a \(f\left(x\right)-h\left(x\right)=g\left(x\right)\)

\(h\left(x\right)=f\left(x\right)-g\left(x\right)\)

\(h\left(x\right)=\left(2x^4+5x^3-x+8\right)-\left(x^4-x^2-3x+9\right)\)

\(h\left(x\right)=2x^4+5x^3-x+8-x^4+x^2+3x-9\)

\(h\left(x\right)=3x^4+5x^3+x^2+2x-1\)

b \(h\left(x\right)-g\left(x\right)=f\left(x\right)\)

\(h\left(x\right)=f\left(x\right)+g\left(x\right)\)

\(h\left(x\right)=2x^4+5x^3-x+8+x^4-x^2-3x+9\)

\(h\left(x\right)=3x^4+5x^3-x^2-4x+17\)

\(x^4+2x^3+x^2=x^2\left(x^2+2x+1\right)=x^2\left(x+1\right)^2\\ x^2-6x=x\left(x-6\right)\\ x^2-2xy-z^2+y^2=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\)

Bài 2:

a) \(3x^2-7x-10=\left(x+1\right)\left(3x-10\right)\)

b) \(x^2+6x+9-4y^2=\left(x+3\right)^2-\left(2y\right)^2=\left(x+3-2y\right)\left(x+3+2y\right)\)

c) \(x^2-2xy+y^2-5x+5y=\left(x-y\right)^2-5\left(x-y\right)=\left(x-y\right)\left(x-y-5\right)\)

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

e) \(1-2a+2bc+a^2-b^2-c^2=\left(a-1\right)^2-\left(b-c\right)^2=\left(a-1-b+c\right)\left(a-1+b-c\right)\)

f) \(x^3-3x^2-4x+12=\left(x+2\right)\left(x-3\right)\left(x-2\right)\)

g) \(x^4+64=\left(x^2+8\right)^2-16x^2=\left(x^2+8-4x\right)\left(x^2+6+4x\right)\)h) \(x^4-5x^2+4=\left(x+2\right)\left(x+1\right)\left(x-1\right)\left(x-2\right)\)

i) \(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+16=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+16=\left(x^2+8x+7\right)^2+8\left(x^2+8x+7\right)+16=\left(x^2+8x+11\right)^2\)

a: \(3x^2-7x-10\)

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

\(=\left(x+1\right)\left(3x-10\right)\)

b: \(x^2+6x+9-4y^2\)

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

\(=\left(x+3-2y\right)\left(x+3+2y\right)\)

c: \(x^2-2xy+y^2-5x+5y\)

\(=\left(x-y\right)^2-5\left(x-y\right)\)

\(=\left(x-y\right)\left(x-y-5\right)\)

\(\left(x^2+6x-1\right)^2+2x^2+x^4+2\left(x^2+6x-1\right)\left(x^2+1\right)\)

\(\left(x^2+6x-1\right)^2+2\left(x^2+6x-1\right)\left(x^2+1\right)+\left(x^2+1\right)^2-1=\left(x^2+6x-1+x^2+1\right)^2-1=\left(2x^2+6x\right)^2-1=\left(2x^2+6x-1\right)\left(2x^2+6x+1\right)\)

\(\left(x^2+6x-1\right)^2+2\left(x^2+6x-1\right)\left(x^2+1\right)+x^4+2x^2\)

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

\(=\left(x^2+6x-1\right)\left(3x^2+6x+1\right)+x^4+2x^2\)

\(=\left(2x^2+6x-1\right)\left(2x^2+6x+1\right)\)

Lời giải:
a.

$x^3+y^3=(x+y)^3-3xy(x+y)=9^3-3.9.18=243$

$x^4+y^4=(x^2+y^2)^2-2x^2y^2=[(x+y)^2-2xy]^2-2x^2y^2$

$=[9^2-2.18]^2-2.18^2=1377$

Nếu $x\geq y$ thì:

$x^3-y^3=(x-y)(x^2+xy+y^2)$

$=|x-y|[(x+y)^2-xy]=\sqrt{(x+y)^2-4xy}[(x+y)^2-xy]$

$=\sqrt{9^2-4.18}(9^2-18)=189$

Nếu $x< y$ thì $x^3-y^3=-189$

b.

$A=(x+y)^2-6(x+y)+y-5$

$=(-9)^2-6(-9)+y-5=130+y$

Chưa đủ cơ sở để tính biểu thức.

cảm ơn bn

 Ta thấy \(B=\left(x-1\right)\left(x-5\right)\) nên để đa thức A chia hết cho đa thức B thì \(A=\left(x-1\right)\left(x-5\right).C\) với \(C\) là một đa thức bậc 2 hệ số nguyên theo \(x\).

 Điều này tương đương với việc \(A\) có 2 nghiệm là \(x=1,x=5\). Do đó \(A\left(1\right)=0\) \(\Leftrightarrow1^4-7.1^3+10.1^2+\left(a-1\right)+b-a=0\) \(\Leftrightarrow b=-3\)

 Ta viết lại \(A=x^4-7x^3+10x^2+\left(a-1\right)x-3-a\). Ta có \(A\left(5\right)=0\) \(\Leftrightarrow5^4-7.5^3+10.5^2+\left(a-1\right).5-3-a=0\) \(\Leftrightarrow4a-8=0\) \(\Leftrightarrow a=2\).

 Vậy để đa thức A chia hết cho đa thức B thì \(a=2,b=-3\).

A:B=x2-x+11 dư (a+70)x+b-a-55

Để A chia hết cho B thì

(a+70)x+b-a-55=0

b-a-55=0 (a khác -70) tại x=0

Vậy b-a=55 thỏa đề bài

\(1,=x^6+27\\ 2,=8x^3+1\\ 3,=x^6+8\\ 4,=27x^3+8\)

1. (x² + 3)(x⁴ - 3x² + 9)

= x⁶ + 27

2. (2x + 1)(4x² - 2x + 1)

= 8x³ + 1

3. (x² + 2)(x⁴ - 2x² + 4)

= x⁶ + 8

4. (3x + 2)(9x² - 6x + 4)

= 27x³ + 8

\(x^4+x^2+6x-8=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2-x+4\right)=0\)  (\(*\))

Vì \(x^2-x+4>0\) nên:

(\(*\)) \(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)

Vậy \(S=\left\{1;-2\right\}\).

Chọn C