Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) P(x) = 5x4 + 2x2 - 3x3 - 4x4+ 3x3 - x + 5
= ( 5x4 - 4x4 ) + ( 3x3 - 3x3 ) + 2x2 -x + 5
= x4 +2x2 - x +5
Q(x) = x - 5x3 - x2 - x4 + 5x3 - x2 + 3x - 1
= -x4 + ( 5x3 - 5x3 ) - ( x2 + x2 ) + 3x -1
= -x4 - 2x2 + 3x -1
b) P(x) + Q(x) = (x4 + 2x2 - x +5) + (-x4 - 2x2 + 3x -1)
= x4 + 2x2 - x +5 - x4 - 2x2 + 3x -1
= ( x4 -x4 ) + ( 2x2 - 2x2 ) + ( 3x - x ) + ( 5 - 1 )
= 2x + 4
c) Để đa thức có nghiệm thì A(x) = 0
hay P(x) + Q(x) = 0
2x + 4 = 0
2x = -4
x = -4 : 2 = -2
Vậy x = -2 là nghiệm của đa thức A(x)
tick cho mk nha các bn
a )\(P\left(x\right)=5x^4+2x^2-3x^3-4x^4+3x^3-x+5\)
\(=x^4+2x^2-x+5\).
\(Q\left(x\right)=x-5x^3-x^2-x^4+5x^3-x^2+3x-1\)
\(=-x^4-2x^2+4x-1\)
b ) \(P\left(x\right)+Q\left(x\right)=x^4+2x^2-x+5-x^4-2x^2+4x-1=3x+4\)
c ) \(Ax=3x+4=0\)
\(\Leftrightarrow x=-\dfrac{4}{3}\)
Vậy nghiệm của \(A\left(x\right)=-\dfrac{4}{3}\)
a, Sắp xếp : \(P\left(x\right)=2x^3+5x^2-3x^4+7-4x\)
\(\Rightarrow P\left(x\right)=-3x^4+2x^3-5x^2-4x+7\)
\(Q\left(x\right)=-3+2x^4-x+x^3-5x^2\)
\(\Rightarrow Q\left(x\right)=2x^4+x^3-5x^2-x-3\)
b, Ta có :* Đặt \(V\left(x\right)=P\left(x\right)+Q\left(x\right)\)
hay \(V\left(x\right)=2x^3+5x^2-3x^4+7-4x-3+2x^4-x+x^3-5x^2\)
\(=3x^3-x^4+4-5x\)
Vậy \(V\left(x\right)=3x^3-x^4+4-5x\)
Ta có : * Đặt \(K\left(x\right)=P\left(x\right)-Q\left(x\right)\)
hay \(2x^3+5x^2-3x^4+7-4x-\left(-3+2x^4-x+x^3-5x^2\right)\)
\(=2x^3+5x^2-3x^4+7-4x+3-2x^4+x-x^3+5x^2\)
\(=x^3+10x^2-5x^4+10-3x\)
Vậy \(K\left(x\right)=x^3+10x^2-5x^4+10-3x\)
F(\(x\)) = - 2\(x\)3 + 7 - 6\(x\) + 5\(x^4\) - 2\(x^3\)
F(\(x\)) = (-2\(x^3\) - 2\(x^3\)) + 7 - 6\(x\) + 5\(x^4\)
F(\(x\)) = -4\(x^3\) + 7 - 6\(x\) + 5\(x^4\)
F(\(x\)) = 5\(x^4\) - 4\(x^3\) - 6\(x\) + 7
G(\(x\)) = 5\(x^2\) + 9\(x\) - 2\(x^4\) - \(x^2\) + 4\(x^3\) - 12
G(\(x\)) = (5\(x^2\) - \(x^2\)) + 9\(x\) - 2\(x^4\) + 4\(x^3\) - 12
G(\(x\)) = 4\(x^2\) + 9\(x\) - 2\(x^4\) + 4\(x^3\) - 12
G(\(x\)) = -2\(x^4\) + 4\(x^3\) +4\(x^2\) + 9\(x\) - 12
b, F(\(x\)) + G(\(x\)) = 5\(x^4\) - 4\(x^3\) - 6\(x\) + 7 + ( -2\(x^4\) + 4\(x^3\)+4\(x^2\)+9\(x\)-12)
F(\(x\)) + G(\(x\)) = 5\(x^4\)- 4\(x^3\) - 6\(x\)+ 7 - 2\(x^4\) + 4\(x^3\) + 4\(x^2\) + 9\(x\) - 12
F(\(x\)) + G(\(x\)) = (5\(x^{4^{ }}\) -2\(x^4\)) -(4\(x^3\) - 4\(x^3\)) + 4\(x^2\) + (9\(x\)-6\(x\)) - ( 12 - 7)
F(\(x\)) + G(\(x\)) = 3\(x^4\) + 4\(x^2\) + 3\(x\) - 5
a) ta có Q=-2x^3+2x^2+12+5^-9x
Q=-2x^3+(2x^2+5x^2)-9x+12
Q=-2x^3+7x^2-9x+12
a) P(x) =5x3 - 5x + 9 +x
=5x3 + (-5x + x) + 9
= 5x3 - 4x + 9
Sắp xếp: tương tự như trên.
Mk đang bận chút mk làm tiếp.
a) \(P\left(x\right)=-3x^3+9-5x+7x^4-2x^2\)
\(=7x^4-3x^3-2x^2-5x+9\)
\(Q\left(x\right)=6x^2+5x-2x^4+8x^3-11\)
\(=-2x^4+8x^3+6x^2+5x-11\)
b) Ta có: \(P\left(x\right)+Q\left(x\right)\)
\(=\left(7x^4-3x^3-2x^2-5x+9\right)+\left(-2x^4+8x^3+6x^2+5x-11\right)\)
\(=7x^4-3x^3-2x^2-5x+9-2x^4+8x^{^3}+6x^2+5x-11\)
\(=\left(7x^4-2x^4\right)-\left(3x^3-8x^3\right)-\left(2x^2-6x^2\right)-\left(5x-5x\right)+\left(9-11\right)\)
\(=5x^4+5x^{^3}+4x^2 -2\)
Lại có: \(P\left(x\right)-Q\left(x\right)\)
\(=\left(7x^4-3x^3-2x^2-5x+9\right)-\left(-2x^4+8x^3+6x^2+5x-11\right)\)
\(=7x^4-3x^3-2x^2-5x+9+2x^4-8x^3-6x^2-5x+11\)
\(=\left(7x^4+2x^4\right)-\left(3x^3+8x^3\right)-\left(2x^2+6x^2\right)-\left(5x+5x\right)+\left(9+11\right)\)
\(=9x^4-11x^3-8x^2-10x+20\)
P(x)=-5x4+2x3-6x2-5x-3
Q(x)=5x4-2x3+5x2+5x+7
b/Có:Q-A=-P
<=>A=Q+P
<=>A=5x4-2x3+5x2+5x+7+(-5x4)+2x3-6x2-5x-3
<,=>A=(5x4-5x4)-(2x3-2x3)+(5x2-6x2)+(5x-5x)+(7-3)
<=>A=-x2+4
c/Có:A=0
<=>A=-x2+4=0
<=>x2=4
<=>x=+-2