Đáp án:
$\begin{array}{l}
B4:\\
c)648\\
d)\dfrac{2}{3}\\
B5:\\
a)\dfrac{3}{5}\\
b)\dfrac{{150}}{{31}}\\
c){2^{51}} - 2\\
d)2 - \dfrac{1}{{{2^{10}}}}
\end{array}$
Giải thích các bước giải:
$\begin{array}{l}
B4:\\
c)\dfrac{{{{5.27}^3}{{.4}^5}}}{{{6^5}}}:\left( {\dfrac{{{5^5}{{.2}^4}}}{{{{10}^4}}}.\dfrac{{{2^6}{{.3}^4}}}{{{6^4}}}} \right)\\
= \dfrac{{5.{{\left( {{3^3}} \right)}^3}.{{\left( {{2^2}} \right)}^5}}}{{{{\left( {2.3} \right)}^5}}}:\left( {\dfrac{{{5^5}{{.2}^4}{{.2}^6}{{.3}^4}}}{{{{\left( {2.5} \right)}^4}.{{\left( {2.3} \right)}^4}}}} \right)\\
= \dfrac{{{{5.3}^9}{{.2}^{10}}}}{{{2^5}{{.3}^5}}}:\dfrac{{{5^5}{{.2}^{10}}{{.3}^4}}}{{{2^8}{{.3}^4}{{.5}^4}}}\\
= \dfrac{{{{5.3}^9}{{.2}^{10}}}}{{{2^5}{{.3}^5}}}.\dfrac{{{2^8}{{.3}^4}{{.5}^4}}}{{{5^5}{{.2}^{10}}{{.3}^4}}}\\
= \dfrac{{{5^5}{{.3}^{13}}{{.2}^{18}}}}{{{5^5}{{.3}^9}{{.2}^{15}}}}\\
= {3^4}{.2^3}\\
= 648\\
d)\dfrac{{5 - \dfrac{5}{3} - \dfrac{5}{9} - \dfrac{5}{{27}}}}{{8 - \dfrac{8}{3} - \dfrac{8}{9} - \dfrac{8}{{27}}}}:\dfrac{{15 - \dfrac{{15}}{{11}} + \dfrac{{15}}{{121}}}}{{16 - \dfrac{{16}}{{11}} + \dfrac{{16}}{{121}}}}\\
= \dfrac{{5\left( {1 - \dfrac{1}{3} - \dfrac{1}{9} - \dfrac{1}{{27}}} \right)}}{{8\left( {1 - \dfrac{1}{3} - \dfrac{1}{9} - \dfrac{1}{{27}}} \right)}}:\dfrac{{15\left( {1 - \dfrac{1}{{11}} + \dfrac{1}{{121}}} \right)}}{{16\left( {1 - \dfrac{1}{{11}} + \dfrac{1}{{121}}} \right)}}\\
= \dfrac{5}{8}:\dfrac{{15}}{{16}}\\
= \dfrac{5}{8}.\dfrac{{16}}{{15}}\\
= \dfrac{2}{3}\\
B5:\\
a)\dfrac{7}{{10.11}} + \dfrac{7}{{11.12}} + \dfrac{7}{{12.13}} + ... + \dfrac{7}{{69.70}}\\
= 7\left( {\dfrac{1}{{10.11}} + \dfrac{1}{{11.12}} + \dfrac{1}{{12.13}} + ... + \dfrac{1}{{69.70}}} \right)\\
= 7\left( {\dfrac{1}{{10}} - \dfrac{1}{{11}} + \dfrac{1}{{11}} - \dfrac{1}{{12}} + \dfrac{1}{{12}} - \dfrac{1}{{13}} + ... + \dfrac{1}{{69}} - \dfrac{1}{{70}}} \right)\\
= 7\left( {\dfrac{1}{{10}} - \dfrac{1}{{70}}} \right)\\
= 7.\dfrac{{7 - 1}}{{70}}\\
= \dfrac{3}{5}\\
b)\dfrac{{{5^2}}}{{1.6}} + \dfrac{{{5^2}}}{{6.11}} + ... + \dfrac{{{5^2}}}{{26.31}}\\
= 5\left( {\dfrac{5}{{1.6}} + \dfrac{5}{{6.11}} + ... + \dfrac{5}{{26.31}}} \right)\\
= 5\left( {\dfrac{1}{1} - \dfrac{1}{6} + \dfrac{1}{6} - \dfrac{1}{{11}} + ... + \dfrac{1}{{26}} - \dfrac{1}{{31}}} \right)\\
= 5\left( {1 - \dfrac{1}{{31}}} \right)\\
= 5.\dfrac{{30}}{{31}}\\
= \dfrac{{150}}{{31}}
\end{array}$
c) Đặt $A = 2 + {2^2} + {2^3} + ... + {2^{50}}$
$\begin{array}{l}
\Rightarrow 2A = {2^2} + {2^3} + {2^4} + ... + {2^{51}}\\
\Rightarrow 2A - A = \left( {{2^2} + {2^3} + {2^4} + ... + {2^{51}}} \right) - \left( {2 + {2^2} + {2^3} + ... + {2^{50}}} \right)\\
\Rightarrow A = {2^{51}} - 2
\end{array}$
d) Đặt $B = 1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + ... + \dfrac{1}{{1024}}$
$\begin{array}{l}
\Rightarrow B = 1 + \dfrac{1}{2} + {\left( {\dfrac{1}{2}} \right)^2} + {\left( {\dfrac{1}{2}} \right)^3} + ... + {\left( {\dfrac{1}{2}} \right)^{10}}\\
\Rightarrow \dfrac{1}{2}B = \dfrac{1}{2} + {\left( {\dfrac{1}{2}} \right)^2} + {\left( {\dfrac{1}{2}} \right)^3} + {\left( {\dfrac{1}{2}} \right)^4} + ... + {\left( {\dfrac{1}{2}} \right)^{11}}\\
\Rightarrow \dfrac{1}{2}B - B = \left( {\dfrac{1}{2} + {{\left( {\dfrac{1}{2}} \right)}^2} + {{\left( {\dfrac{1}{2}} \right)}^3} + {{\left( {\dfrac{1}{2}} \right)}^4} + ... + {{\left( {\dfrac{1}{2}} \right)}^{11}}} \right) - \left( {1 + \dfrac{1}{2} + {{\left( {\dfrac{1}{2}} \right)}^2} + {{\left( {\dfrac{1}{2}} \right)}^3} + ... + {{\left( {\dfrac{1}{2}} \right)}^{10}}} \right)\\
\Rightarrow - \dfrac{1}{2}B = {\left( {\dfrac{1}{2}} \right)^{11}} - 1\\
\Rightarrow B = 2 - \dfrac{1}{{{2^{10}}}}
\end{array}$
