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a) \(\frac{x^4-4x^2+3}{x^4+6x^2-7}=\frac{x^4-3x^2-x^2+3}{x^4+7x^2-x^2-7}=\frac{x^2\left(x^2-3\right)-\left(x^2-3\right)}{x^2\left(x^2+7\right)-\left(x^2+7\right)}=\frac{\left(x^2-1\right)\left(x^2-3\right)}{\left(x^2-1\right)\left(x^2+7\right)}=\frac{x^2-3}{x^2+7}\)
b) \(\frac{x^4+x^3-x-1}{x^4+x^3+2x^2+x+1}=\frac{x^3\left(x+1\right)-\left(x+1\right)}{\left(x^4+2x^2+1\right)+\left(x^3+x\right)}\)
\(=\frac{\left(x+1\right)\left(x^3-1\right)}{\left(x^2+1\right)^2+x\left(x^2+1\right)}=\frac{\left(x+1\right)\left(x-1\right)\left(x^2+x+1\right)}{\left(x^2+1\right)\left(x^2+1+x\right)}\)
\(=\frac{x^2-1}{x^2+1}\)
c) \(\frac{x^3+3x^2-4}{x^3-3x+2}=\frac{x^3+4x^2-x^2-4}{x^3-4x+x+2}=\frac{x^2\left(x^2+4\right)-\left(x^2+4\right)}{x\left(x^2-4\right)+\left(x+2\right)}\)
\(=\frac{\left(x^2-1\right)\left(x-2\right)\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)+\left(x+2\right)}=\frac{\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left(x^2-2x+1\right)}\)
\(=\frac{\left(x-1\right)\left(x+1\right)\left(x-2\right)}{\left(x-1\right)^2}=\frac{x^2-x-2}{x-1}\)
a) ( x - 2 )3 - x( x + 1 )( x - 1 ) + 6x( x - 3 )
= x3 - 6x2 + 12x - 8 - x( x2 - 1 ) + 6x2 - 18x
= x3 - 6x - 8 - x3 + x
= -5x - 8
b) ( x + 1 )3 - ( x - 1 )3 - 6( x - 1 )2
= x3 + 3x2 + 3x + 1 - ( x3 - 3x2 + 3x - 1 ) - 6( x2 - 2x + 1 )
= x3 + 3x2 + 3x + 1 - x3 + 3x2 - 3x + 1 - 6x2 + 12x - 6
= 12x - 4
c) ( 2x + 1 )( 4x2 - 2x + 1 ) + ( 2 - 3x )( 4 + 6x + 9x2 ) - 9
= ( 2x )3 + 13 + 23 - ( 3x )3 - 9
= 8x3 + 1 + 8 - 27x3 - 9
= -19x3
d) ( x + 1 )3 + ( x - 1 )3 + x3 - 3x( x - 1 )( x + 1 )
= x3 + 3x2 + 3x + 1 + x3 - 3x2 + 3x - 1 + x3 - 3x( x2 - 1 )
= 3x3 + 6x - 3x2 + 3x
= 9x
Lời giải:
a) ĐKXĐ: $x\neq \pm 1$
\(\frac{x^4-4x^2+3}{x^4+6x^2-7}=\frac{x^2(x^2-1)-3(x^2-1)}{x^2(x^2-1)+7(x^2-1)}=\frac{(x^2-3)(x^2-1)}{(x^2-1)(x^2+7)}=\frac{x^2-3}{x^2+7}\)
b) ĐKXĐ: Với mọi $x\in\mathbb{R}$
\(\frac{x^4+x^3-x-1}{x^4+x^4+2x^2+x+1}=\frac{(x^4-x)+(x^3-1)}{(x^4+x^3+x^2)+(x^2+x+1)}=\frac{x(x^3-1)+(x^3-1)}{x^2(x^2+x+1)+(x^2+x+1)}\)
\(=\frac{(x^3-1)(x+1)}{(x^2+1)(x^2+x+1)}=\frac{(x-1)(x^2+x+1)(x+1)}{(x^2+1)(x^2+x+1)}=\frac{x^2-1}{x^2+1}\)
c) ĐK: $x\neq 1;-2$
\(\frac{x^3+3x^2-4}{x^3-3x+2}=\frac{x^2(x-1)+4(x^2-1)}{x^2(x-1)+x(x-1)-2(x-1)}=\frac{(x-1)(x^2+4x+4)}{(x-1)(x^2+x-2)}\)
\(=\frac{(x-1)(x+2)^2}{(x-1)(x-1)(x+2)}=\frac{x+2}{x-1}\)
d) ĐK: $x^2+3x-1\neq 0$
\(\frac{x^4+6x^3+9x^2-1}{x^4+6x^3+7x^2-6x+1}=\frac{(x^2+3x)^2-1}{(x^2+3x)^2-2x^2-6x+1}\)
\(=\frac{(x^2+3x-1)(x^2+3x+1)}{(x^2+3x)^2-2(x^2+3x)+1}=\frac{(x^2+3x-1)(x^2+3x+1)}{(x^2+3x-1)^2}=\frac{x^2+3x+1}{x^2+3x-1}\)