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a: \(P\left(x\right)=-5x^4+2x^2-8x+\dfrac{1}{2}\)
\(Q\left(x\right)=4x^4+2x^3-5x^2-6x+\dfrac{3}{2}\)
b: \(A\left(x\right)=-5x^4+2x^2-8x+\dfrac{1}{2}+4x^4+2x^3-5x^2-6x+\dfrac{3}{2}=-x^4+2x^3-3x^2-14x+2\)
\(B\left(x\right)=-5x^4+2x^2-8x+\dfrac{1}{2}-4x^4-2x^3+5x^2+6x-\dfrac{3}{2}=-9x^4-2x^3+7x^2-2x-1\)
a: P(x)=6x^4+5x^3-3x^2+5x-10
Q(x)=5x^4+5x^3+2x^2-4x+4
b: P(x)+Q(x)
=6x^4+5x^3-3x^2+5x-10+5x^4+5x^3+2x^2-4x+4
=11x^4+10x^3-x^2+x-6
P(x)-Q(x)
=6x^4+5x^3-3x^2+5x-10-5x^4-5x^3-2x^2+4x-4
=x^4-5x^2+9x-14
Ta có: \(P\left(x\right)=-2x^4-7x+\frac{1}{2}-3x^4+2x^2-x\)
\(=-5x^4+2x^2-8x+\frac{1}{2}\)
Ta có: \(Q\left(x\right)=3x^3+4x^4-5x^2-x^3-6x+\frac{3}{2}\)
\(=4x^4+2x^3-5x^2-6x+\frac{3}{2}\)
Ta có: R(x)=P(x)-Q(x)
\(=-5x^4+2x^2-8x+\frac{1}{2}-4x^4-2x^3+5x^2+6x-\frac{3}{2}\)
\(=-9x^4-2x^3+7x^2-2x-1\)
Thay x=-1 vào đa thức \(R\left(x\right)=-9x^4-2x^3+7x^2-2x-1\), ta được:
\(R\left(-1\right)=-9\cdot\left(-1\right)^4-2\cdot\left(-1\right)^3+7\cdot\left(-1\right)^2-2\cdot\left(-1\right)-1\)
\(=-9\cdot1+2+7+2-1\)
\(=-9+10=1\)
Vậy: x=-1 không là nghiệm của đa thức R(x)=P(x)-Q(x)
a: P(x)=x^4-2x^4-5x^3-7x^2+2x-1
=-x^4-5x^3-7x^2+2x-1
Q(x)=3x^4-2x^4+5x^3+6x^2-2x+5
=x^4+5x^3+6x^2-2x+5
`P(x)=`\( 2x^4 + 3x^3 + 3x^2 - x^4 - 4x + 2 - 2x^2 + 6x\)
`= (2x^4-x^4)+3x^3+(3x^2-2x^2)+(-4x+6x)+2`
`= x^4+3x^3+x^2+2x+2`
`Q(x)=`\(x^4 + 3x^2 + 5x - 1 - x^2 - 3x + 2 + x^3\)
`= x^4+x^3+(3x^2-x^2)+(5x-3x)+(-1+2)`
`= x^4+x^3+2x^2+2x+1`
`P(x)+Q(x)=(x^4+3x^3+x^2+2x+2)+(x^4+x^3+2x^2+2x+1)`
`=x^4+3x^3+x^2+2x+2+x^4+x^3+2x^2+2x+1`
`=(x^4+x^4)+(3x^3+x^3)+(x^2+2x^2)+(2x+2x)+(2+1)`
`= 2x^4+4x^3+3x^2+4x+3`
`@`\(\text{dn inactive.}\)
P(x)=x^4+3x^3+x^2+2x+2
Q(x)=x^4+x^3+2x^2+2x+1
P(x)+Q(x)=2x^4+4x^3+3x^2+4x+3
a: P(x)=5x^3+3x^2-2x-5
\(Q\left(x\right)=5x^3+2x^2-2x+4\)
b: P(x)-Q(x)=x^2-9
P(x)+Q(x)=10x^3+5x^2-4x-1
c: P(x)-Q(x)=0
=>x^2-9=0
=>x=3; x=-3
d: C=A*B=-7/2x^6y^4
`@` `\text {Ans}`
`\downarrow`
`a)`
`P(x) =`\(3x^2+7+2x^4-3x^2-4-5x+2x^3\)
`= (3x^2 - 3x^2) + 2x^4 + 2x^3 - 5x + (7-4)`
`= 2x^4 + 2x^3 - 5x + 3`
`Q(x) =`\(3x^3+2x^2-x^4+x+x^3+4x-2+5x^4\)
`= (5x^4 - x^4) + (3x^3 + x^3) + 2x^2 + (x + 4x)- 2`
`= 4x^4 + 4x^3 + 2x^2 + 5x - 2`
`b)`
`P(-1) = 2*(-1)^4 + 2*(-1)^3 - 5*(-1) + 3`
`= 2*1 + 2*(-1) + 5 + 3`
`= 2 - 2 + 5 + 3`
`= 8`
___
`Q(0) = 4*0^4 + 4*0^3 + 2*0^2 + 5*0 - 2`
`= 4*0 + 4*0 + 2*0 + 5*0 - 2`
`= -2`
`c)`
`G(x) = P(x) + Q(x)`
`=> G(x) = 2x^4 + 2x^3 - 5x + 3 + 4x^4 + 4x^3 + 2x^2 + 5x - 2`
`= (2x^4 + 4x^4) + (2x^3 + 4x^3) + 2x^2 + (-5x + 5x) + (3 - 2)`
`= 6x^4 + 6x^3 + 2x^2 + 1`
`d)`
`G(x) = 6x^4 + 6x^3 + 2x^2 + 1`
Vì `x^4 \ge 0 AA x`
`x^2 \ge 0 AA x`
`=> 6x^4 + 2x^2 \ge 0 AA x`
`=> 6x^4 + 6x^3 + 2x^2 + 1 \ge 0`
`=> G(x)` luôn dương `AA` `x`
Ta có: \(P\left(x\right)=-5x^4+3x^3-2x^2+\dfrac{1}{2}x-1\)
\(Q\left(x\right)=6x^4+3x^3-4x^2+\dfrac{1}{2}x-4\)
\(\Rightarrow A\left(x\right)=P\left(x\right)-Q\left(x\right)=-11x^4+2x^2+3\)
Bài 4:
P(x)+Q(x)-R(x)
\(=6x^3-2x^2+3x-2-2x^3+3x^2-x+4-4x^3+2x-1\)
\(=x^2+4x+1\)
R(x)-P(x)-Q(x)
\(=-\left[P\left(x\right)+Q\left(x\right)-R\left(x\right)\right]\)
\(=-\left(x^2+4x+1\right)\)
\(=-x^2-4x-1\)
Bài 3:
a: \(P\left(x\right)=-2x^4-7x+\dfrac{1}{2}-6x^4+2x^2-x\)
\(=\left(-2x^4-6x^4\right)+2x^2+\left(-7x-x\right)+\dfrac{1}{2}\)
\(=-8x^4+2x^2-8x+\dfrac{1}{2}\)
\(Q\left(x\right)=3x^3-x^4-5x^2+x^3-6x+\dfrac{3}{4}\)
\(=-x^4+\left(3x^3+x^3\right)+\left(-5x^2\right)-6x+\dfrac{3}{4}\)
\(=-x^4+4x^3-5x^2-6x+\dfrac{3}{4}\)
b: P(x)+Q(x)
\(=-8x^4+2x^2-8x+\dfrac{1}{2}-x^4+4x^3-5x^2-6x+\dfrac{3}{4}\)
\(=-9x^4+4x^3-3x^2-14x+\dfrac{5}{4}\)
P(x)-Q(x)
\(=-8x^4+2x^2-8x+\dfrac{1}{2}+x^4-4x^3+5x^2+6x-\dfrac{3}{4}\)
\(=-7x^4-4x^3+7x^2-2x-\dfrac{1}{4}\)