1) Sửa đề: Cho \(\left(x+\sqrt{x^2+2}\right)\left(y+\sqrt{y^2+2}\right)=2\)
Tính \(S=x\sqrt{y^2+2}+y\sqrt{x^2+2}\)
Nhận xét:
\(S^2=x^2\left(y^2+2\right)+y^2\left(x^2+2\right)+2xy\sqrt{\left(x^2+2\right)\left(y^2+2\right)}\)
\(=x^2y^2+\left(x^2y^2+2x^2+2y^2+4\right)+2xy\sqrt{\left(x^2+2\right)\left(y^2+2\right)}-4\)
\(=x^2y^2+\left(x^2+2\right)\left(y^2+2\right)+2xy\sqrt{\left(x^2+2\right)\left(y^2+2\right)}-4\)
\(=\left(xy+\sqrt{\left(x^2+2\right)\left(y^2+2\right)}\right)^2-4\)
\(\Rightarrow\)\(xy+\sqrt{\left(x^2+2\right)\left(y^2+2\right)}=\pm\sqrt{S^2+4}\)
\(\left(x+\sqrt{x^2+2}\right)\left(y+\sqrt{y^2+2}\right)=2\)
\(\Leftrightarrow xy+\sqrt{\left(x^2+2\right)\left(y^2+2\right)}+\sqrt{y^2+2}+y\sqrt{x^2+2}=2\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{S^2+4}+S=2\\-\sqrt{S^2+4}+S=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{S^2+4}=2-S\left(S\le2\right)\\\sqrt{S^2+4}=S-2\left(S\ge2\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}S^2+4=S^2+4S+4\\S^2+4=S^2+4S+4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}S=0\left(nhận\right)\\S=0\left(loại\right)\end{matrix}\right.\)
2)\(\left(a+b+c\right)^2=0\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)
\(\Leftrightarrow\left(ab+bc+ca\right)=-0,5\Rightarrow\left(ab+bc+ca\right)^2=0,25\)
\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=0,25\)
\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2=0,25\)
\(\left(a^2+b^2+c^2\right)^2=1\)
\(\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=1\)
\(\Leftrightarrow a^4+b^4+c^4=0,5\)
