Ta có: \(x^4+y^4+\frac{x^4y^4}{\left(x^2+y^2\right)^2}\)

\(=\left(x^4+2x^2y^2+y^4\right)-2x^2y^2+\frac{x^4y^4}{\left(x^2+y^2\right)}\)

\(=\left(x^2+y^2\right)^2-2x^2y^2+\left(\frac{x^2y^2}{x^2+y^2}\right)^2\)

\(=\left(x^2+y^2-\frac{x^2y^2}{x^2+y^2}\right)^2\)

Thay vào ta tính được:

\(P=\sqrt{\frac{x^2y^2}{x^2+y^2}+\frac{x^2y^2}{\left(x+y\right)^2}+\sqrt{\left(x^2+y^2-\frac{x^2y^2}{x^2+y^2}\right)^2}}\)

Mà \(x^2+y^2-\frac{x^2y^2}{x^2+y^2}=\frac{\left(x^2+y^2\right)^2-x^2y^2}{x^2+y^2}=\frac{x^4+x^2y^2+y^4}{x^2+y^2}>0\left(\forall x,y\right)\)

Khi đó:

\(P=\sqrt{\frac{x^2y^2}{x^2+y^2}+\frac{x^2y^2}{\left(x+y\right)^2}+x^2+y^2-\frac{x^2y^2}{x^2+y^2}}\)

\(P=\sqrt{x^2+y^2+\frac{x^2y^2}{\left(x+y\right)^2}}\)

\(P=\sqrt{\left(x^2+2xy+y^2\right)-2xy+\frac{x^2y^2}{\left(x+y\right)^2}}\)

\(P=\sqrt{\left(x+y\right)^2-2xy+\left(\frac{xy}{x+y}\right)^2}\)

\(P=\sqrt{\left(x+y-\frac{xy}{x+y}\right)^2}\)

\(P=\left|x+y-\frac{xy}{x+y}\right|=\left|\frac{x^2+xy+y^2}{x+y}\right|=\frac{x^2+xy+y^2}{x+y}\)

Vậy \(P=\frac{x^2+xy+y^2}{x+y}\)

a) \(ĐKXĐ:x,y\ne0;x\ne\pm y\)

Ta có : \(A=\frac{y-x}{xy}:\left[\frac{y^2}{\left(x-y\right)^2}-\frac{2x^2y}{\left(x^2-y^2\right)^2}+\frac{x^2}{y^2-x^2}\right]\)

\(=\frac{y-x}{xy}:\left[\frac{y^2.\left(x+y\right)^2}{\left(x-y\right)^2.\left(x+y\right)^2}-\frac{2x^2y}{\left(x-y\right)^2.\left(x+y\right)^2}-\frac{x^2.\left(x^2-y^2\right)}{\left(x^2-y^2\right).\left(x^2-y^2\right)}\right]\)

\(=\frac{y-x}{xy}:\left[\frac{y^2.\left(x^2+2xy+y^2\right)-2x^2y-x^2.\left(x^2-y^2\right)}{\left(x-y\right)^2.\left(x+y\right)^2}\right]\)

\(=\frac{y-x}{xy}:\left[\frac{x^2y^2+y^4+2xy^3-2x^2y-x^4+x^2y^2}{\left(x-y\right)^2\left(x+y\right)^2}\right]\)

Đề này lỗi mình nghĩ vậy vì trên tử kia không đẹp lắm.....

\(A=\frac{\left(x+y\right)\left(x-y\right)\left(x^2+xy+y^2\right)\sqrt{4x-1-2\sqrt{4x-1}+1}}{-\left(\sqrt{4x-1}-1\right).y^2\left(x^2+xy+y^2\right)}=\frac{\left(x^2-y^2\right)\sqrt{\left(\sqrt{4x-1}-1\right)^2}}{-\left(\sqrt{4x-1}-1\right).y^2}\)

Do \(x>1\Rightarrow4x-1>1\Rightarrow\sqrt{4x-1}>1\Rightarrow\sqrt{4x-1}-1>0\)

\(\Rightarrow A=\frac{\left(x^2-y^2\right)\left(\sqrt{4x-1}-1\right)}{-\left(\sqrt{4x-1}-1\right).y^2}=\frac{x^2-y^2}{-y^2}=1-\left(\frac{x}{y}\right)^2\)

\(A=-8\Rightarrow1-\left(\frac{x}{y}\right)^2=-8\Rightarrow\left(\frac{x}{y}\right)^2=9\)

Do \(\left\{{}\begin{matrix}x>1\\y< 0\end{matrix}\right.\) \(\Rightarrow\frac{x}{y}< 0\Rightarrow\frac{x}{y}=-3\)

1, với x > 0 ; x khác 1 ; 4 

a, \(P=\left(\dfrac{x+\sqrt{x}-x-2}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)+\sqrt{x}-4}{x-1}\right)\)

\(=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}:\dfrac{x-4}{x-1}=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)

b, Ta có P > 0 => \(\sqrt{x}-1>0\Leftrightarrow x>1\)

Kết hợp đk vậy x > 1 ; x khác 4