Question

Cate puteti!Urgent vrg!​

Answer 1

1) x=$\sqrt{2} +\sqrt{3}$

y=$\sqrt{2} -\sqrt{3}$

x+y= $\sqrt{2} +\sqrt{3} +\sqrt{2} -\sqrt{3}$ = $2\sqrt{2}$

x-y=  $\sqrt{2} +\sqrt{3} -\sqrt{2} -\sqrt{3}$=0

$(x+y)^{2} =$ $(\sqrt{2}+\sqrt{3})  ^{2} +2(\sqrt{2} +\sqrt{3} )(\sqrt{2} -\sqrt{3} ) + ( \sqrt{2}-\sqrt{3}  )^{2}$

$5+2\sqrt{6} -2+5-2\sqrt{6}$=8

2) (Acolo cred că e $x^{3} -x$)

a) $\sqrt{2} ^{3} -\sqrt{2} = 2\sqrt{2} -\sqrt{2} =\sqrt{2}$

b) x(x-1)(x+1)

c) $x^{3} -x=0 => x^{3} =3 =>x=\sqrt[3]{3}$

3) a) $x^{2} -2x-4x+8(=) x(x-2)*4(x-2) (=) (x-2)(x-4)$

b) x-2=0=>x=2

x-4=0=>x=4

4) a) x(x-1)

b) (x-2)(x+2)

c) [x+y+1(x-y+1)]*[x+y+1(x-y+1)]

=>(x+y+1-x+y-1)(x+y+1+x-y+1)

=>2y(2x+2)

=>2y*2(x+1)

=>2*2y*(x+1)=>4y(x+1)

5) a) $x^{2} -3=0=>x^{2} =3=>x=\sqrt{3}$ si x= $-\sqrt{3}$

b) x=+1 si x=-1

c) $x^{2} (x^{2} -4)=0=> x^{2} =0=>x=0$

si $x^{2} -4=0=>x^{2} =4=>x=+2$ si x=-2