Question

решите пожалуйста это.
Спасибо за ранее​

Answer 1

Объяснение:

1.

$a)\ 4\sqrt{2} +\sqrt{50}-\sqrt{18} =4\sqrt{2} +\sqrt{25*2}- \sqrt{9*2}=4\sqrt{2}+5\sqrt{2}-3\sqrt{2}=7\sqrt{2}.\\b)\ \sqrt{3}*(2\sqrt{3} +\sqrt{12} )=2*\sqrt{3*3}+\sqrt{12*3}=2*3+\sqrt{36}=6+6=12.\\c)\ (\sqrt{5} -2)^2=(\sqrt{5})^2-2*\sqrt{5}*2 +2^2=5 -4\sqrt{5} +4.\\d)\ (\sqrt{3}-\sqrt{2})*(\sqrt{3}+\sqrt{2})=(\sqrt{3})^2-(\sqrt{2})^2=3-2=1.$

2.

$a)\ \frac{3-\sqrt{3} }{2\sqrt{3} } =\frac{\sqrt{3}*(\sqrt{3}-1) }{2*\sqrt{3} } =\frac{\sqrt{3}-1 }{2}.\\b)\ \frac{4b-2}{2\sqrt{b}-\sqrt{2} } =\frac{(2\sqrt{b})^2-(\sqrt{2})^2 }{2\sqrt{b}-\sqrt{2} }=\frac{(2\sqrt{b}-\sqrt{2})(2\sqrt{b} +\sqrt{2}) }{2\sqrt{b}-\sqrt{2} }=2\sqrt{b}+\sqrt{2}.$

3.

$x^2=\sqrt{\sqrt{10} -3} *\sqrt{\sqrt{10}+3 }=\sqrt{(\sqrt{10}-3)(\sqrt{10}+3) }=\sqrt{(\sqrt{10})^2-3^2 } =\\=\sqrt{10-9} =\sqrt{1}=1.\ \ \ \ \Rightarrow\\x^2=1\\x^2-1=0\\(x+1)(x-1)=0\\x+1=0\\x_1=-1.\\x-1=0\\x_2=1.$

Ответ: x₁,₂=±1.