Question

50б даю за решение, срочнякккк

Answer 1

$1); ; sqrt{3-2sqrt{2}}cdot (1+sqrt2)= sqrt{1+2-2cdot 1cdot sqrt2} cdot (1+sqrt2)=\\= sqrt{1^2+(sqrt2)^2-2cdot 1cdot sqrt2} cdot (1+sqrt2)=sqrt{(1-sqrt2)^2}cdot (1+sqrt2)=\\=|1-sqrt2|cdot (1+sqrt2)=(sqrt2-1)cdot (sqrt2+1)=(sqrt2)^2-1^2=2-1=1\\\star ; ; 1-sqrt2approx1-1,4=-0,4 textless 0; ; Rightarrow ; ; |1-sqrt2|=-(1-sqrt2)=\\=sqrt2-1; ; star$

$2); ; b=-1 frac{1}{8} \\(b+4)^2-(b-3)(b+3)=b^2+8b+16-(b^2-9)=\\=8b+25=8cdot (-1frac{1}{8})+25=-8cdot frac{9}{8}+25=-9+25=16$