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\(F=\frac{\left(\frac{2}{5}\right)^7.5^7+\left(\frac{9}{4}\right)^9\div\left(\frac{3}{16}\right)^3}{2^7.5^2+512}\)
\(F=\frac{\left(\frac{2.5}{5}\right)^7+\left(\frac{9.16}{4.3}\right)^3}{2^7.5^2+2^9}=\frac{2^7+12^3}{2^7.5^2+2^9}=\frac{2^7+2^6.3^3}{2^7.5^2+2^9}=\frac{2^6.\left(2+3^3\right)}{2^7.\left(5^2+2^2\right)}=\frac{2^6.29}{2^7.29}\)
\(F=\frac{1}{2}\)
1)
a)
\(\frac{-5}{6}.\frac{120}{25}< x< \frac{-7}{15}.\frac{9}{14}\)
\(\frac{-1}{1}.\frac{20}{5}< x< \frac{-1}{5}.\frac{3}{2}\)
\(\frac{-20}{5}< x< \frac{-3}{10}\)
\(\frac{-40}{10}< x< \frac{-3}{10}\)
\(\Rightarrow Z\in\left\{-4;-5;-6;-7;-8;-9;-10;...;-39\right\}\)
B1a)\(11\frac34-\left(6\frac56-4\frac12\right)+1\frac23\)
=\(11\frac34-6\frac56+4\frac12+1\frac23\)
=\(\left(11-6+4+1\right)+\left(\frac34-\frac56+\frac12+\frac23\right)\)
=\(10+\left(\frac{9}{12}-\frac{10}{12}+\frac{6}{12}+\frac{8}{12}\right)\)
=\(10+\left(-\frac{1}{12}+\frac{6}{12}+\frac{8}{12}\right)\)
=10+\(\frac{13}{12}\)
=\(\frac{120}{12}+\frac{13}{12}\)
=\(\frac{133}{12}\)
b)\(2\frac{17}{20}-1\frac{11}{5}+6\frac{9}{20}:3\)
= \(\frac{57}{20}-\frac{16}{5}+\frac{129}{20}\times\frac13\)
=\(\frac{57}{20}-\frac{16}{5}+\frac{129}{60}\)
=\(\frac{171}{60}-\frac{192}{60}+\frac{129}{60}\)
=\(\frac{108}{60}\)
=\(\frac95\)
a) \(2^3+3.\left(\frac{1}{2}\right)^0+\left[\left(-2\right)^2:\frac{1}{2}\right]\)
\(=8+3.1+4:\frac{1}{2}\)
\(=8+3+8=19\)
b)\(\frac{2^{15}.9^4}{6^6.8^3}=\frac{2^{15}.\left(3^2\right)^4}{\left(2.3\right)^6.\left(2^3\right)^3}=\frac{2^{15}.3^8}{2^6.3^6.2^9}\)\(=\frac{2^{15}.3^8}{2^{15}.3^6}=3^2=9\)
c) \(\left(1+\frac{2}{3}-\frac{1}{4}\right).\left(\frac{4}{5}-\frac{3}{4}\right)^2\)
\(=\frac{17}{12}.\frac{1}{400}=\frac{17}{4800}\)
d) \(\left(-\frac{10}{3}\right)^3.\left(\frac{-6}{5}\right)^4=-\frac{100}{27}.\frac{1296}{625}\)\(=\frac{-4.48}{1.25}=-\frac{192}{25}\)