Toán Lớp 8: $\frac{3m^{2}+2m+1}{m^{3}-1}$-$\frac{1-m}{m^{2}+m+1}$-$\frac{2}{m-1}$ với m $\neq$ 1
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Toán Lớp 8: $\frac{3m^{2}+2m+1}{m^{3}-1}$-$\frac{1-m}{m^{2}+m+1}$-$\frac{2}{m-1}$ với m $\neq$ 1
Comments ( 2 )
ĐKXĐ: mne1
=\frac{3m^2+2m+1}{(m-1)(m^2+m+1)}-\frac{1-m}{m^2+m+1}-\frac{2}{m-1}
=\frac{3m^2+2m+1+(m-1)^2-2(m^2+m+1)}{(m-1)(m^2+m+1)}
=\frac{3m^2+2m+1+m^2+1-2m-2m^2-2m-2}{(m-1)(m^2+m+1)}
=\frac{2m^2-2m}{(m-1)(m^2+m+1)}
=\frac{2m(m-1)}{(m-1)(m^2+m+1)}
=\frac{2m}{m^2+m+1}