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\(\frac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^9.6^{19}-7.2^{29}.27^6}\)=2
Chào bạn, bạn hãy theo dõi bài giải của mình nhé!
\(P=\frac{\frac{2}{3}-\frac{1}{4}+\frac{5}{11}}{\frac{5}{12}+1-\frac{7}{11}}=\frac{\frac{88}{132}-\frac{33}{132}+\frac{60}{132}}{\frac{55}{132}+\frac{132}{132}-\frac{84}{132}}=\frac{\frac{115}{132}}{\frac{103}{132}}=\frac{115}{132}:\frac{103}{132}=\frac{115}{132}\cdot\frac{132}{103}=\frac{115\cdot132}{132\cdot103}=\frac{115}{103}\)
Chúc bạn học tốt!
\(M=\left(\dfrac{3}{2}-\dfrac{5}{6}+\dfrac{7}{12}-\dfrac{17}{72}\right)+\left(-\dfrac{9}{20}+\dfrac{11}{30}\right)+\left(\dfrac{-13}{42}+\dfrac{15}{56}\right)\)
\(=\dfrac{108}{72}-\dfrac{60}{72}+\dfrac{42}{72}-\dfrac{17}{72}+\dfrac{-27}{60}+\dfrac{22}{60}+\dfrac{-52}{168}+\dfrac{45}{168}\)
\(=\dfrac{73}{72}-\dfrac{1}{12}-\dfrac{1}{24}=\dfrac{73}{72}-\dfrac{6}{72}-\dfrac{3}{72}=\dfrac{64}{72}=\dfrac{8}{9}\)
\(=\left(1+\frac{1}{2}\right)-1+\frac{1}{6}+\left(\frac{1}{2}+\frac{1}{12}\right)-\frac{1}{2}+\frac{1}{20}+\left(\frac{1}{3}+\frac{1}{30}\right)-\frac{1}{3}+\frac{1}{42}+\left(\frac{1}{4}+\frac{1}{56}\right)-\frac{1}{4}+\frac{1}{72}\)
=\(=\left(1-1+\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}\right)+\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{72}\right)\)
\(=0+\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{8\cdot9}\right)=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}=\left(1-\frac{1}{9}\right)+\left(\frac{1}{2}-\frac{1}{2}\right)+...+\left(\frac{1}{8}-\frac{1}{8}\right)\)\(=\left(\frac{9}{9}-\frac{1}{9}\right)+0+...+0=\frac{8}{9}\)
\(C=\frac{5^2}{1.6}+\frac{5^2}{6.11}+...+\frac{5^2}{26.31}\)
\(=5.\left(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{26.31}\right)\)
\(=5.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{26}-\frac{1}{31}\right)\)
\(=5.\left(1-\frac{1}{31}\right)\)
\(=5.\frac{29}{31}\)
\(=\frac{145}{31}\)
đề sai