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a,
\(Q\left(x\right)=-3x^4+4x^3+2x^2+\dfrac{2}{3}-3x-2x^4-4x^3+8x^4+1+3x\\ =\left(-3x^4-2x^4+8x^4\right)+\left(4x^3-4x^3\right)+2x^2+\left(-3x+3x\right)+\left(\dfrac{2}{3}+1\right)\\ =3x^4+0+2x^2+0+\dfrac{5}{3}\\ =3x^4+2x^2+\dfrac{5}{3}\)
b, Ta có
\(\left\{{}\begin{matrix}x^4\ge0\\x^2\ge0\end{matrix}\right.\\ \Rightarrow3x^4+2x^2\ge0\\ \Rightarrow3x^4+2x^2+\dfrac{5}{3}\ge\dfrac{5}{3}>0\)
\(\Rightarrow Q\left(x\right)\) lớn hẳn hơn 0
\(\Rightarrow Q\left(x\right)\) vô nghiệm
a, \(A\left(x\right)+4x^3-x=-5x^2-2x^3+5+3x^2+2x\\ \Leftrightarrow A\left(x\right)=-5x^2-2x^3+5+3x^2+2x-4x^3+x=\left(-2x^3-4x^3\right)+\left(-5x^2+3x^2\right)+\left(2x+x\right)+5\\ =-6x^3-2x^2+3x+5\)
b, \(B\left(x\right)=A\left(x\right):\left(x-1\right)=\left(-6x^3-2x^2+3x+5\right):\left(x-1\right)=-6x^2-8x-5\)
Thay \(x=-1\) vào \(B\left(x\right)\)
\(\Rightarrow-6.\left(-1\right)^2-8\left(-1\right)-5=-3\ne0\)
\(\Rightarrow x=-1\) không là nghiệm của B(x)
Dễ mà bạn!
a)
M(x)= 5x^3+2x^4-x^2+3x^2-x^3-x^4+1-4x^3
M(x)= 2x^4-x^4+5x^3-4x^3-x^3-3x^2-x^2+1
M(x)= x^4+2x^2+1
b)
M(x)= x^4+2x^2+1
M(1)= 1^4+2.1^2+1
M(1)= 1+2+1
M(1)= 4
M(-1)= (-1)^4+2.(-1)^2+1
M(-1)= 1+2+1
M(-1)= 4
c) Vì x^4+2x^2+1 >= 1
Nên M(x)= x^4+2x^2+1 không có nghiệm
* M(x) = 5x3 + 2x4 - x2 + 3x2 - x3 - x4 + 1 - 4x3
= ( 2x4 - x4 ) + ( 5x3 - x3 - 4x3 ) + ( 3x2 - x2 ) + 1
= x4 + 2x2 + 1
* M(1) = 14 + 2 .12 + 1 = 1 + 2 . 1 + 1 = 4
M(-1) = (-1)4 + 2. (-1)2 + 1 = 1 + 2.1 + 1 = 4
* Ta có \(x^4\ge0\forall x,x^2\ge0\forall x\Rightarrow x^4+x^2+1\ge1>0\)
=> M(x) vô nghiệm
\(a,Q\left(x\right)=-3x^4+4x^3+2x^2+\dfrac{2}{3}-3x-2x^4-4x^3+8x^4+1+3x\\ =\left(-3x^4-2x^4+8x^4\right)+\left(4x^3-4x^3\right)+2x^2+\left(3x-3x\right)+1\\ =3x^4+2x^2+1\\ b,Q\left(x\right)=0\\ \Leftrightarrow3x^4+2x^2+1=0\\ \Delta=b^2-4ac=2^2-4.3.1=-8< 0\)
Vậy Q(x) không có nghiệm
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) \(P\left(x\right)=2x^3-2x+x^2-x^3+3x+2\)\(=\left(2x^3-x^3\right)+x^2+\left(3x-2x\right)+2=x^3+x^2+x+2\)
\(Q\left(x\right)=4x^3-5x^2+3x-4x-3x^3+4x^2+1\)
Q(x) \(=\left(4x^3-3x^3\right)+\left(4x^2-5x^2\right)+\left(3x-4x\right)+1\)\(=x^3-x^2-x+1\)
b) \(P\left(x\right)+Q\left(x\right)=2x^3+3\); \(P\left(x\right)-Q\left(x\right)=2x^2+2x+1\)
a) Sắp xếp theo lũy thừa giảm dần
P(x)=x^5−3x^2+7x^4−9x^3+x^2−1/4x
=x^5+7x^4−9x^3−3x^2+x^2−1/4x
=x^5+7x^4−9x^3−2x^2−1/4x
Q(x)=5x^4−x^5+x^2−2x^3+3x^2−1/4
=−x^5+5x^4−2x^3+x^2+3x^2−1/4
=−x^5+5x^4−2x^3+4x^2−1/4
b)
P(x)+Q(x)
=(x^5+7x^4−9x^3−2x^2−1/4^x)+(−x^5+5x^4−2x^3+4x^2−1/4)
=x^5+7x^4−9x^3−2x^2−1/4x−x^5+5x^4−2x^3+4x^2−1/4
=(x^5−x^5)+(7x^4+5x^4)+(−9x^3−2x^3)+(−2x^2+4x^2)−1/4x−1/4
=12x^4−11x^3+2x^2−1/4x−1/4
P(x)−Q(x)
=(x^5+7x^4−9x^3−2x^2−1/4x)−(−x^5+5x^4−2x^3+4x^2−1/4)
=x^5+7x^4−9x^3−2x^2−1/4x+x^5−5x^4+2x^3−4x^2+1/4
=(x^5+x^5)+(7x^4−5x^4)+(−9x^3+2x^3)+(−2x^2−4x^2)−1/4x+1/4
=2x5+2x4−7x3−6x2−1/4x−1/4
c) Ta có
P(0)=0^5+7.0^4−9.0^3−2.0^2−1/4.0
⇒x=0là nghiệm của P(x).
Q(0)=−0^5+5.0^4−2.0^3+4.0^2−1/4=−1/4≠0
⇒x=0không phải là nghiệm của Q(x).
`@` `\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`
\(P\left(x\right)=3x^2-5x^2+2x-x^2+4-x^4-\frac{1}{2}+x-2x\)
=\(\left(3x^2-5x^2-x^2\right)-x^4+\left(2x+x-2x\right)+\left(4-\frac{1}{2}\right)\)
=\(-3x^2-x^4+x+\frac{7}{2}\)
giảm -> =\(-x^4-3x^2+x+\frac{7}{2}\)
b)\(P\left(1\right)=-1^4-3.1^2+1+\frac{7}{2}\)
=\(-1-3.1+1+\frac{7}{2}\)
=\(-1-3+1+\frac{7}{2}\)
=\(\frac{1}{2}\)
\(P\left(\frac{1}{2}\right)=-\frac{1}{2}^4-3.\frac{1}{2}^2+\frac{1}{2}+\frac{7}{2}\)
=\(-\frac{1}{16}-3.-\frac{1}{4}+\frac{1}{2}+\frac{7}{2}\)
=\(-\frac{1}{16}-\left(-\frac{3}{4}\right)+\frac{1}{2}+\frac{7}{2}\)
=\(\frac{75}{16}\)