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\(A=11\dfrac{3}{13}-\left(2\dfrac{4}{7}+5\dfrac{3}{13}\right)\)
\(A=11\dfrac{3}{13}-5\dfrac{3}{13}-2\dfrac{4}{7}\)
\(A=6-2\dfrac{4}{7}\)
\(A=5\dfrac{7}{7}-2\dfrac{4}{7}\)
\(A=3\dfrac{3}{7}\)
\(B=\left(6\dfrac{4}{9}+3\dfrac{7}{11}\right)-4\dfrac{4}{9}\)
\(B=\left(6\dfrac{4}{9}-4\dfrac{4}{9}\right)+3\dfrac{7}{11}\)
\(B=2+3\dfrac{7}{11}\)
\(B=5\dfrac{7}{11}\)
\(C=\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{7}-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{-5}{7}.\left(\dfrac{2}{11}+1\right)-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{-5}{7}.\dfrac{13}{11}-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{-65}{77}-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{4}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{160}{11}\)
\(D=0,7.2\dfrac{2}{3}.20.0,375.\dfrac{5}{28}\)
\(D=\dfrac{7}{10}.\dfrac{8}{3}.20.\dfrac{375}{1000}.\dfrac{5}{28}\)
\(D=\dfrac{7}{28}=\dfrac{5}{2}\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{3}-0,25-\dfrac{1}{12}\right)\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{12}\right)\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{12}-\dfrac{1}{12}\right)\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right).0\)
\(\Rightarrow E=0\)
a) Ta có: \(\left|5\cdot0.6+\dfrac{2}{3}\right|-\dfrac{1}{3}\)
\(=\left|3+\dfrac{2}{3}\right|-\dfrac{1}{3}\)
\(=3+\dfrac{2}{3}-\dfrac{1}{3}\)
\(=3+\dfrac{1}{3}=\dfrac{10}{3}\)
b) Ta có: \(\left(0.25-1\dfrac{1}{4}\right):5-\dfrac{1}{5}\cdot\left(-3\right)^2\)
\(=\left(\dfrac{1}{4}-\dfrac{5}{4}\right)\cdot\dfrac{1}{5}-\dfrac{1}{5}\cdot9\)
\(=\dfrac{-4}{4}\cdot\dfrac{1}{5}-\dfrac{1}{5}\cdot9\)
\(=\dfrac{1}{5}\cdot\left(-1-9\right)\)
\(=-10\cdot\dfrac{1}{5}=-2\)
c) Ta có: \(\dfrac{14}{17}\cdot\dfrac{7}{5}-\dfrac{-3}{17}:\dfrac{5}{7}\)
\(=\dfrac{14}{17}\cdot\dfrac{7}{5}-\dfrac{-3}{17}\cdot\dfrac{7}{5}\)
\(=\dfrac{7}{5}\cdot\left(\dfrac{14}{17}+\dfrac{3}{17}\right)\)
\(=\dfrac{7}{5}\cdot1=\dfrac{7}{5}\)
d) Ta có: \(\dfrac{7}{16}+\dfrac{-9}{25}+\dfrac{9}{16}+\dfrac{-16}{25}\)
\(=\left(\dfrac{7}{16}+\dfrac{9}{16}\right)-\left(\dfrac{9}{25}+\dfrac{16}{25}\right)\)
\(=\dfrac{16}{16}-\dfrac{25}{25}\)
\(=1-1=0\)
e) Ta có: \(\dfrac{5}{6}+2\sqrt{\dfrac{4}{9}}\)
\(=\dfrac{5}{6}+2\cdot\dfrac{2}{3}\)
\(=\dfrac{5}{6}+\dfrac{4}{3}\)
\(=\dfrac{5}{6}+\dfrac{8}{6}=\dfrac{13}{6}\)
a) Hình như nhầm đề thì phải :v
\(P=\dfrac{\dfrac{2}{3}-\dfrac{1}{4}+\dfrac{5}{11}}{\dfrac{5}{12}+1-\dfrac{6}{11}}\)
\(=\dfrac{\dfrac{5}{12}+\dfrac{5}{11}}{\dfrac{5}{12}+\dfrac{5}{11}}=1\)
b) \(Q=\dfrac{0,125-\dfrac{1}{5}+\dfrac{1}{7}}{0,375-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-0,2}{\dfrac{3}{4}+0,5-\dfrac{3}{10}}\)
\(Q=\dfrac{0,125-\dfrac{1}{5}+\dfrac{1}{7}}{3\left(0,125-\dfrac{1}{5}+\dfrac{1}{7}\right)}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-0,2}{\dfrac{3}{4}+0,5-\dfrac{3}{10}}\)
\(Q=\dfrac{1}{3}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-0,2}{\dfrac{3}{4}+0,5-\dfrac{3}{10}}\)
\(Q=\dfrac{1}{3}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}{\dfrac{3}{2}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{3}\right)}=\dfrac{1}{3}+\dfrac{1}{\dfrac{3}{2}}\)
\(Q=\dfrac{1}{3}+\dfrac{2}{3}=1\)
a,\(P=\dfrac{\dfrac{2}{3}-\dfrac{1}{4}+\dfrac{5}{11}}{\dfrac{5}{12}+1-\dfrac{7}{11}}=\dfrac{\left(\dfrac{2}{3}-\dfrac{1}{4}+\dfrac{5}{11}\right).132}{\left(\dfrac{5}{12}+1-\dfrac{7}{11}\right).132}=\dfrac{88-33+60}{55+132-84}=\dfrac{115}{103}\)
b, Ta có : 0,125 = \(\dfrac{1}{8}\) ; 0,375 = \(\dfrac{3}{8}\) ; 0,2 = \(\dfrac{1}{5}\) ; 0,5 = \(\dfrac{3}{6}\)
\(Q=\dfrac{\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}}{\dfrac{3}{8}-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}{\dfrac{3}{4}+\dfrac{3}{6}-\dfrac{3}{10}}\)
\(Q=\dfrac{\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}}{3\cdot\left(\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}\right)}+\dfrac{2\cdot\left(\dfrac{1}{4}+\dfrac{1}{6}-\dfrac{1}{10}\right)}{3\cdot\left(\dfrac{1}{4}+\dfrac{1}{6}-\dfrac{1}{10}\right)}\)
\(Q=\dfrac{1}{3}+\dfrac{2}{3}=1\)
`=(0,125-1/5+1/7)/(3(0,125-1/5+1/7))+(2(1/4+1/6-0,1))/(3(1/4+1/6-0,1))`
`=1/3+2/3=3/3=1`
Ta có:\(\dfrac{\dfrac{5}{12}+\dfrac{3}{4}-1}{3-\dfrac{5}{6}+\dfrac{2}{3}}=\dfrac{\dfrac{5}{12}+\dfrac{9}{12}-\dfrac{12}{12}}{\dfrac{18}{6}-\dfrac{5}{6}+\dfrac{4}{6}}=\dfrac{1}{6}:\dfrac{17}{6}=\dfrac{1}{17}\)
Ta có\(\dfrac{\dfrac{16}{5}+\dfrac{16}{7}-\dfrac{16}{9}}{\dfrac{17}{5}+\dfrac{17}{7}-\dfrac{17}{9}}=\dfrac{16\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}\right)}{17\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}\right)}=\dfrac{16}{17}\)
Ta có:A=\(\dfrac{1}{17}+\dfrac{16}{17}=\dfrac{17}{17}=1\)
Vậy gt bt A=1
Các bạn không cần trả lời câu hỏi trên của mik vì mik đã hiểu rồi nha . Cho nên đừng trả lời ! OK
1. Tính hợp lí
a) \(0,7+\dfrac{-7}{19}-\left(-0,3\right)\)
\(=\dfrac{7}{10}+\dfrac{-7}{19}+\dfrac{3}{10}\)
\(=\left(\dfrac{7}{10}+\dfrac{3}{10}\right)+\dfrac{-7}{19}\)
\(=1+\dfrac{-7}{19}\)
\(=\dfrac{12}{19}\)
b) \(\dfrac{5}{3}.\left(-2,5\right):\dfrac{5}{6}\)
\(=\dfrac{5}{3}.\dfrac{-5}{2}.\dfrac{6}{5}\)
\(=\left(\dfrac{5}{3}.\dfrac{6}{5}\right).\dfrac{-5}{2}\)
\(=2.\dfrac{-5}{2}\)
\(=-5\)
c) \(0,6.\dfrac{-5}{17}-\dfrac{3}{5}.\dfrac{12}{17}\)
\(=\dfrac{3}{5}.\dfrac{-5}{17}-\dfrac{3}{5}.\dfrac{12}{17}\)
\(=\dfrac{3}{5}.\left(\dfrac{-5}{17}-\dfrac{12}{17}\right)\)
\(=\dfrac{3}{5}.-1\)
\(=\dfrac{-3}{5}\)
d) \(\dfrac{7}{4}.\dfrac{5}{2}-\dfrac{7}{4}.\dfrac{3}{2}\)
\(=\dfrac{7}{4}.\left(\dfrac{5}{2}-\dfrac{3}{2}\right)\)
\(=\dfrac{7}{4}.1\)
\(=\dfrac{7}{4}\)
Chúc bạn học tốt
A = \(\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\) . \(\left(\dfrac{1}{3}-0,25-\dfrac{1}{12}\right)\)
A = \(\left(-6,17+\dfrac{32}{9}-\dfrac{230}{97}\right)\) . \(\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{12}\right)\)
A = \(\left(-6,17+\dfrac{32}{9}-\dfrac{230}{97}\right)\) . \(\left(\dfrac{4}{12}-\dfrac{3}{12}-\dfrac{1}{12}\right)\)
A = \(\left(-6,17+\dfrac{32}{9}-\dfrac{230}{97}\right)\) . 0
A = 0*
*Vì số nào nhân với 0 cũng bằng 0 nên không cần tính kết quả của phép tính\(\left(-6,17+\dfrac{32}{9}-\dfrac{230}{97}\right)\)