a)P(x) = 7x³ - x² + 5x - 2x³ +6 - 8x

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

 Q(x) = -2x + x³ - 4x² + 3 - 5x²

=x^3-9x^2-2x+3

b)

P(x) - Q(x)=4^3+8x^2-x-3

P(x) + Q(x)=6^3-10x^2-5x+9

          \(P\left(x\right)=8x^3\)         +    5x   -1 

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

       + \(R\left(x\right)=8x^3+8x^2+7x\)

Tổng :            16x^3 + 12x^2 +9x + 6

a) Ta có: \(P\left(x\right)=5x^2+3x^3-5x^2+2x^3-2+4x-4x^2+x^3\)

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

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

Ta có: \(Q\left(x\right)=6x-x^3+5-6x^3-6+7x^2-10x^2\)

\(=\left(-x^3-6x^3\right)+\left(7x^2-10x^2\right)+6x+\left(5-6\right)\)

\(=-7x^3-3x^2+6x-1\)

b) Ta có: P(x)+Q(x)

\(=6x^3-4x^2+4x-2-7x^3-3x^2+6x-1\)

\(=-x^3-7x^2+10x-3\)

Ta có: P(x)-Q(x)

\(=6x^3-4x^2+4x-2+7x^3+3x^2-6x+1\)

\(=13x^3-x^2-2x-1\)

P(x)+Q(x)=(5x³-+7x⁴+8x²)+(8x²-5x-3x³+x⁴)

= 5x³-7x⁴+8x²+8x²-5x-3x³+x⁴

=(5x³-3x³)+(-7x⁴+x⁴)+(8x²+8x²)-5x

=2x³-6x⁴+16x²-5x

P(x)-Q(x)=(5x³-+7x⁴+8x²)-(8x²-5x-3x³+x⁴)

= 5x³-+7x⁴+8x²-8x²+5x+3x³-x⁴

=(5x³+3x³)+(-7x^4_x⁴)+(8x^2-8x²)+5x

= 8x³-8x⁴+5x

*Cách 1: Hàng ngang:

P(x) - Q(x) = (5x\(^3\) - \(\dfrac{1}{3}\) + 7x\(^4\) + 8x\(^2\)) - (8x\(^2\) - 5x - 3x\(^3\) + x\(^4\) - \(\dfrac{2}{3}\))

= 5x\(^3\) - \(\dfrac{1}{3}\) + 7x\(^4\) + 8x\(^2\) - 8x\(^2\) + 5x + 3x\(^3\) - x\(^4\) +\(\dfrac{2}{3}\)

= (5x\(^3\) + 3x\(^3\)) + (-\(\dfrac{1}{3}\) + \(\dfrac{2}{3}\)) + (7x\(^4\) - x\(^4\)) + (8x\(^2\) - 8x\(^2\)) + 5x

= 8x\(^3\) + \(\dfrac{1}{3}\) + 6x\(^4\) + 5x

Vậy P(x) - Q(x) = 8x\(^3\) + \(\dfrac{1}{3}\) + 6x\(^4\) + 5x

*Cách 2: Hàng dọc:

P(x) = 7x\(^4\) + 5x\(^3\) + 8x\(^2\) + 0x - \(\dfrac{1}{3}\)

-

Q(x) = x\(^4\) - 3x\(^3\) + 8x\(^2\) - 5x - \(\dfrac{2}{3}\)

Hay: P(x) = 7x\(^4\) + 5x\(^3\) + 8x\(^2\) + 0x - \(\dfrac{1}{3}\)

+

[-Q(x)] = -x\(^4\) + 3x\(^3\) - 8x\(^2\) + 5x + \(\dfrac{2}{3}\)

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P(x) - Q(x) = 6x\(^4\) + 8x\(^3\) + 5x - \(\dfrac{1}{3}\)

Vậy P(x) - Q(x) = 6x\(^4\) + 8x\(^3\) + 5x - \(\dfrac{1}{3}\)

P(x) = 5x\(^3\) - \(\dfrac{1}{3}\) + 7x\(^4\) + 8x\(^2\)

Q(x) = 8x\(^2\) - 5x - 3x\(^3\) + x\(^4\) \(-\dfrac{2}{3}\)

*Cách 1: Hàng ngang:

P(x) + Q(x) = (5x\(^3\) - \(\dfrac{1}{3}\) + 7x\(^4\) + 8x\(^2\)) + (8x\(^2\) - 5x - 3x\(^3\) + x\(^4\) \(-\dfrac{2}{3}\))

= 5x\(^3\) - \(\dfrac{1}{3}\) + 7x\(^4\) + 8x\(^2\) + 8x\(^2\) - 5x - 3x\(^3\) + x\(^4\) - \(\dfrac{2}{3}\)

= ( 5x\(^3\) - 3x\(^3\) ) + ( \(-\dfrac{1}{3}\) - \(\dfrac{2}{3}\) ) + ( 7x\(^4\) + x\(^4\) ) + ( 8x\(^2\) + 8x\(^2\) ) - 5x

= 2x\(^3\) - 1 + 8x\(^4\) + 16x\(^2\) - 5x

Vậy P(x) + Q(x) = 2x\(^3\) - 1 + 8x\(^4\) + 16x\(^2\) - 5x

Sắp xếp lại theo thứ tự giảm dần của biến:

\(P\left(x\right)=-5x^3-\frac{1}{3}+8x^4+x=8x^4-5x^3+x-\frac{1}{3}\)

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

\(\Rightarrow P\left(x\right)+Q\left(x\right)=\left(8x^4-5x^3+x-\frac{1}{3}\right)+\left(x^4-2x^3+x^2-5x-\frac{2}{3}\right)\)

\(=\left(8x^4+x^4\right)-\left(5x^3+2x^3\right)+x^2-\left(5x-x\right)-\left(\frac{1}{3}+\frac{2}{3}\right)\)

\(=9x^4-7x^3+x^2-4x-1\)

Câu b tương tự

Lời giải:

a) $P(x)= 5x+x^3y-2xy+4x^3y+3x^2y-10x$

$=(x^3y+4x^3y)+3x^2y-2xy+(5x-10x)$

$=5x^3y+3x^2y-2xy-5x$

$Q(x)=4x-5x^3y+2x^2y-x^3y+6xy+11x^3-8x$

$=-6x^3y+2x^2y+11x^3+6xy-4x$

$P(x)-Q(x)=11x^3y+x^2y-8xy-x-11x^3$

Bậc của $P(x)-Q(x)$ là $3+1=4$

b)

$P(x)+Q(x)=-x^3y+5x^2y+4xy-9x+11x^3$

$P(x)-Q(x)$ đã thu gọn ở phần a.

a: P(x)=5x^2-4x+7

Sửa đề: Q(x)=-5x^3-x^2+4x-5

Q(x)+P(x)+5x^2-2=0

=>5x^2-4x+7-5x^3-x^2+4x-5+5x^2-2=0

=>5x^3=0

=>x=0