@Câu 73. [id1480] (HSG11 Cao Bằng 2011 - 2012) Tính $B=\underset{x\to 1}{\mathop{\lim }}\,\dfrac{x+{{x}^{2}}+...+{{x}^{30}}-30}{x+{{x}^{2}}+{{x}^{3}}+{{x}^{4}}-4}$ . Ta có $\dfrac{x+{{x}^{2}}+...+{{x}^{30}}-30}{x+{{x}^{2}}+{{x}^{3}}+{{x}^{4}}-4}$ $=\dfrac{\left( x-1 \right)+\left( {{x}^{2}}-1 \right)+...+\left( {{x}^{30}}-1 \right)}{\left( x-1 \right)+\left( {{x}^{2}}-1 \right)+\left( {{x}^{3}}-1 \right)+\left( {{x}^{4}}-1 \right)}$ $\dfrac{\left( x-1 \right)\left[ 1+\left( x+1 \right)+...+\left( {{x}^{29}}+{{x}^{28}}+...+x+1 \right) \right]}{\left( x-1 \right)\left[ 1+\left( x+1 \right)+\left( {{x}^{2}}+x+1 \right)+\left( {{x}^{3}}+{{x}^{2}}+x+1 \right) \right]}$ $=\dfrac{1+\left( x+1 \right)+...+\left( {{x}^{29}}+{{x}^{28}}+...+x+1 \right)}{1+\left( x+1 \right)+\left( {{x}^{2}}+x+1 \right)+\left( {{x}^{3}}+{{x}^{2}}+x+1 \right)}$ Khi đó $B=\underset{x\to 1}{\mathop{\lim }}\,\dfrac{x+{{x}^{2}}+...+{{x}^{30}}-30}{x+{{x}^{2}}+{{x}^{3}}+{{x}^{4}}-4}$ $=\underset{x\to 1}{\mathop{\lim }}\,\dfrac{1+\left( x+1 \right)+...+\left( {{x}^{29}}+{{x}^{28}}+...+x+1 \right)}{1+\left( x+1 \right)+\left( {{x}^{2}}+x+1 \right)+\left( {{x}^{3}}+{{x}^{2}}+x+1 \right)}$ $=\dfrac{1+2+...+30}{1+2+3+4}=\dfrac{93}{2}$ . Vậy $B=\dfrac{93}{2}$.

@Câu 73. [id1480] (HSG11 Cao Bằng 2011 - 2012) Tính $B=\underset{x\to 1}{\mathop{\lim }}\,\dfrac{x+{{x}^{2}}+...+{{x}^{30}}-30}{x+{{x}^{2}}+{{x}^{3}}+{{x}^{4}}-4}$ . Ta có $\dfrac{x+{{x}^{2}}+...+{{x}^{30}}-30}{x+{{x}^{2}}+{{x}^{3}}+{{x}^{4}}-4}$ $=\dfrac{\left( x-1 \right)+\left( {{x}^{2}}-1 \right)+...+\left( {{x}^{30}}-1 \right)}{\left( x-1 \right)+\left( {{x}^{2}}-1 \right)+\left( {{x}^{3}}-1 \right)+\left( {{x}^{4}}-1 \right)}$ $\dfrac{\left( x-1 \right)\left[ 1+\left( x+1 \right)+...+\left( {{x}^{29}}+{{x}^{28}}+...+x+1 \right) \right]}{\left( x-1 \right)\left[ 1+\left( x+1 \right)+\left( {{x}^{2}}+x+1 \right)+\left( {{x}^{3}}+{{x}^{2}}+x+1 \right) \right]}$ $=\dfrac{1+\left( x+1 \right)+...+\left( {{x}^{29}}+{{x}^{28}}+...+x+1 \right)}{1+\left( x+1 \right)+\left( {{x}^{2}}+x+1 \right)+\left( {{x}^{3}}+{{x}^{2}}+x+1 \right)}$ Khi đó $B=\underset{x\to 1}{\mathop{\lim }}\,\dfrac{x+{{x}^{2}}+...+{{x}^{30}}-30}{x+{{x}^{2}}+{{x}^{3}}+{{x}^{4}}-4}$ $=\underset{x\to 1}{\mathop{\lim }}\,\dfrac{1+\left( x+1 \right)+...+\left( {{x}^{29}}+{{x}^{28}}+...+x+1 \right)}{1+\left( x+1 \right)+\left( {{x}^{2}}+x+1 \right)+\left( {{x}^{3}}+{{x}^{2}}+x+1 \right)}$ $=\dfrac{1+2+...+30}{1+2+3+4}=\dfrac{93}{2}$ . Vậy $B=\dfrac{93}{2}$.