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Bài 1:
a) \(\frac{16}{15}.\frac{\left(-5\right)}{14}.\frac{54}{24}.\frac{56}{21}\)
\(=\frac{4.2.2}{5.3}.\frac{\left(-5\right)}{2.7}.\frac{3.3}{4}.\frac{8}{3}\)
\(=\frac{4.2.2.\left(-5\right).3.3.8}{5.3.2.7.4.3}\)
\(=\frac{-16}{7}\)
b) \(\frac{7}{3}.\frac{\left(-5\right)}{2}.\frac{15}{21}.\frac{4}{\left(-5\right)}\)
\(=\frac{7}{3}.\frac{\left(-5\right)}{2}.\frac{5}{7}.\frac{2.2}{\left(-5\right)}\)
\(=\frac{7.\left(-5\right).5.2.2}{3.2.7.\left(-5\right)}\)
\(=\frac{10}{3}\)
Bài 2:
a) \(\frac{21}{24}.\frac{11}{9}.\frac{5}{7}=\frac{7}{8}.\frac{11}{9}.\frac{5}{7}=\frac{11.5}{8.9}=\frac{55}{72}\)
b) \(\frac{5}{23}.\frac{17}{26}+\frac{5}{23}.\frac{9}{26}\)
\(=\frac{5}{23}.\left(\frac{17}{26}+\frac{9}{26}\right)=\frac{5}{23}.1=\frac{5}{23}\)
c) \(\left(\frac{3}{29}-\frac{1}{5}\right).\frac{29}{3}=\frac{3}{29}.\frac{29}{3}-\frac{1}{5}.\frac{29}{3}\)
\(=1-1\frac{14}{15}=\frac{14}{15}\)
Bài 3:
a) x/5 = 2/5
=> x =2
b) -4/x = 20/14 = 10/7
=> -4/x = 10/7
=> x.10 = (-4).7
x.10 = - 28
x= -28 :10
x= -2,8
c) 4/7 = 12/x = 12/ 21
=> 12/x = 12/21
=> x = 21
d) 3/7 = x / 21 = 9/21
=> x/21 = 9/21
=> x= 9
-37 + 14 +26 + 37
=-37 + 37 + 14 + 36
=0 + 50
=50
-24 + 6 + 10 + 24
=(-24 + 24) +(6+10)
= 0+16
= 16
a; - \(\dfrac{10}{13}\) + \(\dfrac{5}{17}\) - \(\dfrac{3}{13}\) + \(\dfrac{12}{17}\) - \(\dfrac{11}{20}\)
= - (\(\dfrac{10}{13}\) + \(\dfrac{3}{13}\)) + (\(\dfrac{5}{17}\) + \(\dfrac{12}{17}\)) - \(\dfrac{11}{20}\)
= - 1 + 1 - \(\dfrac{11}{20}\)
= 0 - \(\dfrac{11}{20}\)
= - \(\dfrac{11}{20}\)
b; \(\dfrac{3}{4}\) + \(\dfrac{-5}{6}\) - \(\dfrac{11}{-12}\)
= \(\dfrac{9}{12}\) - \(\dfrac{10}{12}\) + \(\dfrac{11}{12}\)
= \(\dfrac{10}{12}\)
= \(\dfrac{5}{6}\)
c; [13.\(\dfrac{4}{9}\) + 2.\(\dfrac{1}{9}\)] - 3.\(\dfrac{4}{9}\)
= [\(\dfrac{52}{9}\) + \(\dfrac{2}{9}\)] - \(\dfrac{4}{3}\)
= \(\dfrac{54}{9}\) - \(\dfrac{4}{3}\)
= \(\dfrac{14}{3}\)
a) \(\frac{x^7}{3^5}=9\)
=> \(\frac{x^7}{3^5}=3^2\)
=> x7 = 32 . 35 = 37
=> x = 3
b) \(\frac{32}{x^7}=4\)
=> \(\frac{2^5}{x^7}=2^2\)
=> \(2^2\cdot x^7=2^5\)
=> \(x^7=\frac{2^5}{2^2}=2^3\)
=> không tìm được x
2. \(\frac{2^{30}}{3^{20}}-\left(\frac{2}{3}\right)^{20}\cdot2^{10}\)
\(=\frac{2^{30}}{3^{20}}-\frac{2^{20}}{3^{20}}\cdot2^{10}\)
\(=\frac{2^{30}}{3^{20}}-\frac{2^{30}}{3^{20}}=0\)