Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(B=\frac{3+\frac{8}{12}+\frac{9}{13}-\frac{12}{17}}{4+\frac{8}{12}+\frac{12}{13}-\frac{16}{17}}+\frac{4+\frac{16}{60}-\frac{24}{213}-\frac{32}{11}}{5+\frac{20}{61}-\frac{36}{213}-\frac{40}{11}}\)
\(\Leftrightarrow B=\frac{3\left(1+\frac{8}{12}+\frac{3}{13}-\frac{4}{17}\right)}{4\left(1+\frac{8}{12}+\frac{3}{13}-\frac{4}{17}\right)}+\frac{4\left(1+\frac{4}{60}-\frac{6}{213}-\frac{8}{11}\right)}{5\left(1+\frac{4}{60}+\frac{6}{213}-\frac{8}{11}\right)}\)
\(\Leftrightarrow B=\frac{3}{4}+\frac{4}{5}\)
\(\Leftrightarrow B=\frac{15}{20}+\frac{16}{20}\)
\(\Leftrightarrow B=\frac{31}{20}\)
\(3\frac{14}{19}+\frac{13}{17}+\frac{35}{43}+6\)
\(=\frac{71}{19}+\frac{13}{17}+\frac{35}{43}+6\)
\(=\frac{1454}{323}+\frac{35}{43}+6\)
\(=5,...+6\)
\(=11,...\)
\(Bai2a\)\(A=\frac{\sqrt{3}-\sqrt{6}}{1-\sqrt{2}}-\frac{2+\sqrt{8}}{1+\sqrt{2}}\)
\(=\frac{\sqrt{3}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}-\frac{2\left(1+\sqrt{2}\right)}{1+\sqrt{2}}\)
\(=\sqrt{3}-2\)
\(VayA=\sqrt{3}-2\)
\(C=9\cdot\left(\frac{\frac{26299}{2023}-\frac{3757}{2023}-\frac{91}{2023}}{\frac{8092}{2023}-\frac{1156}{2023}-\frac{28}{2023}}\right):\left(\frac{\frac{840255}{21545}+\frac{168051}{21545}+\frac{6045}{21545}+\frac{27105}{21545}}{\frac{344720}{21545}+\frac{68944}{21545}+\frac{2480}{21545}+\frac{11120}{21545}}\right)\cdot\frac{158158}{164164}\)
\(C=9\cdot\left(\frac{\frac{22451}{2023}}{\frac{6908}{2023}}\div\frac{\frac{1041456}{21545}}{\frac{427264}{21545}}\right)\cdot\frac{158158}{164164}\)
\(C=9\cdot\left(\frac{13}{4}\div\frac{39}{16}\right)\cdot\frac{158158}{164164}\)
\(C=9\cdot\frac{4}{3}\cdot\frac{2\cdot79\cdot2\cdot79}{2\cdot82\cdot2\cdot82}=9\cdot\frac{4}{3}\cdot\frac{79}{82}\)
\(C=\frac{474}{41}\)
A = \(\frac{\frac{\frac{5+3.3-1.12}{12}}{3.6-5+2.2}+}{6}+\frac{16\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}\right)}{17\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}\right)}=\frac{\frac{\frac{5+9-12}{12}}{18-5+4}}{6}+\frac{16}{17}=\frac{2}{12}.\frac{6}{17}+\frac{16}{17}=\frac{1}{17}.\frac{6}{17}=1\)
A=\(\frac{\frac{5}{12}+\frac{9}{12}-\frac{12}{12}}{\frac{18}{6}-\frac{5}{6}+\frac{4}{6}}+\frac{16.\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}\right)}{17.\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}\right)}\)
=\(\frac{\frac{1}{6}}{\frac{17}{6}}+\frac{16}{17}\)
=\(\frac{1}{17}+\frac{16}{17}\)
=1
p) \(\frac{-5}{9}+\frac{8}{15}+\frac{-2}{11}+\frac{4}{-9}+\frac{7}{15}\)
\(=\left(\frac{-5}{9}+\frac{-4}{9}\right)+\left(\frac{8}{15}+\frac{7}{15}\right)+\frac{-2}{11}\)
\(=-1+1+\frac{-2}{11}\)
\(=-\frac{2}{11}\)
q) \(\frac{5}{13}+\frac{-5}{17}+\frac{-20}{41}+\frac{8}{13}+\frac{-21}{41}\)
\(=\left(\frac{5}{13}+\frac{8}{13}\right)+\left(\frac{-20}{41}+\frac{-21}{41}\right)+\frac{-5}{17}\)
\(=1+\left(-1\right)+\frac{-5}{17}=\frac{-5}{17}\)
r) \(\frac{1}{5}+\frac{-2}{9}+\frac{-7}{9}+\frac{4}{5}+\frac{16}{17}\)
\(=\left(\frac{1}{5}+\frac{4}{5}\right)+\left(\frac{-2}{9}+\frac{-7}{9}\right)+\frac{16}{17}\)
\(=1+\left(-1\right)+\frac{16}{17}=\frac{16}{17}\)
\(I=\frac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^9.6^{19}-7.2^{29}.27^6}=\frac{5.2^{30}.3^{27}-2^2.3^{20}.2^{27}}{5.2^9.2^{19}.3^{19}-7.2^{29}.3^{18}}\)
\(=\frac{5.2^{30}.3^{27}-3^{30}.2^{29}}{5.2^{28}.3^{19}-7.2^{29}.3^{18}}\)
\(=\frac{2^{29}.3^{27}.\left(5.2-3^3\right)}{2^{28}.3^{18}.\left(5.3-2.7\right)}\)
\(=\frac{2^{29}.3^{27}.-17}{2^{18}.3^{18}}\)
\(=\frac{2^9.3^9.-17}{1}\)
Ta có \(H=\frac{\left(3.4.2^{16}\right)}{11.2^{13}.4^{11}-16^9}\)
\(=\frac{3.4.2^{16}}{11.2^{13}.2^{22}-2^{36}}\)
\(=\frac{3.2^{18}}{11.2^{35}-2^{36}}\)
\(=\frac{3.2^{18}}{2^{35}.\left(11-2\right)}\)
\(=\frac{3.2^{18}}{2^{35}.3^2}\)
\(=\frac{1}{2^{17}.3}\)
\(=\frac{9}{16}.\left(\frac{13}{4}+\frac{17}{4}\right)-\frac{9}{4}.\frac{5}{16}\)
\(=\frac{9}{16}.\frac{30}{4}-\frac{9}{4}.\frac{5}{16}\)
\(=\left(\frac{9}{16}-\frac{5}{16}\right).\left(\frac{30}{4}-\frac{9}{4}\right)\)
\(=\frac{1}{4}.\frac{21}{4}=\frac{21}{16}\)
Hok tốt~