toán 9

L

leminhnghia1

Giải:

$\begin{cases} & \sqrt{x}+\sqrt{y}=4 \ (1) \\ & \sqrt{x+6}+\sqrt{y+6}=14 \ (2) \end{cases}$

$(1) <=> x+y+2\sqrt{xy}=16 => x+y=16-2\sqrt{xy}$

$(2) <=> x+y+12+2\sqrt{xy+6(x+y)+36}=196$

$<=> 16-2\sqrt{xy}+2\sqrt{xy+96-12\sqrt{xy}+36}=168$

$<=> 2\sqrt{xy-12\sqrt{xy}+132}=168+2\sqrt{xy}$

Bình phương rồi tìm sẽ tìm đc $xy$.......
 
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