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\(\left(\frac{-4}{13}+\frac{4}{13}+\frac{-12}{13}\right)\times\left(\frac{5}{17}+\frac{4}{17}\right)\)
\(=\)\(\frac{-12}{13}\times\left(\frac{5}{17}+\frac{4}{17}\right)\)
\(=\)\(\frac{-12}{13}\times\frac{9}{17}=\frac{-106}{221}\)
a) \(\frac{1}{12}+\frac{3}{15}+\frac{11}{12}+\frac{1}{71}-\frac{12}{10}=\left(\frac{1}{12}+\frac{11}{12}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)+\frac{1}{71}\)
\(=\frac{12}{12}+0+\frac{1}{71}=1+\frac{1}{71}=1\frac{1}{71}=\frac{72}{71}\)
b) \(\frac{2}{3}-4\left(\frac{1}{2}+\frac{3}{4}\right)=\frac{2}{3}-4.\frac{5}{4}=\frac{2}{3}-5=\frac{2}{3}-\frac{15}{3}=-\frac{13}{3}\)
c) \(\frac{-4}{13}.\frac{3}{17}+\frac{-12}{13}.\frac{4}{7}+\frac{4}{13}=\frac{4}{13}.\frac{-3}{17}+\frac{4}{13}.\frac{-12}{17}+\frac{4}{13}.1\)
\(=\frac{4}{13}\left(\frac{-3}{17}+\frac{-12}{17}+1\right)=\frac{4}{13}\left(\frac{-15}{17}+\frac{17}{17}\right)=\frac{4}{13}.\frac{2}{17}=\frac{8}{221}\)
d) \(\frac{10^3+2.5+5^3}{55}=\frac{1000+10+125}{55}=\frac{1135}{55}=\frac{227}{11}\)
a)\(\frac{3^6.45^4-15^{13}.5^{-9}}{27^4.25^3+45^6}=\frac{3^6.\left(3^2.5\right)^4-\left(3.5\right)^{13}.5^{-9}}{\left(3^3\right)^4.\left(5^2\right)^3+\left(3^2.5\right)^6}=\frac{3^6.3^8.5^4-3^{13}.5^{13}.5^{-9}}{3^{12}.5^6+3^{12}.5^6}\)
\(=\frac{3^{14}.5^4-3^{13}.5^4}{3^{12}.5^6+3^{12}.5^6}=\frac{3^{13}.5^4.\left(3-1\right)}{3^{12}.5^6\left(1+1\right)}=\frac{3^{13}.5^4}{3^{12}.5^6}=\frac{3}{5^2}=\frac{3}{25}\)
b)\(\frac{4^6.9^5+6^9.120}{-8^4.3^{12}-6^{11}}=\frac{\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^9.2^3.3.5}{-\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}=\frac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{-2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{-2^{12}.3^{12}-2^{11}.3^{11}}=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{-\left(2^{12}.3^{12}+2^{11}.3^{11}\right)}=\frac{2^{12}.3^{10}\left(1+5\right)}{-\left[2^{11}.3^{11}\left(2.3+1\right)\right]}=\frac{2.6}{-\left(3.7\right)}=\frac{4}{-7}\)
Phạm Hồng Anh k tớ nhé
a) \(\frac{6^{15}+6^{13}}{6^{12}}=\frac{6^{13}.\left(6^2+1\right)}{6^{12}}=6.37=222\)
b) \(\frac{2^6.5^7}{10^6}=\frac{2^6.5^6.5}{10^6}=\frac{10^6.5}{10^6}=5\)
a, Ta có: \(A=\frac{6^{15}+6^{13}}{6^{12}}=\frac{6^{12}\left(6^3+6\right)}{6^{12}}=6^3+6\) \(=222\)
b, Ta có: \(B=\frac{2^6.5^7}{10^6}=\frac{2^6.5^6.5}{10^6}=\frac{\left(2.5\right)^6.5}{10^6}\) \(=\frac{10^6.5}{10^6}=5\)
\(a.\)
\(\frac{15}{33}+\frac{7}{20}+\frac{18}{33}+\frac{13}{20}\)
\(=\left(\frac{15}{33}+\frac{18}{33}\right)+\left(\frac{13}{20}+\frac{7}{20}\right)\)
\(=\frac{33}{33}+\frac{20}{20}\)
\(=1+1=2\)
\(b.\)
\(2\frac{1}{2}+\frac{4}{7}:\left(-\frac{8}{21}\right)\)
\(=\frac{5}{2}+\frac{4}{7}:\left(-\frac{8}{21}\right)\)
\(=\frac{5}{2}+\frac{4}{7}.\left(-\frac{21}{8}\right)\)
\(=\frac{5}{2}+\frac{1}{1}.\left(-\frac{3}{2}\right)\)
\(=\frac{5}{2}-\frac{3}{2}\)
\(=1\)
\(c.\)
\(\left(-\frac{1}{2}\right)^3+\frac{1}{2}:5\)
\(=-\frac{1}{8}+\frac{1}{2}.\frac{1}{5}\)
\(=-\frac{1}{8}+\frac{1}{10}\)
\(=-\frac{1}{40}\)
a) \(\frac{15}{33}+\frac{7}{20}+\frac{18}{33}+\frac{13}{20}=\left(\frac{15}{33}+\frac{18}{33}\right)+\left(\frac{7}{20}+\frac{13}{20}\right)\) = 1 + 1 = 2
b) \(2\frac{1}{2}+\frac{4}{7}:\left(\frac{-8}{21}\right)=\frac{5}{2}+\frac{4}{7}:\left(\frac{-8}{21}\right)=\frac{5}{2}+\frac{-3}{2}\) = 1
c) \(\left(\frac{-1}{2}\right)\)3 + \(\frac{1}{2}\) : 5 = \(\frac{-1}{8}+\frac{1}{10}\) = \(\frac{-1}{40}\)
Chúc bạn học tốt!
\(=\dfrac{-80}{99}+\dfrac{70}{324}=-\dfrac{1055}{1762}\)
a) \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
Mà \(\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)\ne0\)
nên x + 1 = 0 => x = -1
Vậy x = -1
b) \(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(1+\frac{x+4}{2000}+1+\frac{x+3}{2001}=1+\frac{x+2}{2002}+1+\frac{x+1}{2003}\)
\(\frac{2004+x}{2000}+\frac{2004+x}{2001}=\frac{2004+x}{2002}+\frac{2004+x}{2003}\)
\(\frac{2004+x}{2000}+\frac{2004+x}{2001}-\frac{2004+x}{2002}-\frac{2004+x}{2003}=0\)
\(\left(2004+x\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
Mà \(\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)\ne0\)
nên 2004 + x = 0 => x = -2004
Vậy x = -2004
=))
\(\frac{13}{20}\cdot\left(-7\frac{1}{2}\right)-\frac{13}{20}\cdot12\frac{1}{2}=\frac{13}{20}\left(-7\frac{1}{2}-12\frac{1}{2}\right)=\frac{13}{20}\cdot\left(-20\right)=-13\)
\(\frac{13}{20}\) . ( -7 \(\frac{1}{2}\)) - \(\frac{13}{20}\). 12\(\frac{1}{2}\)
=\(\frac{13}{20}\). ( -7 \(\frac{1}{2}\)-12\(\frac{1}{2}\))
=\(\frac{13}{20}\).\(\frac{-19}{2}\)
=\(\frac{-247}{40}\)
\(\frac{3}{14}:\frac{1}{28}-\frac{13}{21}:\frac{1}{28}-8=\left(\frac{3}{14}-\frac{13}{21}\right):\frac{1}{28}-8=-\frac{17}{42}.28-8=-\frac{34}{3}-8=-\frac{58}{3}\)
\(\frac{3}{14}:\frac{1}{28}-\frac{13}{21}:\frac{1}{28}-8\)
\(=\left(\frac{3}{14}-\frac{13}{21}\right):\frac{1}{28}-8\)
\(=-\frac{17}{42}:\frac{1}{28}-8\)
\(=-\frac{34}{3}-8\)
\(=-\frac{58}{3}\)
\(\frac{8^{13}}{4^{10}}=\frac{\left(2^3\right)^{13}}{\left(2^2\right)^{10}}=\frac{2^{39}}{2^{20}}=2^{19}=524288\)
Trả lời:
\(\frac{8^{13}}{4^{10}}=\frac{\left(2^3\right)^{13}}{\left(2^2\right)^{10}}=\frac{2^{39}}{2^{20}}=2^{19}\)
=)