Question

[poppy Trang]

Cho 0<x,y<1 thỏa mãn : \(\left(\dfrac{x}{1-x}+\dfrac{y}{1-y}\right)^{2018}=1\)

Tính giá trị biểu thức A=(x+y+\(\sqrt{x^2-xy+y^2}\))²⁰¹⁹

Answer 1

[Nguyễn Việt Lâm]

\(0< x,y< 1\Rightarrow\dfrac{x}{1-x}+\dfrac{y}{1-y}>0\)

\(\left(\dfrac{x}{1-x}+\dfrac{y}{1-y}\right)^{2018}=1\Rightarrow\dfrac{x}{1-x}+\dfrac{y}{1-y}=1\)

\(\Rightarrow x-xy+y-xy=1-x-y+xy\Rightarrow2\left(x+y\right)-1=3xy\) (1)

\(A=\left(x+y+\sqrt{\left(x+y\right)^2-3xy}\right)^{2019}=\left(x+y+\sqrt{\left(x+y\right)^2-2\left(x+y\right)+1}\right)^{2019}\)

\(A=\left(x+y+\sqrt{\left(x+y-1\right)^2}\right)^{2019}=\left(x+y+\left|x+y-1\right|\right)^{2019}\)

Ta xét dấu \(x+y-1\) để phá trị tuyệt đối:

Từ (1) ta cũng có \(2x-1=3xy-2y=y\left(3x-2\right)\Rightarrow y=\dfrac{2x-1}{3x-2}\)

Mà \(0< y< 1\Rightarrow0< \dfrac{2x-1}{3x-2}< 1\Rightarrow0< x< \dfrac{1}{2}\)

\(x+y-1=x+\dfrac{2x-1}{3x-2}-1=\dfrac{3x^2-3x+1}{3x-2}< 0\) \(\forall x:0< x< \dfrac{1}{2}\)

\(\Rightarrow\left|x+y-1\right|=1-x-y\)

\(\Rightarrow A=\left(x+y+1-x-y\right)^{2019}=1^{2019}=1\)