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\(A = {1\over2}-{3\over4}+{5\over6}-{7\over12}={6\over12}-{9\over12}+{10\over12}-{7\over12}\)\(={0\over12}=0\)
\(A=\frac{-5}{12}+\frac{4}{37}+\frac{17}{12}-\frac{41}{37}=(\frac{-5}{12}+\frac{17}{12})+(\frac{4}{37}-\frac{41}{37})=\frac{12}{12}+\frac{-37}{37}=1+(-1)=0\)
\(B=\frac{1}{2}-\frac{43}{101}+\frac{-1}{3}-\frac{1}{6}=\frac{-43}{101}+(\frac{1}{2}+\frac{-1}{3}-\frac{1}{6})=\frac{-43}{101}+(\frac{3}{6}+\frac{-2}{6}-\frac{1}{6})=\frac{-43}{101}+0=\frac{-43}{101}\)
\(A=\frac{-5}{12}+\frac{4}{37}+\frac{17}{12}-\frac{41}{37}.\)
\(A=\left(\frac{-5}{12}+\frac{17}{12}\right)-\left(\frac{41}{37}-\frac{4}{37}\right)\)
\(A=1-1=0\)
\(B=\frac{1}{2}-\frac{43}{101}+\left(\frac{-1}{3}\right)-\frac{1}{6}\)
\(B=\left(\frac{1}{2}+\left(\frac{-1}{3}\right)-\frac{1}{6}\right)-\frac{43}{101}\)
\(A=0-\frac{43}{101}=\frac{-43}{101}\)
\(C=\frac{-5}{6}\cdot\frac{12}{-7}\cdot-\frac{21}{15}\)
\(C=\frac{-5}{2.3}\cdot\frac{3.2.2}{-7}\cdot\frac{3.\left(-7\right)}{3.5}\)
\(C=\frac{-2}{1}=-2\)
a) \(\frac{17}{9}-\frac{17}{9}:\left(\frac{7}{3}+\frac{1}{2}\right)\)
= \(\frac{17}{9}-\frac{17}{9}:\frac{17}{6}\)
= \(\frac{17}{9}-\frac{2}{3}\)
= \(\frac{11}{9}\)
b) \(\frac{4}{3}.\frac{2}{5}-\frac{3}{4}.\frac{2}{5}\)
= \(\frac{2}{5}.\left(\frac{4}{3}-\frac{3}{4}\right)\)
= \(\frac{2}{5}.\frac{7}{12}\)
= \(\frac{7}{30}\)
Mình lười làm quá, hay mình nói kết quả cho bn thôi nha
c) -6
d) 3
e) 3
g) 12
h) \(\frac{23}{18}\)
i) \(\frac{-69}{20}\)
k) \(\frac{-1}{2}\)
l) \(\frac{49}{5}\)
b) \(\frac{\frac{2}{3}+\frac{5}{7}+\frac{4}{21}}{\frac{5}{6}+\frac{11}{7}-\frac{7}{21}}\)
\(=\frac{\frac{29}{21}+\frac{4}{21}}{\frac{101}{42}-\frac{7}{21}}\)
\(=\frac{\frac{11}{7}}{\frac{29}{14}}\)
\(=\frac{22}{29}.\)
Chúc bạn học tốt!
\(E=\frac{1}{13}+\left(\frac{-5}{18}-\frac{1}{13}+\frac{12}{17}\right)-\left(\frac{12}{17}-\frac{5}{18}+\frac{7}{5}\right)\)
\(=\frac{1}{13}+\frac{-5}{18}-\frac{1}{13}+\frac{12}{17}-\frac{12}{17}+\frac{5}{18}-\frac{7}{5}\)
\(=\left(\frac{1}{13}-\frac{1}{13}\right)+\left(\frac{-5}{18}+\frac{5}{18}\right)+\left(\frac{12}{17}-\frac{12}{17}\right)-\frac{7}{5}\)
\(=0+0+0-\frac{7}{5}\)
\(=-\frac{7}{5}\)
\(E=\frac{1}{13}+\left(\frac{-5}{18}-\frac{1}{13}+\frac{12}{17}\right)-\left(\frac{12}{17}-\frac{5}{18}+\frac{7}{5}\right)\)
\(=\frac{1}{13}+\frac{-5}{18}-\frac{1}{13}+\frac{12}{17}-\frac{-12}{17}+\frac{5}{18}-\frac{7}{5}\)
\(=\left(\frac{1}{13}-\frac{1}{13}\right)+\left(\frac{-5}{18}+\frac{5}{18}\right)+\left(\frac{12}{17}-\frac{-12}{17}\right)-\frac{7}{5}\)
\(=0+0+\frac{24}{17}-\frac{7}{5}\)
\(=\frac{24}{17}-\frac{7}{5}\)
\(=\frac{120}{85}-\frac{119}{85}=\frac{1}{85}\)
c )
\(1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{2}}}}=1+\frac{1}{1+\frac{1}{1+\frac{1}{\frac{3}{2}}}}=1+\frac{1}{1+\frac{1}{1+\frac{5}{3}}}=1+\frac{1}{1+\frac{1}{\frac{8}{3}}}=1+\frac{1}{\frac{11}{8}}=\frac{19}{11}\)
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) \(\frac{3}{5}.x-\frac{1}{5}=\frac{4}{5}\)
\(\Leftrightarrow\frac{3}{5}.x=\frac{4}{5}+\frac{1}{5}\)
\(\Leftrightarrow\frac{3}{5}.x=1\)
\(\Leftrightarrow x=1:\frac{3}{5}\)
\(\Leftrightarrow x=\frac{5}{3}\)
Vậy : \(x=\frac{5}{3}\)
b) \(\frac{4}{7}+\frac{5}{7}:x=1\)
\(\Leftrightarrow\frac{5}{7}:x=1-\frac{4}{7}\)
\(\Leftrightarrow\frac{5}{7}:x=\frac{3}{7}\)
\(\Leftrightarrow x=\frac{5}{7}:\frac{3}{7}\)
\(\Leftrightarrow x=\frac{5}{3}\)
Vậy : \(x=\frac{5}{3}\)
c) \(-\frac{12}{7}.\left(\frac{3}{4}-x\right).\frac{1}{4}=-1\)
\(\Leftrightarrow\frac{-12.1}{7.4}.\left(\frac{3}{4}-x\right)=-1\)
\(\Leftrightarrow-\frac{3}{7}.\left(\frac{3}{4}-x\right)=-1\)
\(\Leftrightarrow\frac{3}{4}-x=-1:\left(-\frac{3}{7}\right)\)
\(\Leftrightarrow\frac{3}{4}-x=\frac{7}{3}\)
\(\Leftrightarrow x=\frac{3}{4}-\frac{7}{3}=-\frac{19}{12}\)
Vậy : \(x=-\frac{19}{12}\)
d) \(x:\frac{17}{8}=-\frac{2}{5}.-\frac{9}{17}+3\)
\(\Leftrightarrow x:\frac{17}{8}=\frac{273}{85}\)
\(\Leftrightarrow x=\frac{273}{85}.\frac{17}{8}\)
\(\Leftrightarrow x=\frac{273}{40}\)
Vậy : \(x=\frac{273}{40}\)
\(\)
\(\frac{12}{17}.\frac{5}{7}-\frac{-12}{17}.\frac{1}{7}+\frac{1}{17.7}.12\)
\(=\frac{12}{17}.\frac{5}{7}+\frac{12}{17}.\frac{1}{7}+\frac{1}{17}.\frac{1}{7}.12\)
\(=\frac{12}{17}.\frac{5}{7}-\frac{-12}{17}.\frac{1}{7}+\frac{12}{17}.\frac{1}{7}\)
\(=\frac{12}{17}.\left(\frac{5}{7}+\frac{1}{7}+\frac{1}{7}\right)\)
\(=\frac{12}{17}.1\)
\(=\frac{12}{17}\)