Question

[Thảo Đỗ Phương]

mọi người giúp mìn với!!!

Answer 1

[『Kuroba ム Tsuki Ryoo ︵²ᵏ⁷』]

`a,`

\(\text {P(x)+Q(x)=}\)\(\left(x^3-\dfrac{1}{2}x^3+8x-1\right)+\left(-x^3+\dfrac{3}{2}x^2-10x+4\right)\)

`= x^3-1/2x^3+8x-1-x^3+3/2x^2-10x+4`

`= (x^3-x^3-1/2x^3)+3/2x^2+(8x-10x)+(-1+4)`

`= -1/2x^3+3/2x^2-2x+3`

`b,`

\(\text {P(x)-Q(x)=}\)\(\left(x^3-\dfrac{1}{2}x^3+8x-1\right)-\left(-x^3+\dfrac{3}{2}x^2-10x+4\right)\)

`= x^3-1/2x^3+8x-1+x^3-3/2x^2+10x-4`

`= (x^3+x^3-1/2x^3)-3/2x^2+(8x+10x)+(-1-4)`

`= 3/2x^3-3/2x^2+18x-5`

`c,`

`\text {M(x)-P(x)=Q(x)}`

`->` `\text {M(x)=P(x)+Q(x)}`

`M(x)=-1/2x^3+3/2x^2-2x+3=0`

\(\text{Vì }\dfrac{3}{2}x^2\ge0\text{ }\)\(\forall\)\(\text{x}\)

`->`\(-\dfrac{1}{2}x^3+\dfrac{3}{2}x^2-2x+3>0\text{ }\forall\text{ }x\)

`-> \text {Đa thức vô nghiệm.}`

`@`\(\text {Dnammv}\)

Answer 2

[Thảo Đỗ Phương]

Cảm ơn bạn nhìu nha !!!