Question

cho x,ythoa man 0<x<1 ,0<y<1 va \(\frac{x}{1-x}+\frac{y}{1-y}=1\)

tinh gia tri cua bieu thuc P=\(x+y+\sqrt{x^2-xy+y^2}\)

Answer 1

từ cái đầu=>x-xy+y-xy=(1-x)(1-y)

<=>x+y-2xy=xy-x-y+1

<=>2(x+y)=3xy+1

\(\Leftrightarrow x+y=\frac{3xy+1}{2}\)

\(\sqrt{x^2-xy+y^2}=\sqrt{\left(x+y\right)^2-3xy}=\sqrt{\frac{9x^2y^2+6xy+1}{4}-3xy}=\sqrt{\frac{9x^2y^2-6xy+1}{4}}=\sqrt{\left(\frac{3xy-1}{2}\right)^2}\)với 3xy-1>0

\(\Rightarrow P=\frac{3xy+1}{2}+\frac{3xy-1}{2}=3xy\)

với 3xy-1<(=)0

\(\Rightarrow P=\frac{3xy+1}{2}+\frac{1-3xy}{2}=1\)