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

a) 

Ta thay x=-1 vào P(x) và kiểm tra giá trị thu đc :

P(-1) = 3(-1)⁴ - 4(-1)² + 2(-1) - 5

= 3(1) - 4(1) - 2 - 5

= 3 - 4 - 2 - 5

= -8

Vì giá trị P(-1) khác 0, nên x=-1 ko phải là nghiệm của P(x)

b) Q(x) = x² + x⁶ - (1 - 3x⁴ + 2x + x⁶)

= x² + x⁶ - 1 + 3x⁴ - 2x - x⁶

= -1 + 3x⁴ + x² - 2x⁶

Thu gọn : Q(x) = -2x⁶ + 3x⁴ + x² - 1

c) 

+ Tính P(x)+Q(x):

P(x) + Q(x)

= (3x⁴ - 4x² + 2x - 5) + (-2x⁶ + 3x⁴ + x² - 1)

= -2x⁶ + (3x⁴ + 3x⁴) + (-4x² + x²) + (2x - 5 - 1)

= -2x⁶ + 6x⁴ - 3x² + 2x - 6

+ Tính Q(x)-P(x):

Q(x) - P(x)

= (-2x⁶ + 3x⁴ + x² - 1) - (3x⁴ - 4x² + 2x - 5)

= -2x⁶ + (3x⁴ - 3x⁴) + (x² + 4x²) + (2x - 2x) + (-1 + 5)

= -2x⁶ + 5x² + 4

Thay x=-1 vào P ta được:

P(-1)= 3(-1)⁴-4(-1)²+2(-1)-5 =  - 8 

Vậy x = -10 phải là nghiệm của P(x)

b) Q(x)= x²+x⁶-(1-3x⁴+2x+x⁶)

Q(x)= x²+x⁶-1+3x⁴-2x-x⁶= 3x⁴+x²-2x-1

P(x) + Q(x)= ( x^5 - 2x^2 + 7x^4 - 9x^3 - 1/4x) + ( 5x^4 - x^5 + 4x^2 - 2x^3 - 1/4)

                 = x^5 - 2x^2 + 7x^4 - 9x^3 - 1/4x + 5x^4 - x^5 + 4x^2 - 2x^3 - 1/4

                 = ( x^5 - x^5 ) - ( 2x^2 + 4x^2) + ( 7x^4 + 5x^4) - ( 9x^3 - 2x^3) - 1/4x - 1/4

                 =                            6x^2 + 12x^4 - 6x^3 - 1/4x - 1/4

P(x) - Q(x)=  ( x^5 - 2x^2 + 7x^4 - 9x^3 -1/4x) - ( 5x^4 - x^5 + 4x^2 - 2x^3 -1/4)

                 = x^5 - 2x^2 + 7x^4 - 9x^3 - 1/4x - 5x^4 + x^5 - 4x^2 + 2x^3 + 1/4

                 = ( x^5 + x^5) - ( 2x^2 - 4x^2) + ( 7x^4 - 5x^4) - ( 9x^3 + 2x^3) - 1/4x + 1/4

                 = 2x^5 - (-2)x^2 + 2x^4 - 11x^3 - 1/4x + 1/4

P(x)=x^5+  7x^4- 9x^3+ 2x^2-1/4x-0

Q(x)=(-x^5+5x^4- 2x^3+ 4x^2+0x-1/4

      =      12x^4-11x^3+ 6x^2-1/4x-1/4

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) = x^5 - 2x^2 + 7x^4 - 9x^3 - 1/4x

=x⁵+7x⁴-9x³-2x²-1/4x

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

=-x⁵+5x⁴-2x³+4x²-1/4

P(x)+Q(x)=x⁵+7x⁴-9x³-2x²-1/4x -x⁵+5x⁴-2x³+4x²-1/4

=x⁵-x⁵+7x⁴+5x⁴-9x³-2x³-2x²+4x²-1/4x-1/4

=12x⁴-11x³+2x²-1/4x-1/4

P(x)-Q(x)=x⁵+7x⁴-9x³-2x²-1/4x +x⁵-5x⁴+2x³-4x²+1/4

=x⁵+x⁵+7x⁴-5x⁴-9x³+2x³-2x²-4x²-1/4x-1/4

=2x⁵+2x⁴-7x³-6x²-1/4x-1/4

\(P\left(-1\right)=\left(-1\right)^4+2\cdot\left(-1\right)^2+1=1+2+1=4\)

\(P\left(\dfrac{1}{2}\right)=\left(\dfrac{1}{2}\right)^4+2\cdot\left(\dfrac{1}{2}\right)^2+1=\dfrac{1}{16}+\dfrac{1}{2}+1=\dfrac{9}{16}\)

\(Q\left(-2\right)=\left(-2\right)^4+4\cdot\left(-2\right)^3+2\cdot\left(-2\right)^2-4\cdot\left(-2\right)+1=16-32+8+8+1=1\)

\(Q\left(1\right)=1^4+4\cdot1^3+2\cdot1^2-4\cdot1+1=1+4+2-4+1=5+3-4=2-4=-2\)

a) M(x) = A(x) - 2B(x) + C(x)

\(\Leftrightarrow\)M(x) = 2x⁵ - 4x³ + x² - 2x + 2 - 2(x⁵ - 2x⁴ + x² - 5x + 3) + x⁴ + 4x³ + 3x² - 8x + \(4\frac{3}{16}\)

\(\Leftrightarrow\)M(x) = 2x⁵ - 4x³ + x² - 2x + 2 - 2x⁵ - 4x⁴ - 2x² + 10x - 6 + x⁴ + 4x³ + 3x² - 8x + \(4\frac{3}{16}\)

\(\Leftrightarrow\)M(x) = (2x⁵ - 2x⁵) + (-4x³ + 4x³) + (x² - 2x² + 3x²) + (-2x + 10x - 8x) + (2 - 6 + \(4\frac{3}{16}\))

\(\Leftrightarrow\)M(x) = 2x² + \(\frac{3}{16}\)

b) Thay \(x=-\sqrt{0,25}\)vào M(x), ta được:

\(M\left(x\right)=2\left(-\sqrt{0,25}\right)^2+\frac{3}{16}\)

\(M\left(x\right)=2.0,25+\frac{3}{16}\)

\(M\left(x\right)=0,5+\frac{3}{16}\)

\(M\left(x\right)=\frac{11}{16}\)

c) Ta có : \(x^2\ge0\)

\(\Leftrightarrow2x^2+\frac{3}{16}\ge\frac{3}{16}\)

Vậy để \(M\left(x\right)=0\Leftrightarrow x\in\varnothing\)