Ai làm giúp mình bài này với . Cảm ơn ạ A(x)=-3x+(1- 2x^2) B(x)=3x^3 – (2x^2- 6) C(x)=-1- (2x- 5x^3) a) Sắp xếp A(x), B(x), C(x) theo thứ tự từ bé đến

Đáp án:

$a) A(x)=-2x^2-3x+1$
$B(x)=3x^3-2x^2+6$
$C(x)=5x^3-2x-1$
$b) A(1)=-4\\
A(-2)=-1\\
B(0)=6\\
B(\frac{1}{2})=\frac{47}{8}\\
C(\frac{-1}{2})=\frac{-5}{8}\\
C(-1)=-4\\
c) A(x)+B(x)=3x^3-4x^2-3x+7\\
B(x)-C(x)=-2x^3-2x^2+2x+7\\
C(x)-A(x)=5x^3+2x^2+x-2\\
B(x)+C(x)=8x^3-2x^2-2x+5$

Giải thích các bước giải:

$a) A(x)=-3x+(1-2x^2)=-3x+1-2x^2=-2x^2-3x+1$
Bậc là 3
$B(x)=3x^3-(2x^2-6)=3x^3-2x^2+6$
Bậc 3
$C(x)=-1-(2x-5x^3)=-1-2x+5x^3=5x^3-2x-1$
Bậc 3
$b) A(1)=-2.1^2-3.1+1=-2-3+1=-4\\
A(-2)=-2.(-2)^2-3.(-2)+1=(-2).4+6+1=-1\\
B(0)=3.0^3-2.0^2+6=6\\
B(\frac{1}{2})=3.\frac{1}{2}^3-2.\frac{1}{2}^2+6=\frac{3}{8}-\frac{1}{2}+6\\
=\frac{3}{8}-\frac{4}{8}+\frac{6.8}{8}\\
=\frac{47}{8}\\
C(\frac{-1}{2})=5.\left ( \frac{-1}{2} \right )^3-2.(\frac{-1}{2})-1\\
=\frac{-5}{8}+1-1=\frac{-5}{8}\\
C(-1)=5.(-1)^3-2.(-1)-1=-5+2-1=-4\\
c) A(x)+B(x)=-2x^2-3x+1+3x^3-2x^2+6\\
=3x^3+(-2x^2-2x^2)+(1+6)=3x^3-4x^2-3x+7\\
B(x)-C(x)=3x^3-2x^2+6-(5x^3-2x-1)\\
=3x^3-2x^2+6-5x^3+2x+1\\
=(3x^3-5x^3)-2x^2+2x+(1+6)\\
=-2x^3-2x^2+2x+7\\
C(x)-A(x)=5x^3-2x-1-(-2x^2-3x+1)\\
=5x^3-2x-1+2x^2+3x-1\\
=5x^3+2x^2+(-2x+3x)+(-1-1)\\
=5x^3+2x^2+x-2\\
B(x)+C(x)=3x^3-2x^2+6+5x^3-2x-1\\
=(3x^3+5x^3)-2x^2-2x+(6-1)\\
=8x^3-2x^2-2x+5$