3. Giải
(x+y)^2=x^2+y^2+2xy => xy= -3
x^3+y^3=(x+y)^3-3xy(x+y) = 26
2) Ta có: x^3+y^3 = (x+y)(x^2-xy+y^2) (1)
(x+y)^2=a^2
=> x^2 +2xy +y^2=a^2
=> b+2xy=a^2
=> xy=\(\frac{a^2-b}{2}\)
Thay (1) vào đó ta có:
x^3+y^3= (x+y)(x^2-xy+y^2) = a(b-\(\frac{a^2-b}{2}\)) = \(a\left(\frac{2b-a^2+b}{2}\right)=a.\frac{3b-a^2}{2}\)
\(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=2\left(10-xy\right)\)
Ta có: \(x^2+y^2=\left(x+y\right)^2-2xy=2^2-2xy=4-2xy=10\Rightarrow2xy=-6\Rightarrow xy=-3\)
Vậy: \(x^3+y^3=2\left(10+3\right)=2.13=26\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(x+y\right)^2=x^2+y^2+2xy\Rightarrow4=10+2xy\Leftrightarrow xy=-3\)
\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=2^3+3.3.2=26\)
b) \(\left(x-y\right)^2=x^2+y^2-2xy\Rightarrow m^2=n-2xy\Leftrightarrow xy=\frac{n-m^2}{2}\)
\(x^3-y^3=\left(x-y\right)^3+3xy\left(x-y\right)=m^3+3.m.\frac{n-m^2}{2}=\frac{3mn}{2}-\frac{m^3}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x+y=a\left(1\right)\)
\(x-y=b\left(2\right)\)
\(\left(1\right)+\left(2\right)\Rightarrow2x=a+b\Rightarrow x=\dfrac{a+b}{2}\)
\(\left(1\right)\Rightarrow y=a-x\Rightarrow y=a-\dfrac{a+b}{2}\Rightarrow y=\dfrac{a-b}{2}\)
\(xy=\dfrac{\left(a+b\right)}{2}.\dfrac{\left(a-b\right)}{2}=\dfrac{a^2-b^2}{4}\)
\(x^3-y^3=\left(\dfrac{a+b}{2}\right)^3-\left(\dfrac{a-b}{2}\right)^3=\dfrac{\left(a+b\right)^3}{8}-\dfrac{\left(a-b\right)^3}{8}\)
\(=\dfrac{\left(a+b\right)^3-\left(a-b\right)^3}{8}\)
\(=\dfrac{\left(a+b-a+b\right)\left[\left(a+b\right)^2+\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]}{8}\)
\(=\dfrac{2b\left[a^2+b^2+2ab+a^2-b^2+a^2+b^2-2ab\right]}{8}\)
\(=\dfrac{b\left[3a^2+b^2+2ab\right]}{4}\)
\(\left\{{}\begin{matrix}x+y=a\\x-y=b\end{matrix}\right.\) tính \(x^3\) - y3 theo \(a\) và \(b\)
⇒ \(\left\{{}\begin{matrix}x+y+x-y=a+b\\x-y=b\end{matrix}\right.\)
⇔ \(\left\{{}\begin{matrix}2x=a+b\\y=x-b\end{matrix}\right.\)
⇔ \(\left\{{}\begin{matrix}x=\left(a+b\right):2\\y=\left(a-b\right):2\end{matrix}\right.\) ⇒ \(xy\) = \(\dfrac{a+b}{2}\)\(\times\)\(\dfrac{a-b}{2}\) = \(\dfrac{a^2-b^2}{4}\)
\(x^{3^{ }}\) - y3 = (\(x\) - y)(\(x^2\) + \(x\)y + y2) = \(\left(x-y\right)\)\(\left(\left[x+y\right]^2-xy\right)\) (1)
Thay \(x-y\) = a; \(x\) + y = b và \(xy\) = \(\dfrac{a^2-b^2}{4}\) vào (1) ta có:
\(x^3\) - y3 = b.(a2 - \(\dfrac{a^2-b^2}{4}\)) = b.\(\dfrac{3a^2+b^2}{4}\) = \(\dfrac{3a^2b+b^3}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có:
x + y = 2
=> ( x + y)2 = 4
=> x2 + 2xy + y2 = 4
=> 10 + 2xy = 4
=> 2xy = 4 - 10 = -6
=> xy = -6/2 = -3
Ta có:
A = x3 + y3
A = (x + y)(x2 - xy + y2)
A = 2(10 + 3)
A = 26
b) Ta có:
x + y = a
=> (x + y)2 = a2
=> x2 + 2xy + y2 = a2
=> b + 2xy = a2
=> xy = (a2 - b)/2
Ta có:
B = x3 + y3
B = (x + y)(x2 + xy + y2)
B = a[b + (a2 - b )/2]
B = ab + (a3 - b)/2
cho x+y=2(=)(x+y)^2=4(=)x^2+y^2+2xy=4
(=)10+2xy=4(=)2xy=-6(=)xy=-3
mà x^3+y^3=(x+y)(x^2+y^2-xy)
=2(10+3)=26
vậy x^3+y^3=26
x^3+y^3=(x+y)^3-3xy(x+y)=a^3-3*\(\frac{\left(x+y\right)^2-x^2-y^2}{^{^{ }}2}\)*a=a^3-3*\(\frac{a^2-b}{2}\)*a