Question

Cho x+y=5,xy=6. Tính giá trị biểu thức:

A=\(x^2+y^2\)

B=\(x^3+y^3\)

C=x\(^2\)-y\(^2\)

D=\(\frac{y}{x}+\frac{x}{y}\)

Answer 1

\(A=\left(x+y\right)^2-2xy=25-12=13\)

\(B=\left(x+y\right)\left(x^2+y^2-xy\right)=\left(x+y\right)\left[\left(x+y\right)^2-3xy\right]=5\left(25-18\right)=35\)

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

\(C^2=\left[\left(x+y\right)^2-2xy\right]^2-4\left(xy\right)^2=\left(25-12\right)^2-4.36=25\Rightarrow C=\pm5\)

\(D=\frac{x^2+y^2}{xy}=\frac{\left(x+y\right)^2-2xy}{xy}=\frac{25-12}{6}=\frac{13}{6}\)

Answer 2

x + y = 5 ⇔ x = 5-y

x.y =6⇔ x(5 - x)=6

⇔ -x² + 5x - 6 = 0 ⇒\(\left\{{}\begin{matrix}x=2\Rightarrow y=3\\x=3\Rightarrow y=2\end{matrix}\right.\)

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