Question

Помогите пожалуйста!!! 50 баллов!

Answer 1

$1)\sqrt[10]{(1-x)^{10} }=|1-x|=\boxed{x-1}\\\\\\2)(\frac{1}{4})^{-\frac{1}{2}}*25^{\frac{1}{2} }-81^{\frac{1}{2} }*125^{-\frac{1}{3}}}=\sqrt{4} *\sqrt{25}-\sqrt{81}*\sqrt[3]{\frac{1}{125}} =\\\\=2*5-9*\frac{1}{5}=10-1,8=\boxed{8,2}$

3) ОДЗ :

1) x + 3 ≥ 0      ⇒      x ≥ - 3

2) x - 2 ≥ 0      ⇒      x ≥ 2

Окончательно : x ∈ [2 ; + ∞)

$3\sqrt{x+3}-\sqrt{x-2}=7\\\\3\sqrt{x+3}=\sqrt{x-2}+7\\\\(3\sqrt{x+3})^{2} =(\sqrt{x-2}+7)^{2}\\\\9(x+3)=x-2+14\sqrt{x-2}+49\\\\9x+27=x+14\sqrt{x-2}+47\\\\14\sqrt{x-2}=8x-20\\\\(7\sqrt{x-2})^{2}=(4x-10)^{2}\\\\49(x-2)=16x^{2}-80x+100\\\\49x-98=16x^{2}-80x+100\\\\16x^{2}-129x+198=0 D= 16641-12672=3969=63^{2}\\\\x_{1} =\frac{129-63}{32}=2,0625\\\\x_{2}=\frac{129+63}{32}=6\\\\Proverka:\\\\1)3\sqrt{2,0625+3}-\sqrt{2,0625-2}=7\\\\3\sqrt{5,0625}-\sqrt{0,0625}=7$

$3*2,25-0,25=7\\\\6,75-0,25=7\\\\6=7-neverno\\\\2)3\sqrt{6+3} -\sqrt{6-2} =7\\\\3\sqrt{9} -\sqrt{4}=7\\\\3*3-2=7\\\\7=7-verno\\\\Otvet:\boxed{6}$