Question

Utilizand formule de calcul prescurtat calculati:
a) (x+3y)²+(x-3y)(x+3y)=
b)(4x-3)²+5x(x-3)=
c)(1+x)²+(2+x)²+(3x+x)²=
d) (x²+y²)-(x²-y)²=
e)(x+4)²+(x-2)²-2x²=
f)(2x-3)²-(3x-5)²+(2x+5)(2x-5)=
g)(x+5)²-(x-2)(x+2)+2(x+1)²=
h) (3+4x)²+4(x-2)²-(x-3)(x+3)=
Btw puteți să scrieti pe caiet si sa faceti poză dacă cumva vă este mai ușor. Mă rog nu contează.

Answer 1

$a).(x+3y)^2+(x-3y)(x+3y)= \\ =x^2+2*x*3y+(3y)^2+x^2-9y^2=x^2+6xy+9y^2+x^2-9y^2= \\ =2x^2+6xy \\ b).(4x-3)^2+5x(x-3)=(4x)^2-2*4x*3+3^2+5x^2-15x= \\ =16x^2-24x+9+5x^2-15x=21x^2-39x+9$
$c).(1+x)^2+(2+x)^2+(3x+x)^2= \\ =1^2+2*1*x+x^2+2^2+2*2*x+x^2+(3x)^2+2*3x*x+x^2= \\ =1+2x+x^2+4+4x+x^2+9x^2+6x^2+x^2=18x^2+6x+5 \\ d).(x^2+y)^2-(x^2-y)^2= \\ =(x^2)^2+2*x^ 2*y+y^2-[(x^2)^2-2*x^2*y+y^2]= \\ =x^4+2x^2y+y^2-(x^4-2x^2y+y^2)= \\ =x^4+2x^2y+y^2-x^4+2x^2y-y^2=4x^2y$
$e).(x+4)^2+(x-2)^2-2x^2= \\ =x^2+2*x*4+4^2+x^2-2*x*2+2^2-2x^2= \\ =x^2+8x+16+x^2-4x+4-2x^2=4x+20 \\ f).(2x-3)^2-(3x-5)^2+(2x+5)(2x-5)= \\ =(2x)^2+2*2x*3+3^2-[(3x)^2-2*3x*5+5^2]+4x^2-25= \\ =4x^2+12x+9-(9x^2-30x+25)+4x^2-25= \\ =4x^2+12x+9-9x^2+30x-25+4x^2-25=-x^2+42x-41$
$g).(x+5)^2-(x-2)(x+2)+2(x+1)^2= \\ =x^2+2*x*5+5^2-x^2+4+2(x^2+2*x*1+1^2)= \\ =x^2+10x+25-x^2+4+2(x^2+2x+1)= \\ =x^2+10x+25-x^2+4+2x^2+4x+2=2x^2+14x+31 \\ h).(3+4x)^2+4(x-2)^2-(x-3)(x+3)= \\ =3^2+2*3*4x+(4x)^2+4(x^2-2*x*2+2^2)-x^2+9= \\ =9+24x+16x^2+4(x^2-4x+4)-x^2+9= \\ =9+24x+16x^2+4x^2-16x+16-x^2+9=19x^2+8x+34$