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a: \(P\left(x\right)=3x^2-x-1\)
\(Q\left(x\right)=-3x^2-4x-2\)
b: \(G\left(x\right)=3x^2-x-1+3x^2+4x+2=6x^2+3x+1\)
c: Để G(x)-6x-1=0 thì 6x2-3x=0
=>3x(2x-1)=0
=>x=0 hoặc x=1/2
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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) \(P_{\left(x\right)}+Q_{\left(x\right)}=\left(2x^3+x^2+x+2\right)+\left(x^3+x^2++x+1\right)\)
\(=3x^3+2x^2+2x+3\)
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a,P(x)=4x3+x2-3x+7
Q(x)=4x3-x2-x-15
b,P(x)+Q(x)=4x3+x2-3x+7+4x3-x2-x-15
=8x3-4x-8
P(x)-Q(x)=4x3+x2-3x+7-(4x3-x2-x-15)
= 4x3+x2-3x+7-4x3+x2+x+15
=2x2-2x+22
c,
ta có P(x)=Q(x)
<=> P(x)-Q(x)=0
<=>2x2-2x+22=0
<=>x2-x+11=0
<=> x(x-1)=-11
do x\(\in Z\)nên x,x-1\(\inƯ\left(-11\right)\)
bn lập bảng xét các giá trị của x nha
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P(\(x\)) = \(x^4\) - 2\(x^3\) - 3\(x^2\) + 7\(x\) - 2
Q(\(x\)) = \(x^4\) + \(x^3\) - 2\(x\) + 1
P(\(x\)) + Q(\(x\)) = \(x^4\) - 2\(x^3\) - 3\(x^2\) + 7\(x\)- 2 + \(x^4\) + \(x^3\) - 2\(x\) + 7\(x\) - 2
P(\(x\)) + Q(\(x\)) = ( \(x^4\) + \(x^4\)) - (2\(x^3\) - \(x^3\)) - 3\(x^2\) + ( 7\(x\) - 2\(x\)) - (2-1)
P(\(x\)) +Q(\(x\)) =2 \(x^4\) - \(x^3\) - 3\(x^2\)+ 5\(x\) - 1
P(\(x\)) - Q(\(x\)) = \(x^4\) -2 \(x^3\)-3\(x^2\) +7\(x\) - 2 - \(x^4\) - \(x^3\) +2\(x\) - 1
P(\(x\)) -Q(\(x\)) = (\(x^4\) - \(x^4\)) - (2\(x^3\) + \(x^3\)) - 3\(x^2\) + ( \(7x+2x\)) - ( 2 + 1)
P(\(x\)) -Q(\(x\)) = - 3\(x^3\) - 3\(x^2\)+ 9\(x\) - 3
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a) Bậc P(x) = 4 + 3 + 1 = 8
Bậc của Q (x) = 2 + 3 + 1 = 6
b) P(x) + Q ( x) = x4 + x3 -2x + 1 + 2x2 -2x3 + x- 5
= x4 -x3 + 2x2 -x - 4
P(x) - Q (x) = x4 +x3 -2x + 1 - 2x2 -2x3 + x - 5
= x4 + 3x 3 -2x2 - 3x + 6
a) Bậc của đa thức P(x) là: 4+3+1=8
Bậc xủa đa thức Q(x) là: 2+3+1=6
b) P(x)+Q(x)=(x4+x3-2x+1)+(2x2-2x3+x-5)
P(x)+Q(x)=x4+x3-2x+1+2x2-2x3+x-5
P(x)+Q(x)=x4-x3+2x2-x-4
P(x)-Q(x)=(x4+x3-2x+1)-(2x2-2x3+x-5)
P(x)-Q(x)=x4+x3-2x+1-2x2+2x3-x+5
P(x)-Q(x)=x4+3x3-2x2-3x+6
P(x) + Q(x)
P(x)= x3 -2x + 1
Q(x)= x3 + 2x2 +x - 3
P(x)+Q(X) = 2x3+2x2 - 1 - 2
P(X) - Q(X)
P(X)= x3 +2x + 1
[-Q(X)]=x3 - 2x2 - x + 3
P(x) - Q(x) =x3 - 2x2 +1x + 4