Toán Lớp 8: (x + 4)/(x^2 + 4x) – 1/(x^2 + 2x)
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Toán Lớp 8: (x + 4)/(x^2 + 4x) – 1/(x^2 + 2x)
Comments ( 2 )
$\textit{Giải đáp + Lời giải và giải thích chi tiết:}$
(x+4)/(x^2+4x)-1/(x^2+2x) (x \ne 0,x \ne -2, x \ne -4)
=((x+4)(x+2))/(x(x+4)(x+2))-(1(x+4))/(x(x+2)(x+4))
=(x^2+6x+8-x-4)/(x(x+2)(x+4))
=(x^2+5x+4)/(x(x+2)(x+4))
=((x+1)(x+4))/(x(x+2)(x+4))
=(x+1)/(x(x+2))
\frac{x+4}{x^2+4x}-\frac{1}{x^2+2x} MTC: x(x+4)(x+2), x\ne0
= \frac{x+4}{x(x+4)}-\frac{1}{x(x+2)}
= \frac{(x+4)(x+2)}{x(x+4)(x+2)}-\frac{1(x+4)}{x(x+2)(x+4)}
=\frac{x^2+6x+8-x-4}{x(x+4)(x+2)}
= \frac{x^2+5x+4}{x(x+4)(x+2)}
= \frac{(x+1)(x+4)}{x(x+4)(x+2)}
= \frac{x+1}{x(x+2)}