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a)\(\left(\frac{3}{29}-\frac{1}{5}\right)\cdot\frac{29}{3}\)
\(=\left(\frac{3}{29}\cdot\frac{29}{3}\right)-\left(\frac{1}{5}\cdot\frac{29}{3}\right)\)
\(=1-\frac{29}{15}\)
\(=\frac{-14}{15}\)
b)\(\frac{16}{15}\cdot\frac{-5}{14}\cdot\frac{54}{24}\cdot\frac{56}{21}\)
=\(=\frac{16\cdot\left(-5\right)\cdot54\cdot56}{15\cdot14\cdot24\cdot21}\)
\(=\frac{2^4\cdot\left(-5\right)\cdot2\cdot3^3\cdot2^3\cdot7}{3\cdot5\cdot7\cdot2\cdot2^3\cdot3\cdot7}\)
\(=2^4\)
c)\(\frac{37}{7}\cdot\frac{8}{11}+\frac{37}{7}\cdot\frac{5}{11}-\frac{37}{7}\cdot\frac{2}{11}\)
\(=\frac{37}{7}\cdot\left(\frac{8}{11}+\frac{5}{11}-\frac{2}{11}\right)\)
\(=\frac{37}{7}\cdot1\)
\(=\frac{37}{7}\)
Đúng nhớ k nhen!
1) A = \(\frac{-15}{19}.\frac{23}{37}+\frac{14}{37}.\frac{15}{19}=\frac{15}{19}.\frac{-23}{37}+\frac{14}{37}.\frac{15}{19}=\frac{15}{19}.\left(\frac{-23}{37}+\frac{14}{37}\right)=\frac{15}{19}.\frac{-9}{37}=\frac{-135}{703}\)
1.\(\frac{5}{7}.\frac{5}{11}+\frac{5}{7}.\frac{2}{11}-\frac{5}{7}.\frac{14}{11}\)
\(=\frac{5}{7}.\left(\frac{5}{11}+\frac{2}{11}-\frac{14}{11}\right)\)
\(=\frac{5}{7}.\frac{-7}{11}=\frac{5.\left(-7\right)}{7.11}=\frac{5.\left(-1\right)}{1.11}=\frac{-5}{11}\)
\(C=\frac{-3}{7}.\frac{5}{9}+\frac{4}{9}.\frac{-3}{7}+2\frac{3}{7}\)
\(=\frac{-3}{7}.\left(\frac{5}{9}+\frac{4}{9}\right)+2\frac{3}{7}\)
\(=\frac{-3}{7}.1+2\frac{3}{7}=\frac{-3}{7}+2\frac{3}{7}=2\)
Trả lời
a)
\(x^2:\frac{16}{11}=\frac{11}{4}\)
\(\Leftrightarrow x^2=\frac{11}{4}\cdot\frac{16}{11}\)
\(\Leftrightarrow x^2=\frac{16}{4}\)
\(\Leftrightarrow x^2=\left(\frac{4}{2}\right)^2\)
\(\Leftrightarrow x=\frac{4}{2}\)
Vậy x=\(\frac{4}{2}\)
b) (bạn thiếu nhóm \(\frac{1}{10\cdot13}\))
Đặt \(A=\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+\frac{1}{10\cdot13}+\frac{1}{13\cdot16}+\frac{1}{16\cdot19}\)
\(\Rightarrow3A=3\left(\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+\frac{1}{10\cdot13}+\frac{1}{13\cdot16}+\frac{1}{16\cdot19}\right)\)
\(\Rightarrow3A=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+\frac{3}{13\cdot16}+\frac{3}{16\cdot19}\)
\(\Rightarrow3A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+\frac{1}{16}-\frac{1}{19}+\frac{1}{19}\)
\(\Rightarrow3A=1-\frac{1}{19}\Leftrightarrow3A=\frac{18}{19}\)
\(\Rightarrow A=\frac{18}{19}:3\Leftrightarrow A=\frac{6}{19}\)
1) ( \(\frac{55}{3}\): 15 + \(\frac{26}{3}\) . \(\frac{7}{2}\)) : [(\(\frac{37}{3}\) + \(\frac{62}{7}\)) . \(\frac{7}{18}\)] : \(\frac{-1704}{445}\)
= ( \(\frac{55}{3}\). \(\frac{1}{15}\) + \(\frac{91}{3}\)) : [ \(\frac{445}{21}\) . \(\frac{7}{18}\)] . \(\frac{-445}{1704}\)
= ( \(\frac{11}{9}\)+ \(\frac{91}{3}\)) : \(\frac{445}{54}\). \(\frac{-445}{1704}\) = \(\frac{284}{9}\). \(\frac{54}{445}\). \(\frac{-445}{1704}\)= \(\frac{284}{9}\). (\(\frac{54}{445}\). \(\frac{-445}{1704}\))
= \(\frac{284}{8}\). \(\frac{-9}{284}\)
= \(\frac{-9}{8}\)
\(x^2:\frac{16}{11}=\frac{11}{4}\)
\(x^2=\frac{16}{11}.\frac{11}{4}\)
\(x^2=4\)
\(=>x=2\)
\(x^2\)\(\div\)\(\frac{16}{11}\)=\(\frac{11}{4}\)
\(x^2\)= \(\frac{11}{4}\)\(\cdot\frac{16}{11}\)
\(x^2\)= 4
\(x^2\)=\(2^2\)
\(\Rightarrow\)\(x=2\)
Vậy \(x=2\)