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\(a,\frac{7}{x}=\frac{x}{28}=>x\cdot x=28\cdot7=>x^2=196=>x^2=14^2\)\(=>x=14\)
\(b,\frac{10+x}{x+17}=\frac{3}{4}=>\left(10+x\right)\cdot4=\left(x+17\right)\cdot3=>40+x4=x3+51\)\(=>x4-x3=51-40=>x=11\)
\(\dfrac{10+x}{17+x}=\dfrac{10\cdot11+50\cdot55+70\cdot77}{11\cdot12+55\cdot66+77\cdot84}\)
\(\Leftrightarrow\dfrac{10+x}{17+x}=\dfrac{10\cdot11+5\cdot10\cdot111+7\cdot10\cdot11}{11\cdot12+5\cdot11\cdot12+7\cdot11\cdot12}\)
\(\Leftrightarrow\dfrac{10+x}{17+x}=\dfrac{10\cdot11\left(1+5+7\right)}{11\cdot12\left(1+5+7\right)}\)
\(\Leftrightarrow\dfrac{10+x}{17+x}=\dfrac{5}{6}\)
\(\Leftrightarrow6\left(10+x\right)=5\left(17+x\right)\)
\(\Leftrightarrow60+6x=85+5x\)
\(\Leftrightarrow6x-5x=85-60\)
\(\Leftrightarrow x=25\) (TM x\(\ne17\))
Vậy ...................................
\(=\dfrac{2^{19}\cdot3^9+2^{18}\cdot3^9\cdot5}{2^{19}\cdot3^9+2^{20}\cdot3^{10}}=\dfrac{2^{18}\cdot3^9\cdot\left(2+5\right)}{2^{19}\cdot3^9\cdot7}=\dfrac{1}{2}\)
\(A=\frac{10.11+50.55+70.77}{11.12+55.60+77.84}\)
\(=\frac{10.11+5.10.5.11+7.10.7.11}{11.12+11.5.12.5+11.7.12.7}\)
\(=\frac{10.11\left(1+25+49\right)}{11.12\left(1+25+49\right)}\)
\(=\frac{10.11}{11.12}=\frac{10}{12}=\frac{5}{6}\)
\(B=\frac{1\times3\times5\times7\times........\times49}{26\times27\times28\times...........\times50}\)
\(=\frac{\left(1\times3\times5\times7\times.........\times49\right).\left(2\times4\times6.........48\times50\right)}{\left(26\times27\times28\times.........\times50\right).\left(2\times4\times6\times...........\times48\times50\right)}\)
\(=\frac{1\times2\times3\times4\times..........\times50}{\left(26\times27\times28\times..............\times50\right)2^{25}\left(1\times2\times3\times4\times............\times25\right)}=\frac{1}{2^{25}}\)
\(C=\frac{1.2.6+2.4.12+4.8.24+7.14.42}{1.6.9+2.12.18+4.24.36+7.42.63}\)
\(=\frac{1.2.6\left(1+8+64+343\right)}{1.6.9\left(1+8+64+343\right)}\)
\(=\frac{1.2.6}{1.6.9}=\frac{2}{9}\)
\(A=\frac{5}{6}\)
\(B=\frac{1}{33554432}\)
\(C=\frac{28}{117}\)
\(\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)
\(=\frac{2^{19}.\left(3^3\right)^3+3.5.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(2^2.3\right)^{10}}\)
\(=\frac{2^{19}.3^9+3.5.2^{18}.3^8}{2^9.3^9.2^{10}+2^{20}.3^{10}}\)
\(=\frac{2^{19}.3^9+2^{19}.3^9.5}{2^{19}.3^9+2^{20}.3^{10}}\)
\(=\frac{2^{19}.3^9.\left(1+5\right)}{2^{19}.3^9\left(1+2.3\right)}\)
\(=\frac{6}{7}\)
A= \(\dfrac{10.11.\left(1+5.5+7.7\right)}{11.12.\left(1+5.5+7.7\right)}=\dfrac{10}{12}=\dfrac{5}{6}\)