Question

EX 10 TE ROG FT EXPLICIT SLAVITESCUDORA​

Answer 1

a) 4^n=(2^n)^2

2^(n+1)=2^n*2

(2^n)^2+2^n*2*1+1^2=(2^n +1)^2

a=2^n, b=1 și am aplicat formula de calcul prescurtat (a+b)^2

b) (3^n)^2+2*3^n*1+1^2=(3^n+1)^2

c) (2^2)^2n=(2^2n)^2

(2^2n)^2+2*2^2n*1+1^2=(2^2n+1)^2

d) (10a+4)(10a+6)+1=10a*10a+6*10a+4*10a+4*6+1=100a^2+60a+40a+24+1=100a^2+100a+25=10a*10a+2*10a*5+5^2=(10a+5)^2

e) (10a+2)(10a+4)+1=100a^2+40a+20a+8+1=100a^2+60a+9=10a*10a+2*10a*3+3^2=(10a+3)^2

f) (10a+7)(10a+5)+1=100a^2+50a+70a+35+2=100a^2+120a+36=10a*10a+2*10*6+6^2=(10a+6)^2

g) x*x+2*x*5+5^2=(x+5)^2

h) Notez x^2+x=y

y(y+2)+1=y^2+2y+1=(y+1)^2=(x^2+x+1)^2
i) x(x+3)(x+1)(x+2)+1=(x^2+3x)(x^2+2x+x+2)+1=(x^2+3x)(x^2+3x+2)+1
Notez x^2+3x=y
=>y(y+2)+1=y^2+2y+1=y*y+2*y*1+1^2=(y+1)^2=(x^2+3x+1)^2

Toate sunt pătrate perfecte deoarece sunt ridicate la pătrat, după cum am demonstrat.
a^2+2ab+b^2=(a+b)^2