a, ĐK : \(x\ne\pm3;\frac{1}{2}\)

\(P=\left(\frac{x-1}{x+3}+\frac{2}{x-3}+\frac{x^2+3}{9-x^2}\right):\left(\frac{2x-1}{2x+1}-1\right)\)

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

\(=\frac{x^2-4x+3+2x+6-x^2-3}{\left(x+3\right)\left(x-3\right)}:\left(-\frac{2}{2x+1}\right)\)

\(=\frac{-2x+6}{\left(x+3\right)\left(x-3\right)}.\frac{-\left(2x+1\right)}{2}=\frac{2x+1}{x+3}\)

b, Ta có : \(\left|x+1\right|=\frac{1}{2}\)

TH1 : \(x+1=\frac{1}{2}\Leftrightarrow x=-\frac{1}{2}\)

Thay vào biểu thức A ta được : \(\frac{-1+1}{-\frac{1}{2}+3}=0\)

TH2 : \(x+1=-\frac{1}{2}\Leftrightarrow x=-\frac{3}{2}\)

Thay vào biểu thức A ta được : \(\frac{-3+1}{-\frac{3}{2}+3}=\frac{-2}{\frac{3}{2}}=-\frac{4}{3}\)

c, Ta có : \(P=\frac{x}{2}\Rightarrow\frac{2x+1}{x+3}=\frac{x}{2}\Rightarrow4x+2=x^2+3x\)

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

b, Ta có : \(\frac{2x+1}{x+3}=\frac{2\left(x+3\right)-5}{x+3}=2-\frac{5}{x+3}\)

\(\Rightarrow x+3\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

a) x\(^2\) - 10x + 9 =0

x\(^2\) - 2x . 5 + 25 = 16

(x - 5)\(^2\) = 4\(^2\)

=> x - 5 = 4

x = 9

Vậy x = 9

b) x\(^2\) - 7x + 6 = 0

x\(^2\) - 2x . 3,5 + 12,25 = 6,25

(x - 3,5)\(^2\) = 2,5\(^2\)

=> x - 3,5 = 2,5

x = 6

Vậy x = 6

c) x\(^2\) + 13x + 12 = 0

x\(^2\) + 2x . 6,5 + 42,25 = 30,25

(x + 6,5)\(^2\) = 5,5\(^2\)

=> x + 6,5 = 5,5

x = -1

Vậy x = -1

d) x\(^2\) - 24x + 23 = 0

x\(^2\) - 2x . 12 + 244 = 121

(x - 12)\(^2\) = 11\(^2\)

=> x - 12 = 11

x = 23

Vậy x = 23

e) 3x\(^2\) + 14x + 8 = 0

3x\(^2\) + 2 . \(\sqrt{3}\)x . \(\frac{7}{\sqrt{3}}\) + \(\frac{49}{3}\) = \(\frac{25}{3}\)

(\(\sqrt{3}\)x + \(\frac{7}{\sqrt{3}}\))\(^2\) = \(\left(\frac{5}{\sqrt{3}}\right)^2\)

=> \(\sqrt{3}\)x + \(\frac{7}{\sqrt{3}}\) = \(\frac{5}{\sqrt{3}}\)

=> \(\sqrt{3}\)x  = \(\frac{-2}{\sqrt{3}}\)

=> x = \(\frac{-2}{3}\)

Đặt a - 2b + c = x 

b - 2c + a = y

c - 2a + b = z

Ta có x + y + z = 0

=> x + y = - z

=> (x + y)³ = (-z)³

=> x³ + y³ + 3xy(x + y) = - z³ 

=> x³ + y³ + z³ = -3xy(x + y) 

=> x³ + y³ + z³ = 3xyz

mà x³ + y³ + z³ = 0 

=> 3xyz = 0 

=> xyz = 0

=> x = 0 hoặc y = 0 hoặc z = 0

Khi x = 0 => a - 2b + c = 0 

=> a + c = 2b

=> a + b + c = 3b \(⋮\)3

Khi y = 0 => b - 2c + a = 0

=> a + b = 2c 

=> a + b + c = 3c \(⋮\)3

Khi z = 0 => c - 2a + b = 0

=> b + c = 2a

=> a + b + c = 3a \(⋮\)3

Vậy a + b + c \(⋮3\)

\(a)\)

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

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

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

\(=x^4-1\)

\(b)\)

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

\(=x^5+1^5\)

\(=x^5+1\)

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

\(=\left[\left(2x-y\right)^3-3x.\left(y-2x\right)^2\right]-\left[4x^2+4xy-y^2\right]\)

\(=\left[\left(2x-y\right)^2.\left(2x-y\right)-3x.\left(y-2x\right)^2\right]-\left(2x-y\right)^2\)

\(=\left[\left(y-2x\right)^2.\left(2x-y-3x\right)\right]-\left(2x-y\right)^2\)

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

\(=\left(y-2x\right)^2.\left(-x-y-1\right)\)

Chúc bạn học tốt 

phân tích đa thức

(2x – y)^3 – 3x(y – 2x)2 – 4x2 + 4xy – y ^2  

= -(y-2x)^2(y+x+1)

đây nha bạn 

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

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

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

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

\(=\left(2x-y\right)^2\left(-x-y-1\right)\)

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

\(\left(4x^2-12x-x+3\right)\left(12x^2-4x+6x-2\right)\)

\(\left(4x^2-13x+3\right)\left(12x^2+2x-2\right)\)

\(48x^4-13x^3+36x^2-156x^3-26x^2+26x-8x^2+26x-6\)

\(48x^4-169x^3+2x^2+52x-6\)

\(x^3y-2x^2y^2+5xy=xy\left(x^2-2xy+5\right)\)