Question

[embe]

Answer 1

[Nguyễn Lê Phước Thịnh]

a: \(A=\dfrac{1}{4}\cdot x\left(-2x^5\right)=\left(-\dfrac{1}{4}\cdot2\right)\cdot x\cdot x^5=-\dfrac{1}{2}x^6\)

bậc là 6

 \(M=\dfrac{3}{12}\cdot x^3\cdot\left(-4x^6\right)\cdot x=\left(\dfrac{3}{12}\cdot\left(-4\right)\right)\cdot x^3\cdot x^6\cdot x=-x^{10}\)

=>Bậc là 10

\(N=\left(-\dfrac{2}{3}x\right)\cdot x^2\cdot\left(-6x^7\right)\)

\(=\left[\left(-\dfrac{2}{3}\right)\cdot\left(-6\right)\right]\cdot x\cdot x^2\cdot x^7\)

\(=4x^{10}\)

=>Bậc là 10

b: \(B=-4x^3+3x^2-5x+3x^3-3+x^3+7\)

\(=\left(-4x^3+3x^3+x^3\right)+\left(3x^2\right)-5x+4\)

\(=3x^2-5x+4\)

=>Bậc là 2

\(C=-2x+x^3-4x^2-5x-x^3+3x-1+2x^2-\dfrac{1}{2}\)

\(=\left(x^3-x^3\right)+\left(-4x^2+2x^2\right)+\left(-2x-5x+3x\right)+\left(-1-\dfrac{1}{2}\right)\)

\(=-2x^2-4x-\dfrac{3}{2}\)

=>Bậc là 2

\(D=2x^5-3x^4+4x^3-5+x-3x+x^3+4x^5-x^4+3\)

\(=\left(2x^5+4x^5\right)+\left(-3x^4-x^4\right)+\left(4x^3+x^3\right)+\left(x-3x\right)+\left(-5+3\right)\)

\(=6x^5-4x^4+5x^3-2x-2\)

bậc là 5

\(E=-x^3+2x-3+3x^2+x^3+3x-5-7x^2\)

\(=\left(-x^3+x^3\right)+\left(3x^2-7x^2\right)+\left(2x+3x\right)+\left(-3-5\right)\)

\(=-4x^2+5x-8\)

Bậc là 2