Question

Hai cu 15 va rog!! Cele incercuite! Mss

Answer 1

Răspuns:

Acesta este exercițiul! Scuze ca nu am scris a, b, c, d..............

$a( \sqrt{3} + x) ^{2} \\d(x + \sqrt{2} ) ^{2} \\ g( \sqrt{7} - x) ^{2} \\ b( \sqrt{5} + \sqrt{3} x) \\ e( \sqrt{5} - \sqrt{3} x) \\ h( 2 \sqrt{5} - z) \\ c( \sqrt{2} - x) \\ f(2x + \sqrt{6} )$

Answer 2

V= radical
*=la putearea....
•= înmulțirea


a) (V3)*2 +2V3x+x*2=(V3+x)*2 aici folosim formula a*2+2ab+b*2=(a+b)*2

b) (V5)*2+2V15x+(3x)*2=V5*2( întreg Radicalul la putere nu numai 5)+2•V5•V3+(V3x)*2=(V5+V3x)*3

c)V2*2-2V2x+x*2=2-2V3x+x*2=(V2-x)*2

d)x*2+2V2+2=x+2xV2+V2*2=(x+V2)*2

e)5-2V15x+3x*2=V5*2•V5•V3x+V3*2•x*2=V5*2-2•V5•V3x+(V3x)*2=(V5-V3x)*2

f) 2x*2+2V6x+3=(V2x)*2+2•V2x•V3+V3*2= (V2x+V3)*2

g) 7-2V7x+x*2=V7*2-2V7x+x*2=(V7-x)*2

h)(2V5)*2-4V5z+z*2=(2V5)*2-2•2V5•z+z*2=(2V5-z)*2


Sper ca te-am ajutat măcar puțin !!!