Question

Cho hai số x,y thỏa mãn \(x+y=2, x^2+y^2=10\)
Tính giá trị biểu thức: \(M=x^3+y^3.\)

Answer 1

x + y = 2

=> ( x + y )² = 4

<=> x² + 2xy + y² = 4

<=> 2xy + 10 = 4

<=> 2xy = -6

<=> xy = -3

Ta có : M = x³ + y³ = ( x + y )( x² - xy + y² ) = 2( 10 + 3 ) = 26

Answer 2

Ta có : \(x+y=2\)

\(\Rightarrow\left(x+y\right)^2=4\)

\(\Rightarrow x^2+y^2+2xy=4\)

Mà \(x^2+y^2=10\)

\(\Rightarrow10+2xy=4\)

\(\Rightarrow2xy=-6\)

\(\Rightarrow xy=-3\)

\(\Rightarrow x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=2\left(10+3\right)=2.13=26\)

Vậy \(x^3+y^3=26\)

Answer 3

Ta có:\(x+y=2\)

\(\Rightarrow\left(x+y\right)^2=4\)

\(\Rightarrow x^2+2xy+y^2=4\)

\(\Rightarrow10+2xy=4\)

\(\Rightarrow2xy=-6\)

\(\Rightarrow xy=-3\)

Ta có:

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

               \(=2\left(10+3\right)\)

              \(=26\)