a: \(\dfrac{10^3+10^3+...+10^3}{5^3+5^3+5^3+5^3+5^3}\)
\(=\dfrac{10\cdot10^3}{5\cdot5^3}=\dfrac{10^4}{5^4}=2^4\)
\(\dfrac{4^x+4^x+4^x+4^x}{2^3+2^3}=\dfrac{4\cdot4^x}{16}=\dfrac{4^x}{4}=4^{x-1}\)
Ta có: \(\dfrac{10^3+10^3+...+10^3}{5^3+5^3+5^3+5^3+5^3}\cdot\dfrac{4^x+4^x+4^x+4^x}{2^3+2^3}=2^{12}\)
=>\(2^4\cdot4^{x-1}=2^{12}\)
=>\(4^{x-1}=2^8=4^4\)
=>x-1=4
=>x=5
b: \(\left(\dfrac{2}{2^2}\cdot\dfrac{6}{3^2}\cdot\dfrac{12}{4^2}\cdot\dfrac{20}{5^2}\cdot...\cdot\dfrac{210}{15^2}\cdot\dfrac{240}{16^2}\right)\cdot15=\dfrac{1}{2^x}\)
=>\(\dfrac{1\cdot2}{2\cdot2}\cdot\dfrac{2\cdot3}{3\cdot3}\cdot\dfrac{3\cdot4}{4\cdot4}\cdot\dfrac{4\cdot5}{5\cdot5}\cdot...\cdot\dfrac{14\cdot15}{15\cdot15}\cdot\dfrac{15\cdot16}{16^2}\cdot15=\dfrac{1}{2^x}\)
=>\(\dfrac{1}{16}\cdot15=\dfrac{1}{2^x}\)
=>\(\dfrac{1}{2^x}=\dfrac{15}{16}\)
=>\(2^x=\dfrac{16}{15}\)
=>\(x=log_2\left(\dfrac{16}{15}\right)=4-log_215\)