Question

алгебра помогите!!!!!!!!!!!!!!!!!!!!!!!!

Answer 1

Ответ:

$C) \ 12\sqrt{3}$

$D) \ x \geq 5$

$A) \ x \leq 7$

Объяснение:

$\dfrac{6}{\sqrt{3}-\sqrt{2}}+\dfrac{6}{\sqrt{3}+\sqrt{2}}=\dfrac{6(\sqrt{3}+\sqrt{2})+6(\sqrt{3}-\sqrt{2})}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}=\dfrac{6(\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2})}{(\sqrt{3})^{2}-(\sqrt{2})^{2}}=$

$=\dfrac{6 \cdot 2\sqrt{3}}{3-2}=\dfrac{12\sqrt{3}}{1}=12\sqrt{3} \ ;$

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$\sqrt{(x-5)^{2}}=x-5;$

$|x-5|=x-5;$

$\displaystyle \left \{ {{x-5=\pm (x-5)} \atop {x-5 \geq 0}} \right. \Leftrightarrow \left \{ {{x-5=\pm (x-5)} \atop {x \geq 5}} \right. ;$

$x-5=-(x+5); \quad x-5=-x+5; \quad x+x=5+5; \quad 2x=10; \quad x=5;$

$x \geq 5;$

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$\sqrt{(x-7)^{2}}=7-x;$

$\sqrt{(7-x)^{2}}=7-x;$

$|7-x|=7-x;$

$\displaystyle \left \{ {{7-x=\pm (7-x)} \atop {7-x \geq 0}} \right. \Leftrightarrow \left \{ {{7-x=\pm (7-x)} \atop {x-7 \leq 0}} \right. \Leftrightarrow \left \{ {{7-x=\pm (7-x)} \atop {x \leq 7}} \right. ;$

$7-x=-(7-x); \quad 7-x=-7+x; \quad x+x=7+7; \quad 2x=14; \quad x=7;$