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`@` `\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`
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
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
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) Các đơn thức đồng dạng trong các đơn thức sau là: \(5x^2yz;-2x^2yz\) ; \(x^2yz\) ; \(0,2x^2yz\)
b) \(M\left(x\right)=3x^2+5x^3-x^2+x-3x-4\)
\(M\left(x\right)=(3x^2-x^2)+5x^3+(x-3x)-4\)
\(M\left(x\right)=2x^2+5x^3-2x-4\)
\(M\left(x\right)=5x^3+2x^2-2x-4\)
c) \(P+Q=\left(x^3x+3\right)+\left(2x^3+3x^2+x-1\right)\)
\(P+Q=x^3x+3+2x^3+3x^2+x-1\)
\(P+Q=\left(x^3+2x^3\right)+\left(x+x\right)+\left(3-1\right)+3x^2\)
\(P+Q=3x^3+2x+2+3x^2\)
a) P(x) = -2x^2 + 4x^4 – 9x^3 + 3x^2 – 5x + 3
=4x^4-9x^3+x^2-5x+3
Q(x) = 5x^4 – x^3 + x^2 – 2x^3 + 3x^2 – 2 – 5x
=5x^4-3x^3+4x^2-5x-2
b)
P(x)
-bậc:4
-hệ số tự do:3
-hệ số cao nhất:4
Q(x)
-bậc :4
-hệ số tự do :-2
-hệ số cao nhất:5
a)
\(Q\left(x\right)=-1x^4+3x^2-5x^3-3-x\)
Sắp xếp: \(Q\left(x\right)=-1x^4-5x^3+3x^2-x-3\)
\(P\left(x\right)=5x^3+2x^2+1x^4+4+x\)
Sắp xếp: \(P\left(x\right)=1x^4+5x^3+2x^2+x+4\)
b)
\(Q\left(x\right)+P\left(x\right)=\left(-1x^4+3x^2-5x^3-x-3\right)+\left(1x^4+5x^3+2x^2+x+4\right)\)
\(=-1x^4-3x^2-5x^3-x-3+1x^4+5x^3+2x^2+x+4\)
\(=\left(-1x^4+1x^4\right)+\left(-3x^2+2x^2\right)+\left(-5x^3+5x^3\right)+\left(-x+x\right)+\left(-3+4\right)\)
\(=-1x^2+1\)
Vậy P(x) + Q(x) = -1x2 + 1
\(Q\left(x\right)+P\left(x\right)=\left(-1x^4+3x^2-5x^3-x-3\right)-\left(1x^4+5x^3+2x^2-x-4\right)\)
\(=-1x^4+3x^2-5x^3-x-3-1x^4-5x^3-2x^2-x-4\)
\(=\left(-1x^4-1x^4\right)+\left(3x^2-2x^2\right)+\left(-5x^3-5x^3\right)+\left(x-x\right)+\left(-3-4\right)\)
\(=-2x^4+x^2-10x^3-7\)
Vậy P(x) - Q(x) = -2x4 + x2 - 10x3 - 7
a)
Q\left(x\right)=-1x^4+3x^2-5x^3-3-xQ(x)=−1x4+3x2−5x3−3−x
Sắp xếp: Q\left(x\right)=-1x^4-5x^3+3x^2-x-3Q(x)=−1x4−5x3+3x2−x−3
P\left(x\right)=5x^3+2x^2+1x^4+4+xP(x)=5x3+2x2+1x4+4+x
Sắp xếp: P\left(x\right)=1x^4+5x^3+2x^2+x+4P(x)=1x4+5x3+2x2+x+4
b)
Q\left(x\right)+P\left(x\right)=\left(-1x^4+3x^2-5x^3-x-3\right)+\left(1x^4+5x^3+2x^2+x+4\right)Q(x)+P(x)=(−1x4+3x2−5x3−x−3)+(1x4+5x3+2x2+x+4)
=-1x^4-3x^2-5x^3-x-3+1x^4+5x^3+2x^2+x+4=−1x4−3x2−5x3−x−3+1x4+5x3+2x2+x+4
=\left(-1x^4+1x^4\right)+\left(-3x^2+2x^2\right)+\left(-5x^3+5x^3\right)+\left(-x+x\right)+\left(-3+4\right)=(−1x4+1x4)+(−3x2+2x2)+(−5x3+5x3)+(−x+x)+(−3+4)
=-1x^2+1=−1x2+1
Vậy P(x) + Q(x) = -1x2 + 1
Q\left(x\right)+P\left(x\right)=\left(-1x^4+3x^2-5x^3-x-3\right)-\left(1x^4+5x^3+2x^2-x-4\right)Q(x)+P(x)=(−1x4+3x2−5x3−x−3)−(1x4+5x3+2x2−x−4)
=-1x^4+3x^2-5x^3-x-3-1x^4-5x^3-2x^2-x-4=−1x4+3x2−5x3−x−3−1x4−5x3−2x2−x−4
=\left(-1x^4-1x^4\right)+\left(3x^2-2x^2\right)+\left(-5x^3-5x^3\right)+\left(x-x\right)+\left(-3-4\right)=(−1x4−1x4)+(3x2−2x2)+(−5x3−5x3)+(x−x)+(−3−4)
=-2x^4+x^2-10x^3-7=−2x4+x2−10x3−7
Vậy P(x) - Q(x) = -2x4 + x2 - 10x3 - 7