Zadanie 1
a)
4-\sqrt{10}>0 ,\ bo \ \sqrt{16}-\sqrt{10}>0 , 2\sqrt{10}-4>0 \ bo \ \sqrt{10}>3
|4-\sqrt{10}|+|2\sqrt{10}-4|=4-\sqrt{10}+2\sqrt{10}-4=\sqrt{10}
b)
2+3\sqrt2>0 ; 2-3\sqrt2<0
3|2+3\sqrt2|-3|2-3\sqrt2|
c)
1-\sqrt2<0 ; 2-2\sqrt3<0 ; \sqrt3-2<0
|1-\sqrt3|-3|2-2\sqrt3|+5|\sqrt3-2|=-(1-\sqrt3)-3[-(2-2\sqrt3)]+5[-(\sqrt3-2)]=
=\sqrt3-1-3(2\sqrt3-2)+5(2-\sqrt3)=\sqrt3-1-6\sqrt3+6+10-5\sqrt3=15-10\sqrt3
=5(3-2\sqrt3)
d)
5-\sqrt5>0\ ; \ \sqrt5-5<0
{|5-\sqrt5|-|\sqrt5-5|=5-\sqrt5-[-(\sqrt5-5)]=5-\sqrt5-(5-\sqrt5)=5-\sqrt5-5+\sqrt5=0}
e)
3-\sqrt5>0\ bo \ \sqrt9>\sqrt5
i \sqrt 5-4<0\ bo \ \sqrt5<\sqrt{16}
4-|3-\sqrt5|+2|\sqrt5-4|=4-(3-\sqrt5)+2[-(\sqrt5-4)]=4-3+\sqrt5+2(4-\sqrt5)=
=1+\sqrt5+8-2\sqrt5=9-\sqrt5
f)
2-5\sqrt2<0 ; 2,5\sqrt2-5<0
{1+\frac{1}{2}|2-5\sqrt2|-|2,5\sqrt2-5|=1+\frac{1}{2}[-(2-5\sqrt2)]-[-(2,5\sqrt2-5)]=1+\frac{1}{2}(5\sqrt2-2)-(5-2,5\sqrt2)=}
=1+2,5\sqrt2-1-5+2,5\sqrt2=5\sqrt2-5=5(\sqrt2-1)