\(f\left(x\right)⋮g\left(x\right)\)

\(\Leftrightarrow x^4-3x^3+4x^2-x^2+3x-4+\left(a-3\right)x+\left(b+4\right)⋮x^2-3x+4\)

\(\Leftrightarrow\left(a,b\right)=\left(3;-4\right)\)

`@` `\text {Ans}`

`\downarrow`

Gửi c!

Bài 1: 

a) \(3x^2\left(2x^3-x+5\right)-6x^5-3x^3+10x^2\)

\(=6x^5-3x^3+10x^2-6x^5-3x^3+10x^2\)

\(=10x^2+10x^2\)

\(=20x^2\)

b) \(-2x\left(x^3-3x^2-x+11\right)-2x^4+3x^3+2x^2-22x\)

\(=-2x^4+6x^3+2x^2-22x-2x^4+3x^3+2x^2-22x\)

\(=-4x^4+9x^3+4x^2-44x\)

Alo đề nghị viết đề một cách chính xác 

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

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

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

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

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

\(\Rightarrow P\left(x\right)+Q\left(x\right)=2x^4+4x^3+3x^2+6x+5\)

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

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

\(\Rightarrow P\left(x\right)-Q\left(x\right)=2x^3-x^2-2x+1\)

\(1,A⋮B\Leftrightarrow x^3-3x^2-ax+3=\left(x-1\right)\cdot a\left(x\right)\)

Thay \(x=1\)

\(\Leftrightarrow1-3-a+3=0\\ \Leftrightarrow a=1\)

\(2,A⋮B\Leftrightarrow3x^3-16x^2+25x+a=\left(x^2-4x+3\right)\cdot b\left(x\right)\\ \Leftrightarrow3x^3-16x^2+25x+a=\left(x-3\right)\left(x-1\right)\cdot b\left(x\right)\)

Thay \(x=1\)

\(\Leftrightarrow3-16+25+a=0\\ \Leftrightarrow a=-12\)

Thay \(x=3\)

\(\Leftrightarrow3\cdot27-16\cdot9+25\cdot3+a=0\\ \Leftrightarrow81-144+75+a=0\\ \Leftrightarrow12+a=0\Leftrightarrow a=-12\)

Vậy \(a=-12\)

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

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

`=9x^4+3x^3-x^2-x-5`

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

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

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

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

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

`=8x^4+2x^3-x^2-5x-6`

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

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

`=10x^4+4x^3-x^2-5x-4`

Ta có 

Phần dư của phép chia f(x) cho g(x) là R = (a – 3)x + b + 4. Để phép chia trên là phép chia hết thì R = 0, Ɐx

ó (a – 3)x + b + 4 = 0, Ɐx ó   a - 3 = 0 b + 4 = 0

ó a = 3 b = - 4 => ab = -12

Đáp án cần chọn là: A

a: \(\Leftrightarrow2x^2+8x+\left(a-8\right)x+4\left(a-8\right)-4a+28⋮x+4\)

hay a=7

Ta có : \(x^4-3x^3+4x^2-3x+10.\)

\(=\left(x^4-2x^3+x^2\right)-\left(x^3-3x^2+3x-1\right)+9\)

\(=x^2\left(x-1\right)^2-\left(x-1\right)^3+9\)

\(=\left(x-1\right)^2\left(x^2-x+1\right)+9\)

Mà \(\left(x-1\right)^2\ge0\)

\(x^2-x+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\)

\(\Rightarrow\left(x-1\right)^2\left(x^2-x+1\right)\ge0\)

\(\Rightarrow\left(x-1\right)^2\left(x^2-x+1\right)+9\ge9\)

Dấu " = " xảy ra \(\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\)

Vậy GTNN cảu \(x^4-3x^3+4x^2-3x+10.\)là 9 <=> \(x=1\)