Mình thu gọn 2 đa thức trước r mới cộng nhé

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

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

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

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

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

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

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

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

\(P\left(x\right)+Q\left(x\right)=6x^4+1+2x^2\)

`P(x)=`\( 2x^4 + 3x^3 + 3x^2 - x^4 - 4x + 2 - 2x^2 + 6x\)

`= (2x^4-x^4)+3x^3+(3x^2-2x^2)+(-4x+6x)+2`

`= x^4+3x^3+x^2+2x+2`

`Q(x)=`\(x^4 + 3x^2 + 5x - 1 - x^2 - 3x + 2 + x^3\)

`= x^4+x^3+(3x^2-x^2)+(5x-3x)+(-1+2)`

`= x^4+x^3+2x^2+2x+1`

`P(x)+Q(x)=(x^4+3x^3+x^2+2x+2)+(x^4+x^3+2x^2+2x+1)`

`=x^4+3x^3+x^2+2x+2+x^4+x^3+2x^2+2x+1`

`=(x^4+x^4)+(3x^3+x^3)+(x^2+2x^2)+(2x+2x)+(2+1)`

`= 2x^4+4x^3+3x^2+4x+3`

`@`\(\text{dn inactive.}\)

P(x)=x^4+3x^3+x^2+2x+2

Q(x)=x^4+x^3+2x^2+2x+1

P(x)+Q(x)=2x^4+4x^3+3x^2+4x+3

P(x) = 2x⁴ + 3x³ + 3x² - x⁴ - 4x + 2 - 2x² + 6x

Q(x) = x⁴ + 3x² + 5x - 1 - x² - 3x + 2 + x³

P(x)+Q(x) = 2x⁴ + 3x³ + 3x² - x⁴ - 4x + 2 - 2x² + 6x + x⁴ + 3x² + 5x - 1 - x² - 3x + 2 + x³

P(x)+Q(x) = (2x⁴-x⁴+x⁴) + (3x³+x³) + (3x²-2x²+3x²-x²) - (4x-6x-5x+3x) +(2-1+2)

P(x)+Q(x) = 4x³+3x²-4x+3

f(x)+g(x)=-3x^2+x-1+x^4-x^3-x^2+3x^4+2x^3-x^4-x^2+x^3-x+5-5x^3+x^2+3x^4

            =(-3x^2-x^2-x^2+x^2)+(x-x)-(1-5)+(x^4+3x^4-x^4+3x^4)-(x^3-2x^3-x^3+5x^3)

            =-4x^2+4+6x^4+3x^3

`@` `\text {Ans}`

`\downarrow`

`a,`

`P(x)+Q(x) = (3x^4-2x^3+3x+11)+(3x^2- x^3-5x+3x+4-x+2x^4)`

`= 3x^4-2x^3+3x+11+3x^2- x^3-5x+3x+4-x+2x^4`

`= (3x^4 + 2x^4) + (-2x^3 - x^3) + 3x^2 + (3x + 3x - 5x - x) + (11+4)`

`= 5x^4 - 3x^3 + 3x^2 + 15`

`b,`

` A(x) = P(x) + B(x)`

Thay `B(x) = 2x^3 - 3x^4 - 2`

`A(x) = P(x) + B (x)`

`=> A (x) = (2x^3 - 3x^4 - 2)+(3x^4 - 2x^3 + 3x + 11)`

`= 2x^3 - 3x^4 - 2+ 3x^4 - 2x^3 + 3x + 11`

`= (2x^3 - 2x^3) + (-3x^4 + 3x^4) + 3x + (-2+11) `

`= 3x + 9`

`A(x) = 3x+9 = 0`

`=> 3x = 0-9`

`=> 3x = -9`

`=> x = -9 \div 3`

`=> x = -3`

Vậy, nghiệm của đa thức là `x = -3.`

a, M(\(x\) )+N(\(x\)) = 3\(x^4\) - 2\(x\)³ + 5\(x^2\) - \(4x\)+ 1 + ( -3\(x^4\) + 2\(x^3\)- 3\(x^2\)+ 7\(x\) + 5)

M(\(x\)) + N(\(x\)) = ( 3\(x^4\)- 3\(x^4\))+( -2\(x^3\) + 2\(x^3\))+(5\(x^2\) - 3\(x^2\))+( 7\(x-4x\)) +(1+5)

M(\(x\)) + N(\(x\)) = 0 + 0 + 2\(x^2\) + 3\(x\) + 6

M(\(x\)) + N(\(x\)) = 2\(x^2\) + 3\(x\) + 6

b, P(\(x\)) = M(\(x\)) + N(\(x\)) = 2\(x^2\) + 3\(x\) + 6

P(-2) = 2.(-2)² + 3.(-2) + 6 = 8 - 6 + 6 = 8