Toán Lớp 8: $\frac{x^2+1}{1}$- $\frac{x^4-3x^2+2}{x^2-1}$
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Toán Lớp 8: $\frac{x^2+1}{1}$- $\frac{x^4-3x^2+2}{x^2-1}$
Comments ( 2 )
(x^2+1)/1-(x^4-3x^2+2)/(x^2-1)(x\ne+-1)
=((x^2+1)(x^2-1))/(x^2-1)-(x^4-3x^2+2)/(x^2-1)
=((x^2+1)(x^2-1)-(x^4-3x^2+2))/(x^2-1)
=((x^2)^2-1^2-x^4+3x^2-2)/(x^2-1)
=(x^4-1-x^4+3x^2-2)/(x^2-1)
=(3x^2-3)/(x^2-1)
=(3(x^2-1))/(x^2-1)
=3
\frac{x^2+1}{1}-\frac{x^4-3x^2+2}{x^2-1} (ĐK:x \ne +-1)
=\frac{(x^2+1)(x^2-1)}{x^2-1}-\frac{x^4-3x^2+2}{x^2-1}
=\frac{x^4-1-x^4+3x^2-2}{x^2-1}
=\frac{3x^2-3}{x^2-1}
=\frac{3(x^2-1)}{x^2-1}
=3