Toán Lớp 8: $\frac{2x+2}{x^2-1}$ + $\frac{2x-4}{x-1}$
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Toán Lớp 8: $\frac{2x+2}{x^2-1}$ + $\frac{2x-4}{x-1}$
Comments ( 2 )
~Bạn tham khảo~
\frac{2x+2}{x^2-1}+\frac{2x-4}{x-1} (Đk: x\ne -2; x \ne 2)
=\frac{2.(x+1)}{(x-1).(x+1)}+\frac{2x-4}{x-1}
=\frac{2}{x-1}+\frac{2x-4}{x-1}
=\frac{2+2x-4}{x-1}
=\frac{2.(1+x-2)}{x-1}
=\frac{2.(x-1)}{x-1}
=\frac{2}{1}
=2
Answer
{2x + 2}/{x^2 – 1} + {2x – 4}/{x – 1}
= {2x + 2}/{x^2 – 1^2} + {2x – 4}/{x – 1}
= {2x + 2}/{(x – 1) . (x + 1)} + {2x – 4}/{x – 1}
= {2x + 2}/{(x – 1) . (x + 1)} + {(2x – 4) . (x + 1)}/{(x – 1) . (x + 1)}
= {2x + 2}/{(x – 1) . (x + 1)} + {2x^2 – 2x – 4}/{(x – 1) . (x + 1)}
= {2x + 2 + 2x^2 – 2x – 4}/{(x – 1) . (x + 1)}
= {2x^2 + (2x – 2x) + (2 – 4)}/{(x – 1) . (x + 1)}
= {2x^2 – 2}/{(x – 1) . (x + 1)}
= {2 . (x^2 – 1)}/{(x – 1) . (x + 1)}
= {2 . (x^2 – 1^2)}/{(x – 1) . (x + 1)}
= {2 . (x – 1) . (x + 1)}/{(x – 1) . (x + 1)}
= 2