Question

Dau 20 puncte. Exercitiile 3,4. Va rog cit mai repede.

Answer 1

3.a)x(x+1)(x+2)=(x(x+1))(x+2)=(x(x+1))(x(x(x+1(2)=x^3+x^2+2x^2+2x=x^3+3x^2+2x.
b)a(a+b)(2a+1)=(a(a+b))(2a+1)=(a(a+b))(2a)+(a(a+b))(1)=2a^3+2a^2b+a^2+ab.
c)(a+1)-b·(b+2)=a+1+−b2+−2b.
d)(a+1)(a+3)(2+3a)=((a+1)(a+3))(2+3a)=((a+1)(a+3))(2)+((a+1)(a+3))(3a)=2a^2+8a+6+3a^3+12a^2+9a.
e)(2a+1)(2a+3)(2a+5)=((2a+1)(2a+3))(2a+5)=((2a+1)(2a+3))(2a)+((2a+1)(2a+3))(5)=8a^3+16a^2+6a+20a^2+40a+15=8a^3+36a^2+46a+15.
f)(a+5)(a+2)+a(1+a)=(a)(a)+(a)(2)+(5)(a)+(5)(2)+(a)(1)+(a)(a)=a^2+2a+5a+10+a+a^2=(a^2+a^2)+(2a+5a+a)+(10)=2a^2+8a+10.
g)3(2a+5)(1+a)−2(a+5)(1+3a)=(3(2a+5))(1)+(3(2a+5))(a)+−6a^2+−32a+−10=6a+15+6a^2+15a+−6a^2+−32a+−10=(6a^2+−6a^2)+(6a+15a+−32a)+(15+−10)=−11a+5.
h)(x2+2)(3+x)−(x+5)(x2+2)=(x^2)(3)+(x^2)(x)+(2)(3)+(2)(x)+−x^3+−5x^2+−2x+−10=3x^2+x3+6+2x+−x^3+−5x^2+−2x+−10=(x^3+−x^3)+(3x^2+−5x^2)+(2x+−2x)+(6+−10)=−2x^2+−4.
4.a)2a(x+y)+b(x+y)=(2a)(x)+(2a)(y)+(b)(x)+(b)(y)=2ax+2ay+bx+by.
b)y(a−b)−(a−b)=y(a−b)+−1(a−b)=y(a−b)+−1a+−1(−b)=y(a−b)+−a+b=(y)(a)+(y)(−b)+−a+b=ay+−by+−a+b.
c)(y+3)-x(y+3)=y+3+−xy+−3x=−xy−3x+y+3.
d)9(p-1)+(p-1)^2=(9)(p)+(9)(−1)+p^2+−2p+1=9p+−9+p^2+−2p+1=(p^2)+(9p+−2p)+(−9+1)=p^2+7p+−8.
e)(a+3)^2-a(a+3)=a^2+6a+9+−a^2+−3a=(a^2+−a^2)+(6a+−3a)+(9)=3a+9.
f)-3b(b-2)+7(b-2)^2=(−3b)(b)+(−3b)(−2)+7b^2+−28b+28=−3b^2+6b+7b^2+−28b+28=(−3b^2+7b^2)+(6b+−28b)+(28)=4b^2+−22b+28.
g)a(x-3)+b(x-3)=(a)(x)+(a)(−3)+(b)(x)+(b)(−3)=ax+−3a+bx+−3b.
h)(p^2-5)-q(p^2-5)=p^2+−5+−p^2q+5q.
i)7(c+2)+(x+2)^2=(7)(c)+(7)(2)+x^2+4x+4=7c+14+x^2+4x+4=(x^2)+(7c)+(4x)+(14+4)=x^2+7c+4x+18