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a: P(x)=6x^3-4x^2+4x-2
Q(x)=-5x^3-10x^2+6x+11
M(x)=x^3-14x^2+10x+9
b: \(C\left(x\right)=7x^4-4x^3-6x+9+3x^4-7x^3-5x^2-9x+12\)
=10x^4-11x^3-5x^2-15x+21
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G(x)=x^4(m-5)-x^3+4x^2-n+3
G(x) có hệ số cao nhất là 3 nên m-5=3 và -n+3=-2
=>m=8 và n=5
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a) Ta có: \(M\left(x\right)=4x^2-4x-3x^3-8\)
\(=-3x^3+4x^2-4x-8\)
Ta có: \(N\left(x\right)=2+3x^3+x-4x^2\)
\(=3x^3-4x^2+x+2\)
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\(a,P\left(x\right)=2x^3-x+x^2-x^3+3x+5\\ =\left(2x^3-x^3\right)+x^2+\left(-x+3x\right)+5\\ =x^3+x^2+2x+5\\ Q\left(x\right)=3x^3+4x^2+3x-4x^3-5x^2+10\\ =\left(3x^3-4x^3\right)+\left(4x^2-5x^2\right)+3x+10\\ =-x^3-x^2+3x+10\\ b,M\left(x\right)=P\left(x\right)+Q\left(x\right)=x^3+x^2+2x+5-x^3-x^2+3x+10\\ =\left(x^3-x^3\right)+\left(x^2-x^2\right)+\left(2x+3x\right)+\left(5+10\right)=5x+15\\ N\left(x\right)=P\left(x\right)-Q\left(x\right)=x^3+x^2+2x+5-\left(-x^3-x^2+3x+10\right)\\ =x^3+x^2+2x+5+x^3+x^2-3x-10\\ =\left(x^3+x^3\right)+\left(x^2+x^2\right)+\left(2x-3x\right)+\left(5-10\right)\\ =2x^3+2x^2-x-5\)
`a,P(x)= 2x^3 -x+x^2 -x^3 +3x+5`
`= (2x^3 -x^3)+x^2+(-x+3x) +5`
`= x^3 +x^2 + 2x+5`
`Q(x)=3x^3 +4x^2+3x-4x^3-5x^2+10`
`= (3x^3-4x^3)+(4x^2-5x^2)+3x+10`
`= -x^3 -x^2+3x+10`
`b,M(x)=P(x)+Q(x)`
`->M(x)=(x^3 +x^2 + 2x+5)+(-x^3 -x^2+3x+10)`
`=x^3 +x^2 + 2x+5+(-x^3) -x^2+3x+10`
`=(x^3 -x^3)+(x^2 -x^2)+(2x+3x)+(5+10)`
`= 5x+15`
`N(x)=P(x)-Q(x)`
`->N(x)=(x^3 +x^2 + 2x+5)-(-x^3 -x^2+3x+10)`
`=x^3 +x^2 + 2x+5-x^3 +x^2-3x-10`
`=(x^3-x^3)+(x^2+x^2)+(2x-3x)+(5-10)`
`=2x^2 -x-5`
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\(\dfrac{2x^5+x^4+3x^3-4x^2-14x+m+1}{x^2-2}\)
\(=\dfrac{2x^5-4x^3+x^4-2x^2+7x^3-14x-2x^2+4+m-3}{x^2-2}\)
\(=2x^2+x^2+7x-2+\dfrac{m-3}{x^2-2}\)
Đây là phép chia hết khi m-3=0
=>m=3
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a: \(M\left(x\right)=4x^3+x^2+4x+5\)
Hệ số tự do là 5
b: M(2)=32+4+8+5=49
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a, Thay x=2 vào A, ta được:
\(A\left(2\right)=3.2^3+5-6.2+5.2^2=37\)
Vậy A= 37 khi x=2.
b,
\(A\left(x\right)+B\left(x\right)=\left(3x^3+5-6x+5x^2\right)+\left(4x^2+6x-2x^7-9\right)\\ =-2x^7+3x^3+9x^2-4\)
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Ta có A(1) = 4 + a, A(2) = 16 + 2a.
Vì A(2) = 4A(1) ⇒ 16 + 2a = 4 (4 + a)
⇒ 16 + 2a = 16 + 4a ⇒ a = 0.
Chọn A