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a)\(\left(17\frac{13}{15}-3\frac37\right)-\left(2\frac{12}{15}-4\right)\)
=\(17\frac{13}{15}-3\frac37-2\frac{12}{15}+4\)
=\(\left(17-2-3+4\right)+\left(\frac{13}{15}-\frac{12}{15}-\frac37\right)\)
=\(16+\left(\frac{1}{15}-\frac37\right)\)
=\(16+\left(\frac{7}{105}-\frac{45}{105}\right)\)
=\(16-\frac{38}{105}\)
=\(\frac{1680}{105}-\frac{38}{105}\)
=\(\frac{1642}{105}\)
b)\(\left(3\frac29\times\frac{15}{23}\times1\frac{7}{29}\right):\frac{5}{23}\)
=\(\left(\frac{29}{9}\times\frac{15}{23}\times\frac{36}{29}\right)\times\frac{23}{5}\)
=\(\left(\frac{29}{9}\times\frac{36}{29}\right)\times\left(\frac{15}{23}\times\frac{23}{5}\right)\)
=4x3
=12
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)\(2\frac{17}{20}-1\frac{11}{5}+6\frac{9}{20}:3\)
= \(\frac{57}{20}-\frac{16}{5}+\frac{129}{20}\times\frac13\)
=\(\frac{57}{20}-\frac{16}{5}+\frac{129}{60}\)
=\(\frac{171}{60}-\frac{192}{60}+\frac{129}{60}\)
=\(\frac{108}{60}\)
=\(\frac95\)
2: \(=\dfrac{0.8}{\dfrac{16}{25}-\dfrac{1}{25}}+\dfrac{\dfrac{71}{75}\cdot\dfrac{7}{4}}{\dfrac{119}{36}\cdot\dfrac{36}{17}}\)
\(=\dfrac{4}{5}\cdot\dfrac{5}{3}+\dfrac{71}{300}=\dfrac{471}{300}=\dfrac{157}{100}\)
3: \(=\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{2}{6}-\dfrac{2}{8}+\dfrac{2}{10}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}\)
=2/7-2/7=0
a)\(\frac{-5}{13}+\left(\frac{3}{5}+\frac{3}{13}-\frac{4}{10}\right)=\frac{-5}{13}-\frac{3}{5}-\frac{3}{13}+\frac{4}{10}=\left(\frac{-5}{13}-\frac{3}{13}\right)+\frac{4}{10}-\frac{3}{5}=\frac{-5-3}{13}+\left(\frac{4}{10}-\frac{6}{10}\right)=\frac{-8}{13}+\frac{-2}{10}=\frac{-80}{130}+\frac{-26}{130}=\frac{-106}{130}=\frac{-53}{65}\)
\(\left(2^{17}+17^2\right).\left(9^{15}-15^9\right).\left(4^2-2^4\right)\)
Xét \(4^2-2^4=\left(2^2\right)^2-2^4=2^4-2^4=0\)
Theo quy tắc, một thừa số trong phép nhân đó bằng 0 thì cả tích đó bằng 0
\(\Rightarrow\left(2^{17}+17^2\right).\left(9^{15}-15^9\right).\left(4^2-2^4\right)=0\)
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