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a,P(x)=4x\(^3\)+2x\(^2\)-2x+7-x\(^2\)-x
=4x\(^3\)+(2x\(^2\)-x\(^2\))+(-2x-x)+7
=4x\(^3\)+x\(^2\)-3x+7
Q(x)=-4x\(^3\)+x-14-2x-x\(^2\)-1
=-4x\(^3\)-x\(^2\)+(x-2x)+(-14-1)
= -4x\(^3\)-x\(^2\) -x -15
b, P(x)+Q(x)=4x\(^3\)+x\(^2\)-3x+7-4x\(^3\)-x\(^2\) -x -15
=\(\left(4x^3-4x^3\right)\)+\(\left(x^2-x^2\right)\)+(-3x-x)+(7-15)
= -4x-8
P(x)-Q(x)=(4x\(^3\)+x\(^2\)-3x+7)-(-4x\(^3\)-x\(^2\) -x -15)
=4x\(^3\)+x\(^2\)-3x+7+4x\(^3\)+x\(^2\) +x +15
=\(\left(4x^3+4x^3\right)\)+\(\left(x^2+x^2\right)\)+(-3x+x)+(7+15)
= \(8x^3\) + \(2x^2\) - 2x + 22
a. P(x) = 4x3 + 2x2 +7-x2 -x = 4x3+x2-x+7
Q(x) = -4x3+x-(-14) -2x-x2-1 = -4x3+x+14-2x-x2-1 = -4x3 -x2 -x+13
Bài 1:
2x\(^2\)+3x\(^2\)-\(\dfrac{3}{2}\)x\(^2\)
=(2+3-\(\dfrac{3}{2}\)).x\(^2\)
Bài 2:
a,P(x)=4x\(^3\)+2x\(^2\)-2x+7-x\(^2\)-x
=4x3+(2x\(^2\)-x\(^2\))+(-2x-x)+7
=4x\(^3\)+x\(^2\)-3x+7
Q(x)=-4x\(^3\)+x-14-2x-x\(^2\)-1
=-4x\(^3\)-x\(^2\)+x+(-14-1)
=-4x\(^3\)-x\(^2\)+x-15
b,P(x)+Q(x):
P(x)=4x\(^3\)+x\(^2\)-3x+7
+
Q(x)=-4x\(^3\)-x\(^2\)+x-15
P(x)+Q(x)= -2x-8
P(x)-Q(x):
P(x)=4x\(^3\)+x\(^2\)-3x+7
-
Q(x)=-4x\(^3\)-x\(^2\)+x-15
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(x)=7x^3+(4x^2-3x^2)-x+5=7x^3+x^2-x+5`
`Q(x)=-7x^3-x^2+2x+(6-8)=-7x^3-x^2+2x-2`
b)
`P(x)+Q(x) = 7x^3+x^2-x+5-7x^3-x^2+2x-2`
`=(7x^3-7x^3)+(x^2-x^2)+(2x-x)+(5-2)`
`=x+3`
`P(x)-Q(x)=7x^3+x^2-x+5-(-7x^3-x^2+2x-2)`
`= 7x^3+x^2-x+5+7x^3+x^2-2x+2`
`=(7x^3+7x^3)+(x^2+x^2)-(x+2x)+(5+2)`
`=14x^3+2x^2-3x+7`
c) `A(x) = P(x)+Q(x)=x+3`
`A(x)=0 <=> x+3=0 <=>x=-3`.
Phần c cho x thuộc gì thế bn
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