Question

[Tiến Nguyễnn]

Cho hai đa thức : A(x)= 2x³ +2x - 3x² + 11  ; B(x) = 2x² + 3x³ - x - 5 
 a ) sắp xếp theo lũy giảm dần của biến 
b ) Tính a(x) + b(x)  
c) Tính a(x) - B(x)

Answer 1

[⭐Hannie⭐]

`a,A(x)=2x^3+2x-3x^2+11`

`=2x^3-3x^2+2x+11`

`B(x)=2^2+3x^3-x-5`

`=3x^3+2x^2-x-5`

`b, A(x)+B(x)=(2x^3-3x^2+2x+11)+(3x^3+2x^2-x-5)`

`=2x^3-3x^2+2x+11+3x^3+2x^2-x-5`

`=(2x^3+3x^3)+(-3x^2+2x^2)+(2x-x)+(11-5)`

`=5x^3 -x^2 +x+6`

`c,A(x)-B(x)=(2x^3-3x^2+2x+11)-(3x^3+2x^2-x-5)`

`=2x^3-3x^2+2x+11- 3x^3 -2x^2+x+5`

`=(2x^3-3x^3)+(-3x^2-2x^2)+(2x+x)+(11+5)`

`=-x^3 -5x^2+3x+16`

Answer 2

[Sahara]

a/\(A\left(x\right)=2x^3+2x-3x^2+11\)
\(=2x^3-3x^2+2x+11\)
\(B\left(x\right)=2x^2+3x^3-x-5\)
\(=3x^3+2x^2-x-5\)
b/\(A\left(x\right)+B\left(x\right)=\left(2x^3-3x^2+2x+11\right)+\left(3x^3+2x^2-x-5\right)\)
\(=2x^3-3x^2+2x+11+3x^3+2x^2-x-5\)
\(=\left(2x^3+3x^3\right)-\left(3x^2-2x^2\right)+\left(2x-x\right)+\left(11-5\right)\)
\(=5x^3-x^2+x+6\)
c/\(A\left(x\right)+B\left(x\right)=\left(2x^3-3x^2+2x+11\right)-\left(3x^3+2x^2-x-5\right)\)
\(=2x^3-3x^2+2x+11-3x^3-2x^2+x+5\)
\(=\left(2x^3-3x^3\right)-\left(3x^2+2x^2\right)+\left(2x+x\right)+\left(11+5\right)\)
\(=-x^3-5x^2+3x+16\)
#DarkPegasus