Question

Помогите решить. Хотя бы три первых номера, срочно нужно.

Answer 1

1)     a)
$3x^2+24xy+48y^2= 3(x^2+8xy+16y^2)= 3(x+4y)^2$

б)
$5x^2y^2-45a^2b^2=5(x^2y^2-9a^2b^2)=5(xy-3ab)(xy+3ab)$

в)
$a+5b+a^2-25b^2=-(5b-a-1)(5b+a)$

г)
$-y^2n+n+3m-y^2n=-2ny^2+n+3m$

д)
$45n+5nt^2-9n+nt^2=36n+6nt^2=6n(6+t^2)$

2) a)
$(x-2)^2+(x-1)(x+1)=x^2-4x+4+x^2-1=2x^2-4x+3$

б)
$(y-4)(y+3)+(y+1)^2-(7-y)(y+7)= \\ =y^2+3y-4y-12+y^2+2y+1-49+y^2= \\ =3y^2+y-60$

в)
$a(a-3)(a+3)-(a+2)(a^2-2a+4)=a(a^2-9)-(a^3+8)= \\ =a^3-9a-a^3-8=-9a-8$

3) a)
$x^3-36x=0 \\ x(x^2-36)=0 \\ x(x-6)(x+6)=0 \\ \\ x=0 \\ x-6=0 \\ x+6=0 \\ \\ x_1=0 \\ x_2=6 \\ x_3=-6$

б)
$25y^2-30y+9=0 \\ (5y-3)^2=0 \\ 5y-3=0 \\ y= \frac{3}{5}$

в)
$y^3-y^2=16y-16 \\ y^3-y^2-16y+16=0 \\ y^2(y-1)-16(y-1)=0 \\ (y^2-16)(y-1)=0 \\ (y-4)(y+4)(y-1)=0 \\ \\ y-4=0 \\ y+4=0 \\ y-1=0 \\ \\ y_1=4 \\ y_2=-4 \\ y_3=1$

4)
$3(4*-0.2)-(-1))^2-2(-0.2+1))(-0.2)+(-1))+4(-0.2+3*(-1)^2 \\ 3(-0.8+1)^2-2(-0.2+1)(-0.2-1)+4(-0.2-3)^2= \\ 3*0.2^2-2*0.8*(-1.2)+4*(-3.2)^2= \\ 3*0.04+1.92+4*10.24= \\ 0.12+1.92+40.96=43$