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 \(x\in\left\{\sqrt{2};-\sqrt{2}\right\}\)

`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) \(P\left(x\right)=5x^3-3x+7-x=5x^3-4x+7\)

\(Q\left(x\right)=-5x^3+2x-3+2x-x^2-2=-5x^3-x^2+4x-5\)

b) \(M\left(x\right)=5x^3-4x+7-5x^3-x^2+4x-5=-x^2+2\)

\(N\left(x\right)=5x^3-4x+7-\left(-5x^3-x^2+4x-5\right)=10x^3+x^2-8x+12\)

a) Ta có: \(P\left(x\right)=5x^3-3x+7-x\)

\(=5x^3-4x+7\)

Ta có: \(Q\left(x\right)=-5x^3+2x-3+2x-x^2-2\)

\(=-5x^3-x^2+4x-5\)

b) Ta có: M(x)=P(x)+Q(x)

\(=5x^3-4x+7-5x^3-x^2+4x-5\)

\(=-x^2+2\)

Ta có: N(x)=P(x)-Q(x)

\(=5x^3-4x+7+5x^3+x^2-4x+5\)

\(=10x^3+x^2-8x+12\)

c) Đặt M(x)=0

\(\Leftrightarrow-x^2+2=0\)

\(\Leftrightarrow-x^2=-2\)

\(\Leftrightarrow x^2=2\)

hay \(x\in\left\{\sqrt{2};-\sqrt{2}\right\}\)

a) P(x)=5x³ - 3x - x + 7

Q(x)=-5x³- x² + 2x + 2x -3 - 2

b) P(x) + Q(x) = ( 5x³- 3x - x + 7)+ ( -5x³- x² + 2x + 2x - 3 - 2 )

                       =5x³ - 3x - x + 7 - 5x³ - x² + 2x + 2x - 3 - 2

                       =(5x³-5x³)+(-x²)+(-3x-x+2x+2x)+(7-3-2)

           => M = -x²+2

P(x)-Q(x)= (5x³-3x-x+7)-(-5x³-x²+2x+2x-3-2)

               = 5x³-3x-x+7+5x³-x²+2x+2x-3-2

               =(5x³+5x³)+(-x²)+(-3x-x+2x+2x)+(7-3-2)

       => N =10x³ -x² +2

c)-x²+2=0

-x²=0+2

-x²=2

=>-x²=\(-\sqrt{2}\)

P(x) = 5x³ - 3x + 7 - x = 5x³ + ( -3x - x ) + 7 = 5x³ - 4x + 7

Q(x) = -5x³ + 2x - 3 + 2x - x² - 2 = -5x³ + ( 2x + 2x ) - x² + ( -3 - 2 ) = -5x³ + 4x - x² - 5

M(x) = P(x) + Q(x) 

= 5x³ - 4x + 7 + ( -5x³ + 4x - x² - 5 )

= ( 5x³ - 5x³ ) + ( 4x - 4x ) - x² + ( 7 - 5 )

= -x² + 2

N(x) = P(x) - Q(x) 

= ( 5x³ - 4x + 7 ) - ( -5x³ + 4x - x² - 5 )

= 5x³ - 4x + 7 + 5x³ - 4x + x² + 5

= ( 5x³ + 5x³ ) + ( -4x - 4x ) + x² + ( 7 + 5 )

= 10x³ - 8x + x² + 12

M(x) = 0 <=> -x² + 2 = 0

              <=> -x² = -2

             <=> x² = 2

             <=> x = \(\pm\sqrt{2}\)

Vậy nghiệm của M(x) là \(\pm\sqrt{2}\)

a, P(x)=5x³+x²-3x+7

Q(x)=-5x³-x²+4x-5(đã thu gọn-bn tự trình bày nha)

b,P(x)=5x³+x²-3x+7

  + 

   Q(x)=-5x³-x²+4x-5

   M(x)=               x-2

P(x)= 5x³ +x²  -3x+7

-

Q(x)=-5x³ - x² + 4x-5

N(x)=10x³+2x²-7x+12

c, x-2=0

       x=0+2

       x=2

=>Nghiệm bằng 2.

Để cho H(x) có nghiệm thì \(-\dfrac{1}{5}x-1=0\)

\(\Leftrightarrow-x-5=0\)

\(\Leftrightarrow-x=5\)

\(\Leftrightarrow x=-5\)

Để cho M(x) có nghiệm thì \(2x^2+4x=0\)

\(\Leftrightarrow2x\left(x+2\right)=0\)

\(\Leftrightarrow x\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

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