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\(xy+\sqrt{\left(1+x^2\right)\left(1+y^2\right)}=a\)
\(\Rightarrow x^2y^2+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}+\left(1+x^2\right)\left(1+y^2\right)=a^2\)
\(\Rightarrow x^2\left(1+y^2\right)+y^2\left(1+x^2\right)+2.x\sqrt{1+y^2}.y\sqrt{1+x^2}+1=a^2\)
\(\Rightarrow\left(x\sqrt{1+y^2}+y\sqrt{1+x^2}\right)^2+1=a^2\)
\(\Rightarrow E^2+1=a^2\)
\(\Rightarrow E=\pm\sqrt{a^2-1}\)
\(E^2=x^2\left(y^2+1\right)+y^2\left(x^2+1\right)+2xy\sqrt{\left(y^2+1\right)\left(x^2+1\right)}\)
\(=2\left(xy\right)^2+x^2+y^2+2xy\sqrt{\left(x^2+1\right)\left(y^2+1\right)}\)
\(a^2=\left(xy\right)^2+2xy\sqrt{\left(x^2+1\right)\left(y^2+1\right)}+\left(x^2+1\right)\left(y^2+1\right)\)
\(=2\left(xy\right)^2+2xy\sqrt{\left(x^2+1\right)\left(y^2+1\right)}+x^2+y^2+1\)
\(\Rightarrow E^2=a^2-1\Rightarrow E=\sqrt{a^2-1}\)
\(E=x\sqrt{1+y^2}+y\sqrt{1+x^2}\)
\(\Leftrightarrow E^2=x^2\left(1+y^2\right)+y^2\left(1+x^2\right)+2xy\sqrt{\left(1+y^2\right)\left(1+x^2\right)}\)
\(=2x^2y^2+x^2+y^2+2xy\left(a-xy\right)\)
\(=2x^2y^2+x^2+y^2+2xya-2x^2y^2\)
\(=x^2+y^2+2xya\)
\(=\left(2xy\right)2+a=a^2+a=E^2\)
\(E=\sqrt{a^2+a}\)
Ta có:
\(x^2+1=x^2+xy+yz+zx\)
\(=x\left(x+y\right)+z\left(x+y\right)=\left(x+y\right)\left(x+z\right)\)
Tương tự:
\(\left\{{}\begin{matrix}y^2+1=\left(y+z\right)\left(y+x\right)\\z^2+1=\left(z+y\right)\left(z+x\right)\end{matrix}\right.\)
\(A=x\sqrt{\dfrac{\left(x+y\right)\left(y+z\right)\left(z+x\right)\left(y+z\right)}{\left(x+y\right)\left(z+x\right)}}+y\sqrt{\dfrac{\left(z+x\right)\left(y+z\right)\left(x+y\right)\left(z+x\right)}{\left(x+y\right)\left(y+z\right)}}+z\sqrt{\dfrac{\left(x+y\right)\left(z+x\right)\left(y+z\right)\left(x+y\right)}{\left(z+x\right)\left(y+z\right)}}\)
\(=x\left|y+z\right|+y\left|z+x\right|+z\left|x+y\right|\)
TH1: x,y,z <0
\(A=-x\left(y+z\right)-y\left(z+x\right)-z\left(x+y\right)=-2\)
TH2: x,y,z>0
\(A=x\left(y+z\right)+y\left(z+x\right)+z\left(x+y\right)=2\)
Ta có \(1+z^2=xy+yz+zx+z^2\)
\(=y\left(x+z\right)+z\left(x+z\right)\)
\(=\left(x+z\right)\left(y+z\right)\)
CMTT, \(1+x^2=\left(x+y\right)\left(x+z\right)\) và \(1+y^2=\left(x+y\right)\left(y+z\right)\)
Do đó \(\sqrt{\dfrac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}\) \(=\sqrt{\dfrac{\left(x+y\right)\left(y+z\right)\left(x+z\right)\left(y+z\right)}{\left(x+y\right)\left(x+z\right)}}\)
\(=\sqrt{\left(y+z\right)^2}\) \(=\left|y+z\right|\)
Tương tự như thế, ta được
\(A=x\left|y+z\right|+y\left|z+x\right|+z\left|x+y\right|\)
Cái này không tính ra số cụ thể được nhé bạn. Nó còn phải tùy vào dấu của \(x+y,y+z,z+x\) nữa.
\(S^2=x^2\left(1+y^2\right)+y^2\left(1+x^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)
\(\Rightarrow S^2=2x^2y^2+x^2+y^2+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)
\(a^2=x^2y^2+\left(1+x^2\right)\left(1+y^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)
\(\Rightarrow a^2=2x^2y^2+x^2+y^2+1+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)
\(\Rightarrow S^2=a^2-1\)
\(\Rightarrow S=\pm\sqrt{a^2-1}\)
Từ giả thiết ta có \(2016=x^2y^2+1+x^2+y^2+x^2y^2+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)
\(\Leftrightarrow x^2+y^2+2x^2y^2+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}=2015\)
Ta có \(S^2=x^2\left(1+y^2\right)+y^2\left(1+x^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)
\(=x^2+y^2+2x^2y^2+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}=2015\)
\(\Rightarrow S=\sqrt{2015}\) (Vì S > 0)
Bài này hình như x,y,z>0
Ta có: \(x\sqrt{\frac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}=x\sqrt{\frac{\left(y^2+xy+yz+zx\right)\left(z^2+xy+yz+zx\right)}{\left(x^2+xy+yz+zx\right)}}=x\sqrt{\frac{\left(y+x\right)\left(y+z\right)\left(z+x\right)\left(z+y\right)}{\left(x+y\right)\left(x+z\right)}}=x\sqrt{\left(y+z\right)^2}\)
Tương tự: \(y\sqrt{\frac{\left(1+z^2\right)\left(1+x^2\right)}{1+y^2}}=y\sqrt{\left(x+z\right)^2}\)
\(z\sqrt{\frac{\left(1+x^2\right)\left(1+y^2\right)}{1+z^2}}=z\sqrt{\left(x+y\right)^2}\)
Cộng từng vế, ta có:
\(A=x\left(y+z\right)+y\left(z+x\right)+z\left(x+y\right)\)
\(\Leftrightarrow A=2\left(xy+yz+zx\right)=2\)
\(\hept{\begin{cases}1+y^2=y^2+xy+yz+zx=\left(x+y\right)\left(y+z\right)\\1+z^2=\left(z+x\right).\left(z+y\right)\\1+x^2=\left(x+y\right)\left(x+z\right)\end{cases}}\)
Thế vào \(A=x\sqrt{\left(y+z\right)^2}+y\sqrt{\left(x+z\right)^2}+z\sqrt{\left(x+y\right)^2}\)
\(=x\left|y+z\right|+y\left|x+z\right|+z\left|x+y\right|\)
\(=2\left(\left|xy\right|+\left|yz\right|+\left|zx\right|\right)\)
Nếu x,y,z\(\ge0\Rightarrow A=2\)
Nếu x,y,z\(< 0\)\(\Rightarrow A=-2\)