a, \(x^2-\frac{x^4}{x^2}-1=x^2-x^2-1=-1\)

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

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

\(=\frac{x^2+6x+9}{x\left(x-3\right)\left(x+3\right)}=\frac{\left(x+3\right)^2}{x\left(x-3\right)\left(x+3\right)}=\frac{x+3}{x\left(x-3\right)}\)

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

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

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

d, viết lại đề đy nhé 

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

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

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

Bài làm

Ta có :\(\frac{x^3+x^2-x-1}{x^3+2x-3}\)( ĐKXĐ : x ≠ 1 )

Để phân thức = 0 thì x³ + x² - x - 1 = 0

<=> x²( x + 1 ) - ( x + 1 ) = 0

<=> ( x + 1 )( x² - 1 ) = 0

<=> ( x + 1 )( x - 1 )( x + 1 ) = 0

<=> ( x + 1 )²( x - 1 ) = 0

<=> \(\orbr{\begin{cases}\left(x+1\right)^2=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\left(nhan\right)\\x=1\left(loai\right)\end{cases}}\)

Vậy x = -1 thì phân thức = 0

x^3+x^2-x-1/x^3+2x-3 = 0
<=> x^3+x^2-x-1=0
<=>x^2(x+1)-(x+1)=0
<=>(x^2-1)(x+1)=0
<=>x={1;-1}

c) ( x² + x + 1 )( x² + x + 2 ) - 12

Đặt t = x² + x + 1

<=> t( t + 1 ) - 12

= t² + t - 12

= t² - 3t + 4t - 12

= t( t - 3 ) + 4( t - 3 )

= ( t - 3 )( t + 4 )

= ( x² + x + 1 - 3 )( x² + x + 1 + 4 )

= ( x² + x - 2 )( x² + x + 5 )

= ( x² - x + 2x - 2 )( x² + x + 5 )

= [ x( x - 1 ) + 2( x - 1 ) ]( x² + x + 5 )

= ( x - 1 )( x + 2 )( x² + x + 5 )

d) ( x + 2 )( x + 3 )( x + 4 )( x + 5 ) - 24

= [ ( x + 2 )( x + 5 ) ][ ( x + 3 )( x + 4 ) ] - 24

= ( x² + 7x + 10 )( x² + 7x + 12 ) - 24

Đặt t = x² + 7x + 10

<=> t( t + 2 ) - 24

= t² + 2t - 24

= t² - 4t + 6t - 24

= t( t - 4 ) + 6( t - 4 )

= ( t - 4 )( t + 6 )

= ( x² + 7x + 10 - 4 )( x² + 7x + 10 + 6 )

= ( x² + 7x + 6 )( x² + 7x + 16 )

= ( x² + 6x + x + 6 )( x² + 7x + 16 )

= [ x( x + 6 ) + ( x + 6 ) ]( x² + 7x + 16 )

= ( x + 6 )( x + 1 )( x² + 7x + 16 )

a, Sửa đề:\(\left(x^2+x\right)^2-2\left(x^2+x\right)-15\)

Đặt \(t=x^2+x\)

\(\Rightarrow t^2-2t-15\)

\(=t^2-5t+3t-15\)

\(=t\left(t-5\right)+3\left(t-5\right)\)

\(=\left(t+3\right)\left(t-5\right)\)

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

b,\(x^2+2xy+y^2-x-y-12\)

\(=\left(x+y\right)^2-\left(x+y\right)-12\)

Đặt \(t=x+y\)

\(\Rightarrow t^2-t-12\)

\(=t^2-4t+3t-12\)

\(=t\left(t-4\right)+3\left(t-4\right)\)

\(=\left(t+3\right)\left(t-4\right)\)

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

c,\(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)

Đặt \(t=x^2+x+1\)

\(\Rightarrow t\left(t+1\right)-12\)

\(=t^2+t-12\)

\(=t^2-3t+4t-12\)

\(=t\left(t-3\right)+4\left(t-3\right)\)

\(=\left(t+4\right)\left(t-3\right)\)

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

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

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

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

d,\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

Đặt \(t=x^2+7x+10\)

\(\Rightarrow t\left(t+2\right)-24\)

\(=t^2+2t-24\)

\(=t^2-4t+6t-24\)

\(=t\left(t-4\right)+6\left(t-4\right)\)

\(=\left(t+6\right)\left(t-4\right)\)

\(=\left(x^2+7x+16\right)\left(x^2+7x+6\right)\)

\(=\left(x^2+7x+16\right)\left(x^2+x+6x+6\right)\)

\(=\left(x^2+7x+16\right)\left[x\left(x+1\right)+6\left(x+1\right)\right]\)

\(=\left(x^2+7x+16\right)\left(x+6\right)\left(x+1\right)\)

help me