Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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
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
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)+(-7x4_x4)+(8x2-8x2)+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 + \(\dfrac{2}{3}\)
___________________________________________
P(x) - Q(x) = 6x\(^4\) + 8x\(^3\) + 5x - \(\dfrac{1}{3}\)
Vậy P(x) - Q(x) = 6x\(^4\) + 8x\(^3\) + 5x - \(\dfrac{1}{3}\)
\(P\left(x\right)=8x^3\) + 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)
\(=6x^3-4x^2+4x-2+7x^3+3x^2-6x+1\)
\(=13x^3-x^2-2x-1\)
Cho hai đa thức
P(x)=3x^3+8x^2-7x-6
Q(x)=4x^3-8x^2+3x+6
a) Tính P(0); Q(-1)
b) Tính P(x)+Q(x); P(x)-Q(x)
1) \(A\left(x\right)=-5x^3+3x^4+\frac{5}{7}-8x^2-10x\)
\(A\left(x\right)=3x^4-5x^3-8x^2-10x+\frac{5}{7}\)
\(B\left(x\right)=-2x^4-\frac{2}{7}+7x^2+8x^3+6x\)
\(B\left(x\right)=-2x^4+8x^3+7x^2+6x-\frac{2}{7}\)
2) \(A\left(x\right)=3x^4-5x^3-8x^2-10x+\frac{5}{7}\)
+
\(B\left(x\right)=-2x^4+8x^3+7x^2+6x-\frac{2}{7}\)
\(A\left(x\right)+B\left(x\right)=x^4+3x^3-x^2-4x+\frac{3}{7}\)
\(A\left(x\right)=3x^4-5x^3-8x^2-10x+\frac{5}{7}\)
-
\(B\left(x\right)=-2x^4+8x^3+7x^2+6x-\frac{2}{7}\)
\(A\left(x\right)-B\left(x\right)=5x^4-13x^3-15x^2-16x+1\)
P(x)=-5x^3-1/3+8x^4+x^2
Q(x)=x^4-2x^3+x^2-5x-2/3
P(x)+Q(x)
=x^4-2x^3+x^2-5x-2/3+8x^4-5x^3+x^2-1/3
=9x^4-7x^3+2x^2-5x-1
P(x)-Q(x)
=x^4-2x^3+x^2-5x-2/3-8x^4+5x^3-x^2+1/3
=-7x^4+3x^3-5x-1/3