b) $\mathrm{B}=\left(\frac{151515}{161616}+\frac{17^{9}}{17^{10}}\right)-\left(\frac{1500}{1600}-\frac{176}{187}\right)$

Q: b) $\mathrm{B}=\left(\frac{151515}{161616}+\frac{17^{9}}{17^{10}}\right)-\left(\frac{1500}{1600}-\frac{176}{187}\right)$

$\begin{array}{l}\mathrm{B}=\left(\frac{15.10101}{16.10101}+\frac{1}{17}\right)-\left(\frac{15.100}{16.100}-\frac{16.11}{17.11}\right) \\\mathrm{B}=\frac{15}{16}+\frac{1}{17}-\frac{15}{16}+\frac{16}{17} \\\mathrm{~B}=\left(\frac{15}{16}-\frac{15}{16}\right)+\left(\frac{1}{17}+\frac{16}{17}\right) \\\mathrm{B}=0+1=1\end{array}$Vậy $B=1$

Question

Accepted Answer

$\begin{array}{l}\mathrm{B}=\left(\frac{15.10101}{16.10101}+\frac{1}{17}\right)-\left(\frac{15.100}{16.100}-\frac{16.11}{17.11}\right) \\mathrm{B}=\frac{15}{16}+\frac{1}{17}-\frac{15}{16}+\frac{16}{17} \\mathrm{~B}=\left(\frac{15}{16}-\frac{15}{16}\right)+\left(\frac{1}{17}+\frac{16}{17}\right) \\mathrm{B}=0+1=1\end{array}$

Vậy $B=1$

Đề thi HSG (CT) 18-19 - H.Thái Thụy - Thái Bình - MĐ 6191