Giải hệ phương trình: $\left\{\begin{array}{l}\sqrt{\text{x}}\text{+}\sqrt{\text{y}}\text{+4}\sqrt{\text{xy}}\text{=16}\\ \text{x+y=10}\end{array}\right.$

$\left\{\begin{array}{l}\sqrt{\text{x}}\text{+}\sqrt{\text{y}}\text{+4}\sqrt{\text{xy}}\text{=16}\\ \text{x+y=10}\end{array}\right.$(I)( Điều kiện:$x;y\ge 0$)

Đặt S= $\sqrt{x}+\sqrt{y}$  ; P = $\sqrt{xy}$   ( $S\ge 0;P\ge 0$ ) hệ (I) có dạng:

$\left\{\begin{array}{l}\text{S+4P=16 }\\ {\text{S}}^{\text{2}}\text{-2P=10 }\end{array}\right.\iff \left\{\begin{array}{l}\text{S+4P=16 }\\ {\text{2S}}^{\text{2}}\text{-4P=20 }\end{array}\right.\iff \left\{\begin{array}{l}\text{S+4P=16 }\\ {\text{2S}}^{\text{2}}\text{ +S-36=0 }\end{array}\right.$

$\iff \left\{\begin{array}{l}\text{S=4(tm);S=}\frac{\text{-9}}{\text{2}}\text{( loai) }\\ \text{P=3 }\end{array}\right.\iff \left\{\begin{array}{l}\text{S=4 }\\ \text{P=3 }\end{array}\right.$

Khi đó $\sqrt{x};\sqrt{y}$ là 2 nghiệm của phương trình:  $t^2–4t+3=0$

Giải phương trình ta được  $t_1=3;t_2=1$( thỏa mãn )

   $TH1:\left\{\begin{array}{l}\sqrt{\text{x}}\text{=3}\\ \sqrt{\text{y}}\text{=1}\end{array}\right.\iff \left\{\begin{array}{l}\text{x=9}\\ \text{y=1}\end{array}\right.$             $TH2:\left\{\begin{array}{l}\sqrt{\text{x}}\text{=1}\\ \sqrt{\text{y}}\text{=3}\end{array}\right.\iff \left\{\begin{array}{l}\text{x=1}\\ \text{y=9}\end{array}\right.$

( thỏa mãn)                                          (thỏa mãn)

Vậy hệ phương trình đã cho có hai nghiệm $\left\{\begin{array}{l}\text{x=9}\\ \text{y=1}\end{array}\right.;\left\{\begin{array}{l}\text{x=1}\\ \text{y=9}\end{array}\right.$