Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(K\left(x\right)=L\left(x\right)\)
\(\Rightarrow x^2-3x+2=x^2+px+q+1\)
\(\Rightarrow-3x+2=px+q+1\)
-Áp dụng PP hệ số bất định:
\(\Rightarrow p=-3;q+1=2\Rightarrow q=1\)
a) \(P\left(x\right)=3x^3-x^2-2x^4+3+2x^3+x+3x^4-x^2-2x^4+3+2x^3+x+3x^4\)
\(=2x^4+7x^3-2x^2+2x+6\)
\(Q\left(x\right)=-x^4+x^2-4x^3-2+2x^2-x-x^3-x^4+x^2-4x^3-2+2x^2-x-x^3\)
\(=-2x^4-10x^3+6x^2-2x-4\)
b) \(P\left(x\right)+Q\left(x\right)=2x^4+7x^3-2x^2+2x+6-2x^4-10x^3+6x^2-2x-4\)
\(=-3x^3+4x^2+2\)
\(a.\)
\(F\left(x\right)=2\left(x^4+x^3\right)+2x-4\left(x^2-x^3-1\right)+4\)
\(=2x^4+2x^3+2x-4x^2+4x^3-1\)
\(=4x^4+6x^3-4x^2+2x-1\)
\(G\left(x\right)=5x^4-4\left(3+x^4\right)-2x^2+4x^3+2\left(x^3-x^2+x\right)\)
\(=5x^4-12-4x^4-2x^2+4x^3+2x^3-2x^2+2x\)
\(=x^4+6x^3-4x^2+2x-12\)
\(b.\)
\(K\left(x\right)=F\left(x\right)+G\left(x\right)\)
\(=\left(4x^4+6x^3-4x^2+2x-1\right)+\left(x^4+6x^3-4x^2+2x-12\right)\)
\(=5x^4+12x^3-8x^2+4x-13\)
\(H\left(x\right)=F\left(x\right)-G\left(x\right)\)
\(=\left(4x^4+6x^3-4x^2+2x-1\right)-\left(x^4+6x^3-4x^2+2x-12\right)\)
\(=4x^4+6x^3-4x^2+2x-1-x^4-6x^3+4x^2-2x+12\)
\(=3x^4+11\)
\(c.\)
Ta có : \(H\left(x\right)=36\)
\(\Rightarrow3x^4+11=36\)
\(\Rightarrow3x^4=25\)
F(x) = 2x5 + 3x3 - 4x4 + 5x - x2 + x3 + x1
F(x) = 2x5 -4x4 + ( 3x3 + x3 ) -x2 + ( 5x+x)
F(x) = 2x5 - 4x4 + 4x3 - x2 + 6x
G(x) = -x2 - x5 + 2x4 - 3x3 + x4 +7
G(x) = -x5 + ( 2x4 + x4) -x2 +7
G ( x) = -x5 + 3x4 -x2 +7
a,F(x)= 2x\(^5\) + 3x\(^3\) - 4x\(^4\) + 5x - x\(^2\) + x\(^3\) + x\(^1\)
=2x\(^5\)- 4x\(^4\) \(+4x^3\)\(-x^2+6x\)
G(x)= -x\(^2\) - x\(^5\) + 2x\(^4\) - 3x\(^3\) + x\(^4\) + 7
=\(-x^5\)\(+3x^4\)\(-3x^3\)\(-x^2\)+7
b,F(x)-G(x)=(2x\(^5\)- 4x\(^4\) \(+4x^3\)\(-x^2+6x\))-\((-x^5+3x^4-3x^3-x^2+7)\)
=\(2x^5-4x^4+4x^3-x^2+6x\) \(+x^5-3x^4\)\(+3x^3\)\(+x^2-7\)
=\(\left(2x^5+x^5\right)\)+\(\left(-4x^4-3x^4\right)\)+\(\left(4x^3+3x^3\right)\)\(\left(-x^2+x^2\right)\)+6x-7
=\(3x^5-7x^4\)\(+7x^3+6x-7\)
Giải:
a)
Vậy ...
b)
Vậy ...
Đây nhé!!!!!