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Madalin0999
a fost răspuns

(3x+1) totul la 2 -(x+3) totul la 2 ?

Răspuns :

Flavistin
(3x+1)²=(3x+1)(3x+1)=3x(3x+1)+1(3x+1)=9x²+3x+3x+1=9x²+6x+1
(x+3)²=(x+3)(x+3)=x(x+3)+3(x+3)=x²+3x+3x+9=x²+6x+9
deci:
(3x+1)²-(x+3)²=(9x²+6x+1)-(x²+6x+9)=8x²-8=8(x²-1)
Am ajuns la concluzie ca: (3x+1)²-(x+3)²=8(x²-1)
[tex](3x+1)^2-(x+3)^2=(3x)^2+2 \cdot 3x \cdot 1+1^2-(x^2+2 \cdot x \cdot 3+3^2)= \\ =9x^2+6x+1-(x^2+6x+9)=9x^2+6x+1-x^2-6x-9= \\ =8x^2-8=8(x^2-1)[/tex]