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Theo đề bài ta có \(M(x) = 2{x^4} - 5{x^3} + 7{x^2} + 3x\)
\(\begin{array}{l}M(x) + Q(x) = 6{x^5} - {x^4} + 3{x^2} - 2\\ \Rightarrow Q(x) = (6{x^5} - {x^4} + 3{x^2} - 2) - (2{x^4} - 5{x^3} + 7{x^2} + 3x)\\ \Rightarrow Q(x) = 6{x^5} - {x^4} + 3{x^2} - 2 - 2{x^4} + 5{x^3} - 7{x^2} - 3x\\Q(x) = 6{x^5} - 3{x^4} + 5{x^3} - 4{x^2} - 3x - 2\end{array}\)
Theo đề bài ta có :
\(\begin{array}{l}N(x) - M(x) = - 4{x^4} - 2{x^3} + 6{x^2} + 7\\ \Rightarrow N(x) = - 4{x^4} - 2{x^3} + 6{x^2} + 7 + 2{x^4} - 5{x^3} + 7{x^2} + 3x\\ \Rightarrow N(x) = - 2{x^4} - 7{x^3} + 13{x^2} + 3x + 7\end{array}\)
a, \(P\left(x\right)=5x^3-3x+7-x=5x^3-4x+7\)
\(Q\left(x\right)=-5x^3+2x-3+2x-x^2-2=-5x^3-x^2+4x-5\)
b, \(M\left(x\right)=5x^3-4x+7-5x^3-x^2+4x-5=-x^2+2\)
c, Đặt \(M\left(x\right)+2=0\Rightarrow-x^2+4=0\Leftrightarrow x^2=4\Leftrightarrow x=\pm2\)
a: \(P\left(x\right)=5x^3-3x+7-x=5x^3-4x+7\)
\(Q\left(x\right)=-5x^3+2x-3+2x-x^2-2=-5x^3-x^2+4x-5\)
b: Ta có: \(M\left(x\right)=P\left(x\right)+Q\left(x\right)\)
\(=5x^3-4x+7-5x^3-x^2+4x-5\)
\(=-x^2+2\)
c: Đặt M(x)+2=0
\(\Leftrightarrow4-x^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
`Q(x)=-5x^3+2x-3+2x-x^2-2`
`=-5x^3+4x-5`
`M(x)=P(x)+Q(x)`
`=5x^3-3x+7-5x^3+4x-5`
`=x+2`
`N(x)=P(x)-Q(x)`
`=5x^3-3x+7+5x^3-4x+5`
`=10x^3-7x+12`
b)Đặt `M(x)=0`
`<=>x+2=0`
`<=>x=-2`
Vậy M(x) có nghiệm `x=-2`
1k like đâu 
a) \(P\left(x\right)=5x^3-3x+7-x\\ =5x^3+\left(-3x-x\right)+7\\ =5x^3-4x+7\\ Q\left(x\right)=-5x^3+2x-3+2x-x^2-2\\ =-5x^3+\left(2x+2x\right)+\left(-3-2\right)+x^2\\ =-5x^3+4x-5+x^2\)
\(M\left(x\right)=P\left(x\right)+Q\left(x\right)\\ =5x^3-4x+7+\left(-5x^3\right)+4x-5-x^2\\ =\left(5x^3-5x^3\right)+\left(-4x+4x\right)+\left(7-5\right)-x^2\\ =2-x^2\\ N\left(x\right)=P\left(x\right)-Q\left(x\right)\\ =5x^3-4x+7-\left(-5x^3+4x-5+x^2\right)\\ =5x^3-4x+7+5x^3-4x+5-x^2\\ =\left(5x^3+5x^3\right)+\left(-4x-4x\right)+\left(7+5\right)+x^{^2}\\ =10x^3-8x+12+x^2\)
a: \(P\left(x\right)=5x^3-4x+7\)
\(Q\left(x\right)=-5x^3-x^2+4x-5\)
b: \(M\left(x\right)=-x^2+2\)
\(N\left(x\right)=10x^3+x^2-8x+12\)
c: Đặt M(x)=0
=>2-x2=0
hay \(x\in\left\{\sqrt{2};-\sqrt{2}\right\}\)
1: P(x)=M(x)+N(x)
=-2x^3+x^2+4x-3+2x^3+x^2-4x-5
=2x^2-8
2: P(x)=0
=>x^2-4=0
=>x=2 hoặc x=-2
3: Q(x)=M(x)-N(x)
=-2x^3+x^2+4x-3-2x^3-x^2+4x+5
=-4x^3+8x+2
\(Câu\text{ }4:\\ Ta\text{ }có:\text{(x^2 – 3x + 2) + (4x^3– x^2+ x – 1)}\\ =x^2-3x+2+4x^3-x^2+x-1\\ =\text{4x}^3+\left(x^2-x^2\right)+\left(-3x+x\right)+\left(2-1\right)\\ =4x^3-2x+1\)
\(Câu\text{ }5:Đặt\text{ }tính\text{ }trừ\text{ }như\text{ }sau:\)
`a)P(x)=5x^3-3x+7-x`
`=5x^3-3x-x+7`
`=5x^3-4x+7`
`Q(x)=-5x^3+2x-3+2x-x^2-2`
`=-5x^3-x^2+2x+2x-3-2`
`=-5^3-x^2+4x-5`
`M(x)=5x^3-4x+7-5x^3-x^2+4x-5`
`=5x^3-5x^3-x^2-4x+4x+7-5`
`=-x^2+2`
`N(x)=5x^3-4x+7+5x^3+x^2-4x+5`
`=5x^3+5x^3+x^2-4x-4x+7+5`
`=10x^3+x^2-8x+12`
Đặt `M(x)=0`
`<=>-x^2+2=0`
`<=>2=x^2`
`<=>x=+-sqrt2`
a: \(M\left(x\right)=-2x^4-3x^2-7x-2\)
\(N\left(x\right)=2x^4+3x^2+4x-5\)
\(P\left(x\right)=M\left(x\right)+N\left(x\right)=-3x-7\)
Đặt P(x)=0
=>-3x-7=0
hay x=-7/3
b: Q(x)=N(x)-M(x)
\(=2x^4+3x^2+4x+5+2x^4+3x^2+7x+2\)
\(=4x^4+6x^2+11x+7\)
Vì M(x) + N(x) = \(3{x^2} - 2x\)
Mà M(x) = \(7{x^3} - 2{x^2} + 8x + 4\)
Ta có: N(x) = M(x) + N(x) – M(x)
= \(3{x^2} - 2x - 7{x^3} + 2{x^2} - 8x - 4\)
\( = - 7{x^3} + 5{x^2} - 10x - 4\)