Toán Lớp 8: $\frac{1}{x+2}$+$\frac{1}{(x+2)(4x+7)}$
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Toán Lớp 8: $\frac{1}{x+2}$+$\frac{1}{(x+2)(4x+7)}$
Comments ( 2 )
Lời giải và giải thích chi tiết:
1/(x+2)+1/((x+2)(4x+7)) (x \ne -2; -7/4)
=(4x+7)/((x+2)(4x+7))+1/((x+2)(4x+7))
=(4x+7+1)/((x+2)(4x+7))
=(4x+8)/((x+2)(4x+7))
=(4(x+2))/((x+2)(4x+7))
=4/(4x+7)
\frac{1}{x+2}+\frac{1}{(x+2)(4x+7)} (ĐK:x \ne -2,x \ne \frac{-7}{4})
=\frac{4x+7}{(x+2)(4x+7)}+\frac{1}{(x+2)(4x+7)}
=\frac{4x+7+1}{(x+2)(4x+7)}
=\frac{4x+8}{(x+2)(4x+7)}
=\frac{4(x+2)}{(x+2)(4x+7)}
=\frac{4}{4x+7}