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a) \(\frac{2^5\cdot2^{12}\cdot2^6}{2^{24}}=\frac{2^{23}}{2^{24}}=\frac{1}{2}\)
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d) \(\dfrac{8^4.3^6}{2^7.65}=\dfrac{\left(2^3\right)^4.3^6}{2^7.65}=\dfrac{2^{12}.3^6}{2^7.65}=\dfrac{2^7.2^5.3^6}{2^7.65}=\dfrac{2^5.3^6}{65}=\dfrac{23328}{65}\)
c) \(\left(\dfrac{3}{5}-\dfrac{3}{4}\right).\left(\dfrac{2}{6}-\dfrac{1}{5}\right)^2=\dfrac{3.4-3.5}{4.5}.\left(\dfrac{2.5-1.6}{6.5}\right)^2\\ =\dfrac{-3}{20}.\left(\dfrac{2}{15}\right)^2=\dfrac{-3}{20}.\dfrac{4}{225}\\ =\dfrac{-3}{4.5}.\dfrac{4}{75.3}=\dfrac{-1}{375}\)
a) \(\frac{11}{24}-\frac{5}{41}+\frac{13}{24}+0,5-\frac{36}{41}\)
= \(\frac{11}{24}-\frac{5}{41}+\frac{13}{24}+\frac{1}{2}-\frac{36}{41}\)
= \(\frac{1}{2}-\left\{\frac{11}{24}+\frac{13}{24}\right\}-\left\{\frac{5}{41}+\frac{36}{41}\right\}\)
=\(\frac{1}{2}-\frac{24}{24}-\frac{41}{41}\)
=\(\frac{1}{2}-1-1\)
=\(\frac{-3}{2}\)
b) \(-12:\left\{\frac{3}{4}-\frac{5}{6}\right\}^2\)
= \(-12:\left\{\frac{9}{12}-\frac{10}{12}\right\}^2\)
= \(-12:\left\{\frac{-1}{12}\right\}^2\)
= \(-12:\frac{1}{144}\)
= \(-12.144\)
= -1728
c) \(\frac{7}{23}.\left[\left(\frac{-8}{6}\right)-\frac{45}{18}\right]\)
= \(\frac{7}{23}.\left[\left(\frac{-24}{18}\right)-\frac{45}{18}\right]\)
= \(\frac{7}{23}.\left(\frac{-23}{6}\right)\)
= \(\frac{-7}{6}\)
d) \(23\frac{1}{4}.\frac{7}{5}-13\frac{1}{4}:\frac{5}{7}\)
= \(23\frac{1}{4}.\frac{7}{5}-13\frac{1}{4}.\frac{7}{5}\)
= \(\left\{23\frac{1}{4}-13\frac{1}{4}\right\}.\frac{7}{5}\)
= \(10.\frac{7}{5}\)
= 14
e) (1+23−14).(0,8−34)2
= (1+23−14).(\(\frac{4}{5}\)−34)2
= \(\left(\frac{12}{12}+\frac{8}{12}-\frac{3}{12}\right).\left(\frac{16}{20}-\frac{15}{20}\right)^2\)
= \(\frac{17}{12}.\left(\frac{1}{20}\right)^2\)
= \(\frac{17}{20}.\frac{1}{400}\)
= \(\frac{17}{8000}\)
a)4.-1/8+1/2=-4/8+1/2
b)1/36.36+(0,6^1)^2/(0,2^3)^2
=1+0,36/0,000064
=1+5625
=5626
a: \(=\dfrac{-3^4\cdot2^8}{2^2\cdot2^2\cdot3^2}=-3^2\cdot2^2=-6^2=-36\)
b: \(=\dfrac{3^6\cdot2^{15}}{3^6\cdot2^6\cdot2^{15}}=\dfrac{1}{2^6}=\dfrac{1}{64}\)
c: \(=\left(\dfrac{0.8}{0.4}\right)^5\cdot\dfrac{1}{0.4}=2^5\cdot\dfrac{1}{0.4}=\dfrac{32}{0.4}=80\)