Bài này giải thế nào nhỉ?

V

vipboycodon

1) $x-5\sqrt{x-3}+3 = 0$
<=> $x+3 = 5\sqrt{x-3}$
<=> $x^2+6x+9 = 25(x-3)$
<=> $x^2+6x+9 = 25x-75$
<=> $x^2-19x+84 = 0$
<=> $(x-7)(x-12) = 0$
<=> $\begin{cases} x = 7 \\ x = 12 \end{cases}$
 
A

angleofdarkness

2)

$(\sqrt{12}-6\sqrt{3}+\sqrt{24})\sqrt{6}-5\sqrt{\dfrac{1}{2}}–12$

$=(2\sqrt{3}-6\sqrt{3}+2\sqrt{6})\sqrt{6}-5{\dfrac{\sqrt{2}}{2}} – 12$

$=(2\sqrt{6}-4\sqrt{3})\sqrt{6}-5{\dfrac{\sqrt{2}}{2}} – 12$

$=(12-12\sqrt{2})-5{\dfrac{\sqrt{2}}{2}} – 12$

$=-(12\sqrt{2}+5{\dfrac{\sqrt{2}}{2}})$

$= -14,5\sqrt{2}.$

 
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