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![](https://rs.olm.vn/images/avt/0.png?1311)
`Q(-2)=(-2)^4+4*(-2)^3+2*(-2)^2-4*(-2)+1`
`= 16+4*(-8)+2*4+8+1`
`= 16-32+8+8+1`
`= -16+8+8+1`
`= -8+8+1=1`
`Q(1)=1^4+4*1^3+2*1^2-4*1+1`
`= 1+4+2-4+1`
`= 2+2+4-4=4`
Q(-2) = (-2)⁴ + 4.(-2)³ + 2.(-2)² - 4.(-2) + 1
= 16 - 32 + 8 + 8 + 1
= 1
--------------------
Q(1) = 1⁴ + 4.1³ + 2.1² - 4.1 + 1
= 1 + 4 + 2 - 4 + 1
= 4
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P\left(-1\right)=\left(-1\right)^4+2.\left(-1\right)^2+1=4\\ P\left(1\right)=1^4+2.1^2+1=4\)
\(P\left(-1\right)=\left(-1\right)^4+2\cdot\left(-1\right)^2+1=4\)
\(P\left(1\right)=P\left(-1\right)=4\)
\(Q\left(2\right)=2^4+4\cdot2^3+2\cdot2^2-4\cdot2+1=49\)
\(Q\left(1\right)=1^4+4\cdot1^3+2\cdot1^2-4\cdot1+1=4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(P\left(x\right)=3x^3-x^2-2x^4+3+2x^3+x+3x^4-x^2-2x^4+3+2x^3+x+3x^4\)
\(=2x^4+7x^3-2x^2+2x+6\)
\(Q\left(x\right)=-x^4+x^2-4x^3-2+2x^2-x-x^3-x^4+x^2-4x^3-2+2x^2-x-x^3\)
\(=-2x^4-10x^3+6x^2-2x-4\)
b) \(P\left(x\right)+Q\left(x\right)=2x^4+7x^3-2x^2+2x+6-2x^4-10x^3+6x^2-2x-4\)
\(=-3x^3+4x^2+2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(M+N=2x^2+x^2-2xy-2xy-3y^2+3y^2+1-1=3x^2-4xy\)
\(M-N=2x^2-x^2-2xy+2xy-3y^2-3y^2+1+1=x^2-6y^2+2\)
b, \(P\left(x\right)+Q\left(x\right)=x^3-4x^3+2x^2-6x+x+2-5=-3x^3+2x^2-5x-3\)
\(P\left(x\right)-Q\left(x\right)=x^3+4x^3-2x^2-6x-x+2+5=5x^3-2x^2-7x+7\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
Vì : P(x) + Q(x) = x5 – 2x2 + 1
Suy ra Q(x) = x5 – 2x2 + 1– P(x).
![](https://rs.olm.vn/images/avt/0.png?1311)
\(Q\left(x\right)=x^4+4x^3+2x^2-4x+1\)
\(\Rightarrow Q\left(-2\right)=\left(-2\right)^4+4\cdot\left(-2\right)^3+2\cdot\left(-2\right)^2-4\cdot\left(-2\right)+1=1\)
\(Q\left(-1\right)=\left(-1\right)^4+4\cdot\left(-1\right)^3+2\cdot\left(-1\right)^2-4\cdot\left(-1\right)+1=4\)
KQ la -4x3+4x
bn có thể giải từng bước cho mik hỉu dk ko.......