Question

Fie a=(√2-1)x(1+√2)^2
b=(-6+2^2)^2 -√(1-√2)^2
a) Demonstrați ca a=1+√2
b) Comparați numerele a si b.
Va rog!!!

Answer 1

a) Demonstrați ca a=1+√2
a=(√2-1)x(1+√2)^2 =(√2-1)x(√2+1)^2 =(√2-1)x(√2+1)x(√2+1) =(2-1)(√2+1)=
=(√2+1)=1+√2

b=(-6+2^2)^2 -√(1-√2)^2=(-6+4)^2 -√(1-√2)^2=(-2)^2 +√2-1=
=4+√2 -1=3+√2


b) Comparați numerele a si b.
b>a

Answer 2

a)
a=(√2-1)x(1+√2)^2;
Prima data calculam (1+√2)^2;

(1+√2)^2=1+2√2+2=3+2√2;

a=(√2-1)×(1+√2)^2;
a=(√2-1)×(3+2√2)
a=3√2+4-3-2√2
a=√2+1;

b)
b=(-6+2^2)^2 -√(1-√2)^2
b=(-6+4)^2-| 1-√2|
b=(-2)^2-[ -(¬/2-1)]
b=4-(-√2+1)
b=4+√2-1=3+√2;

a=√2+1;
b=3+√2;
√2+1<3+√2
Din toate 3 =&gt ca b>a;
Iata niste formule de calcul prescurtat:

(a+b)^2=a^2 +2ab+b^2
(a-b)^2=a^2 -2ab+b^2
(a+b)(a-b)=a^2-b^2;

√(x^2)=|x|;
|x|=x; daca x>0;
|x|=0; daca x=0;
|x|=-x; daca x<0;