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Bài 1 ( a )
\(A_x=-4x^5-x^3+4x^2+5x+9+4x^5-6x^2-2\)
\(=-x^3-2x^2+5x-7\)
\(B_x=-3x^4-2x^3+10x^2-8x+5x^3-7-2x^3+8x\)
\(=-3x^4+x^3+10x^2-7\)
Bài 1 ( b )
\(P_x=\left(-x^3-2x^2+5x-7\right)+\left(3x^4+x^3+10x-7\right)\)
\(=-x^3-2x^2+5x-7+3x^4+x^3+10x-7\)
\(=3x^4-2x^2+15x-14\)
\(Q_x=\left(-x^3-2x^2+5x-7\right)-\left(3x^4+x^3+10x-7\right)\)
\(=-x^3-2x^2+5x-7-3x^4-x^3-10x+7\)
\(=-3x^4-2x^3-5x\)
a) \(A\left(x\right)+B\left(x\right)=4x^5-2x^2-1\)
\(\Rightarrow2x^4-3x^3-4x+\dfrac{1}{2}+B\left(x\right)=4x^5-2x^2-1\)
\(\Rightarrow B\left(x\right)=4x^5-2x^2-1-2x^4+3x^3+4x-\dfrac{1}{2}\)
\(\Rightarrow B\left(x\right)=4x^5-2x^4+3x^3-2x^2+4x-\dfrac{3}{2}\)
b) \(A\left(x\right)-C\left(x\right)=2x^3\)
\(\Rightarrow2x^4-3x^3-4x+\dfrac{1}{2}-C\left(x\right)=2x^3\)
\(\Rightarrow C\left(x\right)=2x^4-3x^3-4x+\dfrac{1}{2}-2x^3\)
\(\Rightarrow C\left(x\right)=2x^4-3x^3-2x^3-4x+\dfrac{1}{2}\)
\(\Rightarrow C\left(x\right)=2x^4-5x^3-4x+\dfrac{1}{2}\)
a) B(x) = 4x5 -2x2 -1 - A(x) = 4x5 -2x2 -1 -2x4 +3x3+4x -1/2
B(x) = 4x5 -2x4 +3x3-2x2 +4x - 1/2
b) tt
a) \(A\left(x\right)+B\left(x\right)=4x^5-2x^2-1\)
\(\Rightarrow B\left(x\right)=4x^5-2x^2-1-A\left(x\right)\)
\(\Rightarrow B\left(x\right)=4x^5-2x^2-1-\left(2x^4-3x^3+\dfrac{1}{2}-4x\right)\)
\(B\left(x\right)=4x^5-2x^2-1-2x^4+3x^3-\dfrac{1}{2}+4x\)
Vậy \(B\left(x\right)=4x^5-2x^4+3x^3-2x^2+4x-\dfrac{3}{2}\)
b) \(A\left(x\right)-C\left(x\right)=2x^3\)
\(\Rightarrow C\left(x\right)=2x^3+A\left(x\right)\)
\(\Rightarrow C\left(x\right)=2x^3+2x^4-3x^3+\dfrac{1}{2}-4x\)
Vậy \(C\left(x\right)=2x^4-x^3-4x+\dfrac{1}{2}\)
a) \(P_{\left(x\right)}=2x^3-2x+x^2+3x+2\)
\(P_{\left(x\right)}=2x^3+x^2+x+2\)
\(Q_{\left(x\right)}=4x^3-3x^2-3x+4x-3x^3+4x^2+1\)
\(Q_{\left(x\right)}=x^3+x^2+x+1\)
b) \(P_{\left(x\right)}+Q_{\left(x\right)}=\left(2x^3+x^2+x+2\right)+\left(x^3+x^2++x+1\right)\)
\(=3x^3+2x^2+2x+3\)
a, \(P\left(x\right)=4x^3+2x-3+2x-2x^2-1\\ =4x^3-2x^2+\left(2x+2x\right)+\left(-3-1\right)\\ =4x^3-2x^2+4x-4\)
Bậc của P(x) là 3
\(Q\left(x\right)=6x^3-3x+5-2x+3x^2\\ =6x^3+3x^2+\left(-3x-2x\right)+5\\ =6x^3+3x^2-5x+5\)
Bậc của Q(x) là 3
b, \(M\left(x\right)=P\left(x\right)+Q\left(x\right)=4x^3-2x^2+4x-4+6x^3+3x^2-5x+5\\ =\left(4x^3+6x^3\right)+\left(-2x^2+3x^2\right)+\left(4x-5x\right)+\left(-4+5\right)\\ =10x^3+x^2-x+1\)
a)
P(x) + O(x) = \(\left(x^3+2x^2-3x+2020\right)+\left(2x^3-3x^2+4x+2021\right)\)
P(x) + O(x) = \(3x^3-x^2+x+4041\)
b)
P(x) - O(x) = \(x^3+2x^2-3x+2020-2x^3+3x^2-4x-2021\)
P(x) - O(x) = \(-x^3+5x^2-7x-1\)
a, M(\(x\) )+N(\(x\)) = 3\(x^4\) - 2\(x\)3 + 5\(x^2\) - \(4x\)+ 1 + ( -3\(x^4\) + 2\(x^3\)- 3\(x^2\)+ 7\(x\) + 5)
M(\(x\)) + N(\(x\)) = ( 3\(x^4\)- 3\(x^4\))+( -2\(x^3\) + 2\(x^3\))+(5\(x^2\) - 3\(x^2\))+( 7\(x-4x\)) +(1+5)
M(\(x\)) + N(\(x\)) = 0 + 0 + 2\(x^2\) + 3\(x\) + 6
M(\(x\)) + N(\(x\)) = 2\(x^2\) + 3\(x\) + 6
b, P(\(x\)) = M(\(x\)) + N(\(x\)) = 2\(x^2\) + 3\(x\) + 6
P(-2) = 2.(-2)2 + 3.(-2) + 6 = 8 - 6 + 6 = 8
Bài 1:
a) Ta có: \(P\left(x\right)=3x^4+2x^2-3x^4-2x^2+2x-5\)
\(=\left(3x^4-3x^4\right)+\left(2x^2-2x^2\right)+2x-5\)
\(=2x-5\)
Bài 1:
b)
\(P\left(-1\right)=2\cdot\left(-1\right)-5=-2-5=-7\)
\(P\left(3\right)=2\cdot3-5=6-5=1\)
A(x)+B(x)= x^3 + 2x^2 -x+1+2x^3 +3x^2 +4x +5
= ( x^3 +2x^3) + ( 2x^2 + 3x^2) + ( -x +4x ) + ( 1 +5)
= 3x^3 + 5x^2 + 3x +6
A(x) - B(x) = x^3 +2x^2 -x+1 - 2x^3 - 3x^2 -4x-5
= (x^3 - 2x^3) + ( 2x^2 - 3x^2) + ( -x -4x ) + ( 1-5)
= -x^3 - x^2 - 5x-4
Đây nha bạn :)