a) Ta có: \(M\left(x\right)=2x^3+3x^2-5x^2+6x^3-10x^2-8\)

\(=8x^3-12x^2-8\)

Bậc của đa thức M là 3

Ta có: \(N\left(x\right)=10x^3+6x^2-10x+8x^2-12\)

\(=10x^3+14x^2-10x-12\)

Bậc của đa thức N là 3

b) Ta có: \(M\left(x\right)=8x^3-12x^2-8\)

\(\Leftrightarrow M\left(x\right)=-8-12x^2+8x^3\)

Ta có: \(N\left(x\right)=10x^3+14x^2-10x-12\)

\(\Leftrightarrow N\left(x\right)=-12-10x+14x^2+10x^3\)

a)\(M\left(x\right)=3x^4-x^3-2x^2+5x+7\)

\(N\left(x\right)=-3x^4+x^3+10x^2+x-7\)

b)\(A\left(x\right)=M\left(x\right)+N\left(x\right)\)

\(=>A\left(x\right)=3x^4-x^3-2x^2+5x+7-3x^4+x^3+10x^2+x-7\)

\(A\left(x\right)=8x^2+6x\)

\(B\left(x\right)=3x^4-x^3-2x^2+5x+7+3x^4-x^3-10x^2-x+7\)

\(B\left(x\right)=6x^4-2x^3-12x^2+x+14\)

c)cho A(x) = 0

\(=>8x^2+6x=0=>x\left(8x+6\right)=0=>\left[{}\begin{matrix}x=0\\8x=-6\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=-\dfrac{3}{4}\end{matrix}\right.\)

a) P(x) = 6x^3 - 4x + 8 - x

P(x)=6x^3+(4-1)x+8

P(x)=6x^3+3x+8

Q(x) =-6x^3 + 3x - 3 + 2x - x^2 - 2

Q(x)=-6x^3+(3+2)x+(-3-2)-x^2

Q(x)=-6x^3+5x+(-5)-x^2

b)M(x)=P(x)+Q(x)

M(x)=(6x^3+3x+8)+ (-6x^3+5x+(-5)-x^2]

M(x)=6x^3+3x+8+(-6)x^3+(-5)-x^2

M(x)=(6-6)x^3+(8+5)+3x-x^2

M(x)=0+13+3x-x^2

M(x)=-x^2+3x+13

vậy đa thức M(x)= -x^2+3x+13

tick mình nha

a, A(x) = -4x⁵ - x³ + 4² + 5x + 7 + 4x⁵ - 6x²

= ( 4x⁵ - 4x⁵) - x³ + ( 4x² - 6x²) + 5x + 7

= -x³ - 2x² +5x +7

B(x) = -3x⁴ - 4x³ + 10x² - 8x + 5x³ -7 +8x

= -3x⁴ + ( 5x³ - 4x³ ) + 10x² + ( 8x - 8x )

= -3x⁴ + x³ + 10x²

b, A(x) = -x³ - 2x² + 5x +7

+

B(x) = -3x⁴ + x³ + 10x²

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P(x) = A(x) +B(x) = -3x⁴ + 8x² + 5x + 7

A(x) = -x³ - 2x² + 5x + 7

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B(x) = -3x⁴ + x³ + 10x²

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Q(x) = A(x) - B(x) = 3x⁴ - 2x³ - 12x² + 5x + 7