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\(A=\left(2-\frac{3}{2}\right).\left(2-\frac{4}{3}\right).\left(2-\frac{5}{4}\right).\left(2-\frac{6}{5}\right)\)
\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}\)
\(A=\frac{1}{5}\)
\(A=\left(2-\frac{3}{2}\right)\times\left(2-\frac{4}{3}\right)\times\left(2-\frac{5}{4}\right)\times\left(2-\frac{6}{5}\right)\)
\(\Rightarrow\)\(A=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times\frac{4}{5}\)
\(\Rightarrow A=\frac{1}{5}\)
\(\left(1-\frac{4}{5}+\frac{3}{10}\right).\left(-\frac{2}{3}\right)^2\)
\(=\left(\frac{10}{10}-\frac{8}{10}+\frac{3}{10}\right).\frac{4}{9}\)
\(=\frac{1}{2}.\frac{4}{9}\)
\(=\frac{2}{9}\)
a) \(4\frac{5}{9}:\left(-\frac{5}{7}\right)+\frac{49}{9}:\left(-\frac{5}{7}\right)=\frac{41}{9}:\left(-\frac{5}{7}\right)+\frac{49}{9}:\left(-\frac{5}{7}\right)\)
\(=\frac{41}{9}\cdot\left(-\frac{7}{5}\right)+\frac{49}{9}\cdot\left(-\frac{7}{5}\right)=\left(\frac{41}{9}+\frac{49}{9}\right)\cdot\left(-\frac{7}{5}\right)=10\cdot\left(-\frac{7}{5}\right)=-14\)
b) \(\left(\frac{-3}{5}+\frac{4}{9}\right):\frac{7}{11}+\left(\frac{-2}{5}+\frac{5}{9}\right):\frac{7}{11}\)
\(=\left(\frac{-3}{5}+\frac{4}{9}+\frac{-2}{5}+\frac{5}{9}\right):\frac{7}{11}\)
\(=\left(\frac{-3}{5}+\frac{-2}{5}+\frac{4}{9}+\frac{5}{9}\right):\frac{7}{11}\)
\(=\left(-1+1\right):\frac{7}{11}=0\cdot\frac{11}{7}=0\)
c) \(\left(\frac{3}{4}\right)^4\cdot\left(\frac{8}{9}\right)^2=\left(\frac{3}{4}\right)^2\cdot\left(\frac{3}{4}\right)^2\cdot\left(\frac{8}{9}\right)^2=\left(\frac{3}{4}\cdot\frac{3}{4}\cdot\frac{8}{9}\right)^2\)
\(=\left(\frac{1}{2}\right)^2=\frac{1}{4}\)
d) \(\left(-\frac{3}{5}\right)^6\cdot\left(-\frac{5}{3}\right)^5=\left(-\frac{3}{5}\right)^5\cdot\left(-\frac{3}{5}\right)\cdot\left(-\frac{5}{3}\right)^5=\left[\left(-\frac{3}{5}\right)\cdot\left(-\frac{5}{3}\right)\right]^5\cdot\left(-\frac{3}{5}\right)\)
\(=1^5\cdot\left(-\frac{3}{5}\right)=1\cdot\left(-\frac{3}{5}\right)=-\frac{3}{5}\)
e) \(\frac{8^{14}}{4^4\cdot64^5}=\frac{\left(2^3\right)^{14}}{\left(2^2\right)^4\cdot\left(2^6\right)^5}=\frac{2^{42}}{2^8\cdot2^{30}}=\frac{2^{42}}{2^{38}}=2^4=16\)
f) \(\frac{9^{10}\cdot27^7}{81^7\cdot3^{15}}=\frac{\left(3^2\right)^{10}\cdot\left(3^3\right)^7}{\left(3^4\right)^7\cdot3^{15}}=\frac{3^{20}\cdot3^{21}}{3^{28}\cdot3^{15}}=\frac{3^{41}}{3^{43}}=3^{-2}=\frac{1}{3^2}=\frac{1}{9}\)
\(=\frac{11}{-5}\cdot\frac{-9}{11}\cdot\frac{15}{-14}\cdot\frac{2}{5}+-\frac{2}{77}\cdot\frac{5}{-3}\)
\(=\frac{9}{5}\cdot-\frac{15}{14}\cdot\frac{2}{5}+\frac{10}{231}\)
\(=-\frac{841}{1155}\)
a) \(C=\frac{\left(\frac{2}{3}\right)^3\times\left(-\frac{3}{4}\right)^2\times\left(-1\right)^5}{\left(\frac{2}{5}\right)^2\times\left(-\frac{5}{12}\right)^2}\)
\(C=\frac{\frac{2^3}{3^3}.\frac{\left(-3\right)^2}{4^2}.\left(-1\right)^5}{\frac{2^2}{5^2}.\frac{\left(-5\right)^2}{12^2}}\)
\(C=\frac{\frac{-\left(2^3.3^2\right)}{3^3.2^4}}{\frac{2^2.5^2}{5^2.2^4.3^2}}\)
\(C=\frac{\frac{-1}{3.2}}{\frac{1}{2^2.3^2}}\)
\(C=\frac{\frac{-1}{6}}{\frac{1}{36}}\)
\(C=-6\)
b) \(D=\frac{6^6+6^3\times3^3+3^6}{-73}\)
\(D=\frac{2^6.3^6+2^3.3^3.3^3+3^6}{-73}\)
\(D=\frac{2^6.3^6+2^3.3^6+3^6}{-73}\)
\(D=\frac{3^6\left(2^6+2^3+1\right)}{-73}\)
\(D=\frac{3^6.73}{\left(-1\right).73}\)
\(D=-3^6=-729\)
\(4.\left(\frac{1}{4}\right)^2+25\left[\left(\frac{3}{4}\right)^3:\left(\frac{5}{4}\right)^3\right]:\left(\frac{3}{2}\right)^3=4.\frac{1}{16}+25\left(\frac{27}{64}.\frac{64}{125}\right).\frac{8}{27}\)
\(=\frac{1}{4}+25.\frac{27}{125}.\frac{8}{27}=\frac{1}{4}+\frac{8}{5}=\frac{37}{20}\)
\(2^3+3\left(\frac{1}{2}\right)^0-1+\left[\left(-2\right)^2:\frac{1}{2}\right]-8=8+3-1+4.2-8=10\)
Bài làm :
a)\(=-\frac{3}{5}+\frac{28}{5}\times\frac{9}{14}=-\frac{3}{5}+\frac{18}{5}=3\)
b)\(=\frac{55}{126}+\frac{5}{42}+\frac{4}{9}=1\)
c)\(=-\frac{51}{13}-\frac{27}{13}=-6\)
d)\(=\frac{7}{3}-11\frac{1}{4}\times\frac{2}{15}=\frac{7}{3}-\frac{3}{2}=\frac{5}{6}\)
e)\(=1\times\frac{8}{3}\times0,25=\frac{2}{3}\)
\(A=\left(2-\dfrac{3}{2}\right)\left(2-\dfrac{4}{3}\right)\left(2-\dfrac{5}{4}\right)\left(2-\dfrac{6}{5}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}\)
\(=\dfrac{1}{5}\)