Question

E urgent, pls, dau coroana

Answer 1

$\it 2)\ 9>8 \Rightarrow \sqrt9>\sqrt8 \Rightarrow 3>\sqrt{4\cdot2} \Rightarrow 3>2\sqrt2 \Rightarrow 3-2\sqrt2>0 \Rightarrow \\ \\ \Rightarrow |3-2\sqrt2|=3-2\sqrt2\\ \\ \sqrt{(2\sqrt2-3)^2}=\sqrt{(3-2\sqrt2)^2}=|3-2\sqrt2|=3-2\sqrt2$

$\it 3)\ \ a\cdotb=\sqrt{2-2\sqrt2}\cdot\sqrt{2+\sqrt2}=\sqrt{(2-\sqrt2)(2+\sqrt2)}=\\ \\ =\sqrt{2^2-(\sqrt2)^2}=\sqrt{4-2}=\sqrt2=\sqrt2$

Answer 2

1. √(2-√5)²-(3+√5)=

=|2-√5|-3-√5=

√5>2⇒ |2-√5|=√5-2

=√5-2-3-√5= -5

2. (2√2-3)²=

=(2√2)²-2×2√2×3+3²=

=8-12√2+9=17-12√2

√(2√2-3)²=

=|2√2-3|=3-2√2

3>2√2⇒ |2√2-3|=3-2√2

3. a) a×b=

=√(2-√2)×√(2+√2)=

=√(2-√2)(2+√2)=

=√[2²-(√2)²]=

=√(4-2)=√2

b) (a+b)²=

=[√(2-√2)+√(2+√2)]²=

=[√(2-√2)]²+2×√(2-√2)×√(2+√2)+[√(2-√2)]²=

=2-√2+2√2+2-√2=4