Bài 1.

a)\(\frac{4x-4}{x^2-4x+4}\div\frac{x^2-1}{\left(2-x\right)^2}=\frac{4\left(x-1\right)}{\left(x-2\right)^2}\div\frac{\left(x-1\right)\left(x+1\right)}{\left(x-2\right)^2}=\frac{4\left(x-1\right)}{\left(x-2\right)^2}\times\frac{\left(x-2\right)^2}{\left(x-1\right)\left(x+1\right)}=\frac{4}{x+1}\)

b) \(\frac{2x+1}{2x^2-x}+\frac{32x^2}{1-4x^2}+\frac{1-2x}{2x^2+x}=\frac{2x+1}{x\left(2x-1\right)}+\frac{-32x^2}{4x^2-1}+\frac{1-2x}{x\left(2x+1\right)}\)

\(=\frac{\left(2x+1\right)\left(2x+1\right)}{x\left(2x-1\right)\left(2x+1\right)}+\frac{-32x^3}{x\left(2x-1\right)\left(2x+1\right)}+\frac{\left(1-2x\right)\left(2x-1\right)}{x\left(2x-1\right)\left(2x+1\right)}\)

\(=\frac{4x^2+4x+1}{x\left(2x-1\right)\left(2x+1\right)}+\frac{-32x^3}{x\left(2x-1\right)\left(2x+1\right)}+\frac{-4x^2+4x-1}{x\left(2x-1\right)\left(2x+1\right)}\)

\(=\frac{4x^2+4x+1-32x^3-4x^2+4x-1}{x\left(2x-1\right)\left(2x+1\right)}=\frac{-32x^3+8x}{x\left(2x-1\right)\left(2x+1\right)}\)

\(=\frac{-8x\left(4x^2-1\right)}{x\left(2x-1\right)\left(2x+1\right)}=\frac{-8x\left(2x-1\right)\left(2x+1\right)}{x\left(2x-1\right)\left(2x+1\right)}=-8\)

c) \(\left(\frac{1}{x+1}+\frac{1}{x-1}-\frac{2x}{1-x^2}\right)\times\frac{x-1}{4x}\)

\(=\left(\frac{1}{x+1}+\frac{1}{x-1}+\frac{2x}{x^2-1}\right)\times\frac{x-1}{4x}\)

\(=\left(\frac{x-1}{\left(x-1\right)\left(x+1\right)}+\frac{x+1}{\left(x-1\right)\left(x+1\right)}+\frac{2x}{\left(x-1\right)\left(x+1\right)}\right)\times\frac{x-1}{4x}\)

\(=\left(\frac{x-1+x+1+2x}{\left(x-1\right)\left(x+1\right)}\right)\times\frac{x-1}{4x}\)

\(=\frac{4x}{\left(x-1\right)\left(x+1\right)}\times\frac{x-1}{4x}=\frac{1}{x+1}\)

Bài 3.

N = ( 4x + 3 )² - 2x( x + 6 ) - 5( x - 2 )( x + 2 )

= 16x² + 24x + 9 - 2x² - 12x - 5( x² - 4 )

= 14x² + 12x + 9 - 5x² + 20

= 9x² + 12x + 29

= 9( x² + 4/3x + 4/9 ) + 25

= 9( x + 2/3 )² + 25 ≥ 25 > 0 ∀ x 

=> đpcm

C1: Gọi đa thức thương là Q(x)

Vì x^4 : x^2 = x^2

=> đa thức có dạng x^2+mx+n

Đề x^4 - 3x^2 + ax+b chia hết x^2 - 3x + 2

=> x^4 - 3x^2 + ax + b = (x^2 - 3x + 2)(x^2 + mx + n)

x^4+ 0x^3 - 3x^2 +ax+b  = x^4 +mx^3 +(x^2)n -3x^3 -3mx^2 - 3xn + 2x^2 + 2mx + 2n

x^4 + 0x^3 -3x^2 + ax+b = x^4 + x^3(m-3) - x^2(3m - n -2) +x(2m - 3n) +2n

<=>| 0 = m-3                     <=> | m = 3

| 3=3m-n-2                                | b= 8

| a=2m-3n                                 | n = 4

| b = 2n                                     | a = -6

Vậy a= -6, b= 8

\(1,\left(x-3\right)\left(x-1\right)-3\left(x-3\right)\)

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

\(=\left(x-3\right)\left(x-4\right)\)

\(2,6x+3-\left(2x-5\right)\left(2x+1\right)\)

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

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

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

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

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

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

\(=\left(3x-2\right)\left(3x-6\right)\)

\(5,\left(x-5\right)^2+\left(x+5\right)\left(x-5\right)-\left(5-x\right)\left(2x+1\right)\)\(=\left(x-5\right)^2+\left(x+5\right)\left(x-5\right)+\left(x-5\right)\left(2x+1\right)\)\(=\left(x-5\right)\left(x-5+x+5+2x+1\right)\)

\(=\left(x-5\right)\left(4x+1\right)\)

6, Tương tự

a: \(=\dfrac{x^4-6x^3+12x^2-14x+3}{x^2-4x+1}\)

\(=\dfrac{x^4-4x^3+x^2-2x^3+8x^2-2x+3x^2-12x+3}{x^2-4x+1}\)

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

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

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

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

\(=2x^2-3x+1\)

b)\(\frac{9x^4-6x^3+15x^2+2x+1}{3x^2-2x+5}=\frac{3x^2.\left(3x^2-2x+5\right)+2x+1}{3x^2-2x+5}=3x^2+\frac{2x+1}{3x^2-2x+5}\)

=> đa thức dư trong phép chia là 2x+1

\(\frac{x^3+2x^2-3x+9}{x+3}=\frac{x^3+9x^2+27x+27-7x^2-30x-18}{x+3}=\frac{\left(x+3\right)^3-7x^2-30x-18}{x+3}\)

\(\left(x+3\right)^2-\frac{7x^2+21x+9x+18}{x+3}=\left(x+3\right)^2-\frac{7x.\left(x+3\right)+9.\left(x+3\right)-9}{x+3}\)

\(=\left(x+3\right)^2-\frac{\left(7x+9\right).\left(x+3\right)-9}{x+3}=\left(x+3\right)^2-\left(7x+9\right)-\frac{9}{x+3}\)

=> đa thức dư trong phép chia là 9

p/s: t mới lớp 7_sai sót mong bỏ qua :>

em chưa cho đa thức f(x) và g(x) nà

a: \(\dfrac{f\left(x\right)}{g\left(x\right)}\)

\(=\dfrac{x^4-9x^3+21x^2+ax+b}{x^2-x-1}\)

\(=\dfrac{x^4-x^3-x^2-8x^3+8x^2+8x+14x^2-14x-14+\left(a+6\right)x+b+14}{x^2-x-1}\)

\(=x^2-8x+14+\dfrac{\left(a+6\right)x+b+14}{x^2-x-1}\)

Để f(x) chia hết cho g(x) thì a+6=0 và b+14=0

=>a=-6 và b=-14

b: \(\dfrac{f\left(x\right)}{g\left(x\right)}=\dfrac{x^4-x^3+5x^2+x^2-x+5+a-5}{x^2-x+5}\)

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

Để f(x) chia hết g(x) thì a-5=0

=>a=5

Lời giải:

a.

$A+B=(5x^2-7x+2)+(4x^2+3x-1)=9x^2-4x+1$
$A-B=(5x^2-7x+2)-(4x^2+3x-1)=x^2-10x+3$

b. 

$A(x)=2x^2-x+m=x(2x-5)+4x+m=x(2x-5)+2(2x-5)+m+10$

$=B(x)(x+2)+m+10$

Để $A(x)\vdots B(x)$ thì $m+10=0\Leftrightarrow m=-10$