a: \(\dfrac{2x^4-x^3-x^2+7x-4}{x^2+x-1}\)

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

=2x^2-3x+4

b: \(=\dfrac{y}{x\left(2x-y\right)}+\dfrac{4x}{y\left(y-2x\right)}\)

\(=\dfrac{y^2-4x^2}{xy\left(2x-y\right)}=\dfrac{-\left(2x-y\right)\left(2x+y\right)}{xy\left(2x-y\right)}=\dfrac{-2x-y}{xy}\)

c: \(=\dfrac{6\left(x+8\right)}{7\left(x-1\right)}\cdot\dfrac{\left(x-1\right)^2}{\left(x-8\right)\left(x+8\right)}=\dfrac{6\left(x-1\right)}{7\left(x-8\right)}\)

\(\frac{4}{x+2}+\frac{3}{x-2}+\frac{-5x-2}{x^2-4}\)ĐK : \(x\ne\pm2\)

\(=\frac{4\left(x-2\right)+3\left(x+2\right)-5x-2}{\left(x+2\right)\left(x-2\right)}=\frac{4x-8+3x+6-5x-2}{\left(x+2\right)\left(x-2\right)}\)

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

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

\(=\dfrac{x\left(x+1\right)-2.5}{5x}=\dfrac{x^2+x-10}{5x}\)

\(b,\dfrac{x+y}{9x}:\dfrac{x+y}{3x}\)

\(=\dfrac{x+y}{9x}.\dfrac{3x}{x+y}=\dfrac{1}{3}\)

a. \(\dfrac{x+1}{5}\)-\(\dfrac{2}{x}\)=\(\dfrac{x\left(x+1\right)-2.5}{5x}\)=\(\dfrac{x^2+x-10}{5x}\)

b. \(\dfrac{x+y}{9x}:\dfrac{x+y}{3x}\)=\(\dfrac{x+y}{9x}.\dfrac{3x}{x+y}=\dfrac{1}{3}\)

c: \(\dfrac{3x+5}{x^2-5x}+\dfrac{25-x}{25-5x}\)

\(=\dfrac{3x+5}{x\left(x-5\right)}+\dfrac{x-25}{5\left(x-5\right)}\)

\(=\dfrac{15x+25+x^2-25x}{5x\left(x-5\right)}=\dfrac{x^2-10x+25}{5x\left(x-5\right)}=\dfrac{x-5}{5x}\)

e: \(\dfrac{4x^2-3x+17}{x^3-1}+\dfrac{2x-1}{x^2+x+1}+\dfrac{6}{1-x}\)

\(=\dfrac{4x^2-3x+17+\left(2x-1\right)\left(x-1\right)-6x^2-6x-6}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\dfrac{-2x^2-9x+11+2x^2-3x+1}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\dfrac{-12\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{-12}{x^2+x+1}\)

1) \(\left(x^3-8\right):\left(x-2\right)=\left[\left(x-2\right)\left(x^2+2x+4\right)\right]:\left(x-2\right)=x^2+2x+4\)

2) \(\left(x^3-1\right):\left(x^2+x+1\right)=\left[\left(x-1\right)\left(x^2+x+1\right)\right]:\left(x^2+x+1\right)=x-1\)

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

4) \(\left(25x^2-4y^2\right):\left(5x-2y\right)=\left[\left(5x-2y\right)\left(5x+2y\right)\right]:\left(5x-2y\right)=5x+2y\)

a) Ta có: \(\left(5x-2y\right)\left(x^2-xy+1\right)\)

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

\(=5x^3-7x^2y+2xy^2+5x-2y\)

b) Ta có: \(\left(x-1\right)\left(x+1\right)\left(x+2\right)\)

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

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

c) Ta có: \(\dfrac{1}{2}x^2y^2\cdot\left(2x+y\right)\left(2x-y\right)\)

\(=\dfrac{1}{2}x^2y^2\left(4x^2-y^2\right)\)

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

1) Ta có: \(x^2-4xy+4y^2\)

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

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

Phép tính trở thành: \(\left(x-2y\right)^2:\left(x-2y\right)=x-2y\)

2) Ta có: \(25x^2+2xy+\dfrac{1}{25}y^2\)

\(=\left(5x\right)^2+2.5x.\dfrac{1}{5}y+\left(\dfrac{1}{5}y\right)^2\)

\(=\left(5x+\dfrac{1}{5}y\right)^2\)

Phép tính trở thành: \(\left(5x+\dfrac{1}{5}y\right)^2:\left(5x+\dfrac{1}{5}y\right)=5x+\dfrac{1}{5}y\)

1) (x² - 4xy + 4y²) : (x - 2y)

= (x - 2y)² : (x - 2y)

= x - 2y

2) (25x² + 2xy + 1/25 y²) : (5x + 1/5 y)

= 5x + 1/5 y)² : (5x + 1/5 y)

= 5x + 1/5 y

Giúp vs ạ

\(A=-3\left(3+1\right)+5\left(1-1\right)+8\left(-1+1-2\right)\)

\(A=-28\)

A=9x-3x²-5x²-5x+8x²-8x-16

   =-4x-16

   =-4(x-4)

Thay x=-1 vào A, ta được:

A(-1)=-4(-1-4)=(-4).(-5)=20

Vậy A(-1)=20