\(P=\dfrac{x^3}{y^2}+\dfrac{y^3}{x^2}+2020=\dfrac{x^5+y^5}{\left(xy\right)^2}+2020=\dfrac{\left(x^3+y^3\right)\left(x^2+y^2\right)-\left(xy\right)^2\left(x+y\right)}{\left(-2\right)^2}\)

\(=\dfrac{\left[\left(x+y\right)^3-3xy\left(x+y\right)\right]\left[\left(x+y\right)^2-2xy\right]-\left(-2\right)^2.5}{4}\)

\(=\dfrac{\left(-8+6.5\right)\left(25+4\right)-20}{4}=...\)

Lời giải:
Đặt $xy=a; x+y=b$ thì theo đề ta có:

$a+b=-1$ và $ab=-12$

Ta cần tính: $A=(x+y)^3-3xy(x+y)=b^3-3ab=b^3-3(-12)=b^3+36$
 

Từ $a+b=-1\Rightarrow a=-b-1$. Thay vào $ab=-12$
$\Rightarrow (-b-1)b=-12$
$\Leftrightarrow (b+1)b=12$

$\Leftrightarrow b^2+b-12=0$

$\Leftrightarrow (b-3)(b+4)=0$
$\Leftrightarrow b=3$ hoặc $b=-4$
Nếu $b=3$ thì $A=3^3+36=63$

Nếu $b=-4$ thì $A=(-4)^3+36=-28$

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

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 \(x^3+y^3=26\)

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:

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

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

              \(=26\)

Ta có \(\left(x+y\right)^2=4\Rightarrow x^2+y^2+2xy=4\Rightarrow xy=\frac{4-10}{2}=-3\)

\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=8-6xy=8-6.\left(-3\right)=26\)

Học tốt!!!!!!

Ta có: x + y = 2

<=> (x + y)² = 2²

<=> x² + y² + 2xy = 4

<=> 10 + 2xy = 4

<=> 2xy = -6

<=> xy = -3

Khi đó: M = x³ + y³ = (x  + y)(x² - xy + y²) = 2(10 + 3) = 2.13 = 26