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a) \(=\frac{6}{7}+\frac{5}{8}:5-\frac{3}{16}.4\)
\(=\frac{6}{7}+\frac{1}{8}-\frac{3}{4}\)
\(=\frac{13}{56}\)
b) \(=\frac{3}{2}+\frac{1}{2}.\frac{7}{18}:\frac{7}{12}\)
\(=\frac{3}{2}+\frac{1}{3}\)
\(=\frac{11}{6}\)
Ta có : \(A=8\frac{2}{7}-\left(3\frac{4}{9}+4\frac{2}{7}\right)\)
\(\Rightarrow A=\frac{58}{7}-\left(\frac{31}{9}+\frac{30}{7}\right)\)
\(\Rightarrow A=\frac{58}{7}-\frac{487}{63}=\frac{5}{9}\)
P/s:Câu B tương tự nhé
Tiếp B của @Phạm Tuấn Đạt
\(B=\left(10\frac{2}{9}+2\frac{3}{5}\right)-6\frac{2}{9}\)
\(\Rightarrow B=\left(\frac{92}{9}+\frac{13}{5}\right)-\frac{56}{9}\)
\(B=\left(\frac{92}{9}-\frac{56}{9}\right)+\frac{13}{5}\)
\(B=\frac{36}{9}+\frac{13}{5}\)
\(B=4+\frac{13}{5}\)
\(B=\frac{20}{5}+\frac{13}{5}=\frac{33}{5}\)
\(A=3+3^3+3^5+...+3^{75}\)
\(< =>9A=3^3+3^5+3^7+...+3^{77}\)
\(< =>8A=3^{77}-3< =>A=\frac{3^{77}-3}{8}\)
Mình cảm ơn bạn Amasterasu nhưng mk k hiểu cho lắm, bạn giúp mình làm cụ thể hơn ở đoạn từ 9A sao ra được 8A như vậy thế?
Ta có : \(\frac{5^4+2^2.5^2-125}{2^2.3}=\frac{5^4+5^2.2^2-5^3}{8.3}=\frac{5^2\left(5^2+2^2-5\right)}{24}=\frac{5^2.24}{24}=5^2=25\)
a,\(\frac{21}{25}.\frac{11}{9}.\frac{5}{7}=\frac{21.11.5}{25.9.7}=\frac{3.7.11.5}{5^2.3^2.7}=\frac{11}{5.3}=\frac{11}{15}\)
b,\(\frac{5}{23}.\frac{17}{26}+\frac{5}{23}.\frac{9}{26}=\frac{5}{23}.\left(\frac{17}{26}+\frac{9}{26}\right)=\frac{5}{23}.1=\frac{5}{23}\)
c, \(\left(\frac{3}{29}-\frac{1}{5}\right).\frac{29}{3}=\frac{3}{29}.\frac{29}{3}-\frac{1}{5}.\frac{29}{3}=1-\frac{29}{15}=-\frac{14}{15}\)
a , \(\frac{21}{25}\times\frac{11}{9}\times\frac{5}{7}\)
\(=\frac{21\times11\times5}{25\times9\times7}\)
\(=\frac{3\times7\times11\times5}{5\times5\times3\times3\times7}\)
\(=\frac{11}{5\times3}\)
\(=\frac{11}{15}\)
b , \(\frac{5}{23}\times\frac{17}{26}+\frac{5}{23}\times\frac{9}{26}\)
\(=\frac{5}{23}\times\left(\frac{17}{26}+\frac{9}{26}\right)\)
\(=\frac{5}{23}\times\frac{26}{26}\)
\(=\frac{5}{23}\times1\)
\(=\frac{5}{23}\)
c , \(\left(\frac{3}{29}-\frac{1}{5}\right)\times\frac{29}{3}\)
\(=\frac{3}{29}\times\frac{29}{3}-\frac{1}{5}\times\frac{29}{3}\)
\(=1-\frac{29}{15}\)
\(\frac{\frac{2}{3}+\frac{2}{5}-\frac{2}{9}}{\frac{4}{3}+\frac{4}{5}-\frac{4}{9}}=\frac{2\left(\frac{1}{3}+\frac{1}{5}-\frac{1}{9}\right)}{4\left(\frac{1}{3}+\frac{1}{5}-\frac{1}{9}\right)}=\frac{2}{4}=\frac{1}{2}\)
\(B=\frac{125^2.72^3:9^3}{4^3.25^5:5^5}=\frac{125^2.\left(72:9\right)^3}{4^3.\left(25:5\right)^5}=\frac{5^6.2^9}{2^6.5^5}=\frac{5.2^3}{1}=5.8=40\)