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Ответ

Автор - SkyBy
 sqrt{x+4}- sqrt{6-x}=2 \<br />
 (sqrt{x+4}- sqrt{6-x})^{2}=4 \<br />
x+4-2 sqrt{x+4} sqrt{6-x}+6-x=4 \<br />
-2 sqrt{x+4} sqrt{6-x}=-6 \<br />
sqrt{x+4} sqrt{6-x}=3 \<br />
sqrt{(x+4)(6-x)}=3 \<br />
sqrt{6x-x^{2}+24-4x}=3 \<br />
2x-x^{2}+24=9 \<br />
-x^{2}+2x+24-9=0 \<br />
x^{2}-2x-15=0 \<br />
x_{1}=5 \<br />
x_{2}=-3

Проверка:

x_{1}=5; \<br />
 sqrt{5+4} - sqrt{6-5} = 2 \<br />
2 = 2

верно;

x_{2}=-3; \<br />
 sqrt{-3+4}- sqrt{6-(-3)} =2 \<br />
-2 = 2

неверно.

Значит, уравнение имеет 1 корень: 5.

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