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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) \(P\left(x\right)+Q\left(x\right)=2x^4+7x^3-2x^2+2x+6-2x^4-10x^3+6x^2-2x-4\)
\(=-3x^3+4x^2+2\)
`@` `\text {Ans}`
`\downarrow`
`a)`
\(P(x) = 5x^3 + 3 - 3x^2 + x^4 - 2x - 2 + 2x^2 + x\)
`= x^4 + 5x^3 + (-3x^2 + 2x^2) + (-2x+x) + (3-2)`
`= x^4 + 5x^3 - x^2 - x + 1`
\(Q(x) = 2x^4 + x^2 + 2x + 2 - 3x^2 - 5x + 2x^3 - x^4\)
`= (2x^4 - x^4) + 2x^3 + (x^2 - 3x^2) + (2x-5x) + 2`
`= x^4 + 2x^3 - 2x^2 - 3x +2`
`b)`
`P(x)+Q(x) = (x^4 + 5x^3 - x^2 - x + 1) + (x^4 + 2x^3 - 2x^2 - 3x +2)`
`= x^4 + 5x^3 - x^2 - x + 1 + x^4 + 2x^3 - 2x^2 - 3x +2`
`= (x^4+x^4)+(5x^3 + 2x^3) + (-x^2 - 2x^2) + (-x-3x) + (1+2)`
`= 2x^4 + 7x^3 - 3x^2 - 4x + 3`
`P(x)-Q(x)=(x^4 + 5x^3 - x^2 - x + 1) - (x^4 + 2x^3 - 2x^2 - 3x +2)`
`= x^4 + 5x^3 - x^2 - x + 1 - x^4 - 2x^3 + 2x^2 + 3x -2`
`= (x^4 - x^4) + (5x^3 - 2x^3) + (-x^2+2x^2)+(-x+3x)+(1-2)`
`= 3x^3 + x^2 + 2x - 1`
`Q(x)-P(x) = (x^4 + 2x^3 - 2x^2 - 3x +2)-(x^4 + 5x^3 - x^2 - x + 1)`
`= x^4 + 2x^3 - 2x^2 - 3x +2-x^4 - 5x^3 + x^2 + x - 1`
`= (x^4-x^4)+(2x^3 - 5x^3)+(-2x^2+x^2)+(-3x+x)+(2-1)`
`= -3x^3 - x^2 - 2x + 1`
`@` `\text {Kaizuu lv u.}`
`a,`
`P(x)=5x^3+3-3x^2+x^4-2x-2+2x^2+x`
`P(x)=x^4+5x^3+(-3x^2+2x^2)+(-2x+x)+(3-2)`
`P(x)=x^4+5x^3-x^2-x+1`
`Q(x)=2x^4+x^2+2x+2-3x^2-5x+2x^3-x^4`
`Q(x)=(2x^4-x^4)+2x^3+(x^2-3x^2)+(2x-5x)+2`
`Q(x)=x^4+2x^3-2x^2-3x+2`
`b,`
`P(x)-Q(x)=(x^4+5x^3-x^2-x+1)-(x^4+2x^3-2x^2-3x+2)`
`P(x)-Q(x)= x^4+5x^3-x^2-x+1-x^4-2x^3+2x^2+3x-2`
`P(x)-Q(x)=(x^4-x^4)+(5x^3-2x^3)+(-x^2+2x^2)+(-x+3x)+(1-2)`
`P(x)-Q(x)=3x^3+x^2+2x-1`
`M(x)=P(x)+Q(x)`
`=x^4-5x+2x^2+1+5x+x^2+5-3x^2+x^4`
`=2x^4+6`
Đặt `M(x)=0`
`<=>2x^4+6=0`
`<=>x^4=-3`(vô lý vì `x^4>=0`)
a) Ta có M(x)=P(x)+Q(x)
=(\(x^4-5x+2x^2+1\))+(\(5x+x^2+5-3x^2+x^4\))
=\(x^4-5x+2x^2+1\)+\(5x+x^2+5-3x^2+x^4\)
=(\(x^4+x^4\))+(-5x+5x)+(\(2x^2\)+\(x^2\)-\(3x^2\))+(1+5)
=\(2x^4\)+6
Vậy M(x)=\(2x^4+6\)
b)Vì 2x\(^4\)\(\ge\) 0 với \(\forall\) x
nên \(2x^4+6\) \(\ge\)0 với \(\forall\)x
\(\Rightarrow\)M(x) \(\ge\) 0 với \(\forall\) x
Vậy M(x) vô nghiệm
a.
\(P(x)=3x^3-x^2-2x^4+3+2x^3+x+3x^4\)
\(=(-2x^4+3x^4)+(3x^3+2x^3)-x^2+x+3\)
\(=x^4+5x^3-x^2+x+3\)
\(Q(x)=-x^4+x^2-4x^3-2+2x^2-x-x^3\)
\(=-x^4+(-4x^3-x^3)+(x^2+2x^2)-x-2\)
\(=-x^4-5x^3+3x^2-x-2\)
b.
\(P(x)+Q(x)=(x^4+5x^3-x^2+x+3)+(-x^4-5x^3+3x^2-x-2)\)
\(=(x^4-x^4)+(5x^3-5x^3)+(-x^2+3x^2)+(x-x)+(3-2)\)
\(=2x^2+1\)
c.\(H(x)=Q(x)+P(x)\)
\(\Rightarrow H(x)=2x^2+1=0\)
\(\Rightarrow2x^2+1=0\)
\(2x^2\) \(=-1\)
\(x^2\) \(=\frac{-1}{2}\)
mà \(x^2\ge0\)
\(\Rightarrow\)Đa thức \(H(x)=P(x)+Q(x)\)ko có nghiệm
học tốt
Nhớ kết bạn với mình đó
1/
a,=>P(x)=2x3-4x2+5x-7-2x3+4x2-x+10=4x+3
=>Q(x)=-9x3-8x2+5x+11+9x3+8x2-2x-7=3x+4
b, Ta có: P(x)=0 => 4x+3=0 => x=-3/4
Q(x)=0 => 3x+4=0 => x=-4/3
c, P(x)+Q(x)=4x+3+3x+4=7x+7
P(x)-Q(x)=4x+3-(3x+4)=4x+3-3x-4=x-1
2/
a, x2-5x-6=0
=>x2-6x+x-6=0
=>x(x-6)+(x-6)=0
=>(x+1)(x-6)=0
=>\(\orbr{\begin{cases}x+1=0\\x-6=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-1\\x=6\end{cases}}}\)
b, (x+1)(x2+1)=0
Vì x2+1>0
=>x+1=0=>x=-1
c, \(-x^2-\frac{2}{5}=0\Rightarrow-x^2=\frac{2}{5}\Rightarrow x^2=\frac{-2}{5}\)
mà x2 lớn hoặc bằng 0 => không có x thỏa mãn
d, \(2x^2-x-6=0\Rightarrow2x^2-4x+3x-6=0\)
=>2x(x-2)+3(x-2)=0
=>(2x+3)(x-2)=0
=>\(\orbr{\begin{cases}2x+3=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\x=2\end{cases}}}\)
3/
a, P(x)=(5x3-x3-4x3)+(2x4-x4)+(-x2+3x2)+1=x4+2x2+1
b, P(1)=14+2.12+1=1+2+1=4
P(-1)=(-1)4+2.(-1)2+1=1+2+1=4
c, Vì \(x^4\ge0;2x^2\ge0\Rightarrow x^4+2x^2\ge0\Rightarrow P\left(x\right)=x^4+2x^2+1\ge1>0\)
Vậy P(x) khoogn có nghiệm