Rồi, nghiệm để làm gì? 

Bài yêu cầu rút gọn và sắp xếp lại phải không bạn?

\(A\left(x\right)=3x^4+10x^2+9\)

\(B\left(x\right)=x^4-5x^2-8\)

a) Ta có: \(B\left(x\right)=-2x^3+2x^2+12+5x^2-9x\)

\(=-2x^3+7x^2-9x+12\)

b) Ta có: A(x)+B(x)

\(=4x^3-7x^2+3x-12-2x^3+7x^2-9x+12\)

\(=2x^3-6x\)

b) Ta có: A(x)-B(x)

\(=4x^3-7x^2+3x-12+2x^3-7x^2+9x-12\)

\(=6x^3-14x^2+12x-24\)

b)

Sửa đề: f(x)=A(x)+B(x)

Ta có: f(x)=A(x)+B(x)

\(=x^5+7x^4-9x^3-2x^2-\dfrac{1}{4}x-x^5+5x^4-2x^3+4x^2-\dfrac{1}{4}\)

\(=12x^4-11x^3+2x^2-\dfrac{1}{4}x-\dfrac{1}{4}\)

a) Ta có: \(A\left(x\right)=x^5-3x^2+7x^4-9x^3+x^2-\dfrac{1}{4}x\)

\(=x^5+7x^4-9x^3+\left(-3x^2+x^2\right)-\dfrac{1}{4}x\)

\(=x^5+7x^4-9x^3-2x^2-\dfrac{1}{4}x\)

Ta có: \(B\left(x\right)=5x^4-x^5+x^2-2x^3+3x^2-\dfrac{1}{4}\)

\(=-x^5+5x^4-2x^3+\left(x^2+3x^2\right)-\dfrac{1}{4}\)

\(=-x^5+5x^4-2x^3+4x^2-\dfrac{1}{4}\)

b) Ta có: G(x)=A(x)-B(x)

\(=x^5+7x^4-9x^3-2x^2-\dfrac{1}{4}x+x^5-5x^4+2x^3-4x^2+\dfrac{1}{4}\)

\(=2x^5+2x^4-7x^3-6x^2-\dfrac{1}{4}x+\dfrac{1}{4}\)

`a)`

`@A(x)=5x^2+2x^3+8-7x`

         `=2x^3+5x^2-7x+8`

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

         `=4x^3+3x^2-2x-1`

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`b)A(-1)=2.(-1)^3+5.(-1)^2-7.(-1)+8`

          `=2.(-1)+5.1+7+8`

          `=-2+5+7+8=18`

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`c)A(x)=B(x)+C(x)`

`=>C(x)=A(x)-B(x)`

`=>C(x)=(2x^3+5x^2-7x+8)-(4x^3+3x^2-2x-1)`

`=>C(x)=2x^3+5x^2-7x+8-4x^3-3x^2+2x+1`

`=>C(x)=-2x^3+2x^2-5x+9`

a)\(A\left(x\right)=2x^3+5x^2-7x+8\)

\(B\left(x\right)=4x^2+3x^2-2x-1\)

b)\(A\left(-1\right)=2.\left(-1\right)^3+5.\left(-1\right)^2-7.\left(-1\right)+8\)

\(A\left(-1\right)=-2+5+7+8=18\)

c)\(A\left(x\right)=B\left(x\right)+C\left(x\right)\)

\(=>C\left(x\right)=A\left(x\right)-B\left(x\right)\)

\(C\left(x\right)=2x^3+5x^2-7x+8-4x^2-3x^2+2x+1\)

\(C\left(x\right)=-x^3+x^2-5x+9\)

$a/$

Sắp xếp theo lũy thừa giảm dần của biến 

`A(x) = 5x^2 + 2x^3 + 8 - 7x = 2x^3 + 5x^2 - 7x+8`

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

$b/$

`A(-1) = 2*(-1)^3 + 5*(-1)^2 - 7*(-1)+8=18`

$c/$

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

`<=> A(x) -B(x) = C(x)`

                        `= ( 2x^3 + 5x^2 - 7x+8 )-(4x^3 + 3x^2 - 2x-1)`

                        `= ( 2x^3 - 4x^3 ) + ( 5x^2 - 3x^2 )+(2x-7x)+(8+1)`

                        `= 2x^2 - 2x^3 - 5x +9` 

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) Ta có: \(M\left(x\right)=3x^3+x^2+4x^4-x-3x^3+5x^4+2x^2-6\)

\(=\left(4x^4+5x^4\right)+\left(3x^3-3x^3\right)+\left(x^2+2x^2\right)-x-6\)

\(=9x^4+3x^2-x-6\)

Ta có: \(N\left(x\right)=-2x^2-x^4+4x^3-x^2-5x^3+3x+5+x\)

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

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

c) Ta có: M(x)+N(x)

\(=9x^4+3x^2-x-6-x^4-x^3-3x^2+4x+5\)

\(=8x^4-x^3+3x-1\)

a) P (x) =11+5x³+3x²-9x⁶-(6x²+5-9x⁶-4x⁴)

             =11+5x³+3x²-9x⁶-6x²-5+9x⁶+4x⁴

             =4x⁴+5x³-3x²+6                           

Q(x)=(3x⁴-5x²)-4x²+x⁴-4x-1

       =3x⁴-5x²-4x²+x⁴-4x-1

       =4x⁴-9x²-4x-1

b) M(x) = 4x⁴+5x³-3x²+6 + 4x⁴-9x²-4x-1

            = 8x⁴+5x³-12x²-4x+5

N(x)= 4x⁴+5x³-3x²+6 - 4x⁴+9x²+4x+1

       = 5x³+6x²+4x+7

\(a,\)Thu gọn và sắp xếp :

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

\(Q\left(x\right)=-7x^5-x^2+8x-5\)

\(b,Q\left(x\right)-P\left(x\right)=-7x^5-x^2+8x-5-x^5+3x^2-4x+5\)

                          \(=-8x^5+2x^2+4x\)