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

Xin lỗi mk nhầm, cáo thứ 3 là -3xy ko phải -xy đâu. kết quả là a^3- 3XY. I am sorry

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: \(x^3+y^3=2\left(10+3\right)=2.13=26\)

\(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=a\left(b-xy\right)\)

x+y=a=>(x+y)²=a²=>x²+2xy+y²=a²=>b+2xy=a²=>xy=(a²-b)/2

x³ + y³ = ( x + y ) ( x² - xy + y² ) = a[ b - (a²-b)/2 ] = ( 3ab - a² )/2.

Dau bang cuoi cung la: (3ab-a³)/2.

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) \(\left(x-y\right)^2=x^2+y^2-2xy\Rightarrow m^2=n-2xy\Leftrightarrow xy=\frac{n-m^2}{2}\)

\(x^3-y^3=\left(x-y\right)^3+3xy\left(x-y\right)=m^3+3.m.\frac{n-m^2}{2}=\frac{3mn}{2}-\frac{m^3}{2}\)

\(x+y=a\left(1\right)\)

\(x-y=b\left(2\right)\)

\(\left(1\right)+\left(2\right)\Rightarrow2x=a+b\Rightarrow x=\dfrac{a+b}{2}\)

\(\left(1\right)\Rightarrow y=a-x\Rightarrow y=a-\dfrac{a+b}{2}\Rightarrow y=\dfrac{a-b}{2}\)

\(xy=\dfrac{\left(a+b\right)}{2}.\dfrac{\left(a-b\right)}{2}=\dfrac{a^2-b^2}{4}\)

\(x^3-y^3=\left(\dfrac{a+b}{2}\right)^3-\left(\dfrac{a-b}{2}\right)^3=\dfrac{\left(a+b\right)^3}{8}-\dfrac{\left(a-b\right)^3}{8}\)

\(=\dfrac{\left(a+b\right)^3-\left(a-b\right)^3}{8}\)

\(=\dfrac{\left(a+b-a+b\right)\left[\left(a+b\right)^2+\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]}{8}\)

\(=\dfrac{2b\left[a^2+b^2+2ab+a^2-b^2+a^2+b^2-2ab\right]}{8}\)

\(=\dfrac{b\left[3a^2+b^2+2ab\right]}{4}\)

\(\left\{{}\begin{matrix}x+y=a\\x-y=b\end{matrix}\right.\) 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.\(\dfrac{3a^2+b^2}{4}\) = \(\dfrac{3a^2b+b^3}{4}\)

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

a) Ta có (x+y)²=x²+2xy+y²=10+2xy=2²=4

có 10+2xy=4 suy ra 2xy=-8 và xy=-4

Có A=x³+y³=(x+y)(x²-xy+y²)=2.(10-xy)=2.(10+4)=2.14=28

b)Ta có (x+y)²=x²+2xy+y²=2xy+b=a²

có 2xy+b=a² suy ra 2xy=a²-b và xy=(a²-b):2

Có B=x³+y³=(x+y)(x²-xy+y²)=a.(b-xy)=a.[b-(a²-b):2]