Đáp án: $N=4$
Giải thích các bước giải:
Ta có:
$xy=2-x-y\to xy+x+y=2\to xy+x+y+1=3\to (x+1)(y+1)=3$
$\to x^2+2=x^2+xy+x+y=x(x+y)+(x+y)=(x+1)(x+y)$
Tương tự $y^2+2=(y+1)(x+y)$
Ta có:
$N=x\sqrt{\dfrac{6+3y^2}{2+x^2}}+y\sqrt{\dfrac{6+3x^2}{2+y^2}}+\sqrt{\dfrac{(2+x^2)(2+y^2)}{3}}$
$\to N=x\sqrt{\dfrac{3(2+y^2)}{2+x^2}}+y\sqrt{\dfrac{3(2+x^2)}{2+y^2}}+\sqrt{\dfrac{(2+x^2)(2+y^2)}{3}}$
$\to N=x\sqrt{\dfrac{3(y+1)(x+y)}{(x+1)(x+y)}}+y\sqrt{\dfrac{3(x+1)(x+y)}{(y+1)(x+y)}}+\sqrt{\dfrac{(x+1)(x+y)(y+1)(x+y)}{3}}$
$\to N=x\sqrt{\dfrac{3(y+1)}{(x+1)}}+y\sqrt{\dfrac{3(x+1)}{(y+1)}}+(x+y)\sqrt{\dfrac{(x+1)(y+1)}{3}}$
$\to N=\dfrac{x(y+1)\sqrt{3}}{\sqrt{(x+1)(y+1)}}+\dfrac{y(x+1)\sqrt{3}}{\sqrt{(x+1)(y+1)}}+(x+y)\sqrt{\dfrac{(x+1)(y+1)}{3}}$
$\to N=\dfrac{x(y+1)\sqrt{3}+y(x+1)\sqrt{3}}{\sqrt{(x+1)(y+1)}}+(x+y)\sqrt{\dfrac{(x+1)(y+1)}{3}}$
$\to N=\dfrac{\sqrt{3}(x(y+1)+y(x+1))}{\sqrt{(x+1)(y+1)}}+(x+y)\sqrt{\dfrac{(x+1)(y+1)}{3}}$
$\to N=\dfrac{\sqrt{3}(x+y+2xy)}{\sqrt{(x+1)(y+1)}}+(x+y)\sqrt{\dfrac{(x+1)(y+1)}{3}}$
$\to N=\dfrac{3(x+y+2xy)}{\sqrt{3(x+1)(y+1)}}+\dfrac{(x+y)(x+1)(y+1)}{\sqrt{3(x+1)(y+1)}}$
$\to N=\dfrac{3(x+y+2xy)+(x+y)(x+1)(y+1)}{\sqrt{3(x+1)(y+1)}}$
$\to N=\dfrac{3(x+y+2(2-x-y))+(x+y)\cdot 3}{\sqrt{3(x+1)(y+1)}}$ vì $(x+1)(y+1)=3$
$\to N=\dfrac{3(4-(x+y))+3(x+y)}{\sqrt{3(x+1)(y+1)}}$
$\to N=\dfrac{12-3(x+y)+3(x+y)}{\sqrt{3(x+1)(y+1)}}$
$\to N=\dfrac{12}{\sqrt{3\cdot 3}}$
$\to N=4$
