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a) Ta có:
C = 5/18 + 8/19 - 7/21 + (-10/36 + 11/19 + 1/3) - 5/8
C = 5/18 + 8/19 - 1/3 - 5/18 + 11/19 + 1/3 - 5/8
C = (5/18 - 5/18) + (8/19 + 11/19) - (1/3 - 1/3) - 5/8
C = 1 - 5/8
c = 3/8
b) F = 15/14 - (17/23 - 80/87 + 5/4) + (17/23 - 15/14 + 1/4)
F = 15/14 - 17/23 + 80/87 - 5/4 + 17/23 - 15/14 + 1/4
F = (15/14 - 15/14) - (17/23 - 17/23) + 80/87 - (5/4 - 1/4)
F = 80/87 - 1
F = -7/87
c) G = 1/25 - 4/27 + (-23/27 + -1/25 - 5/43) + 5/43 - 4/7
G = 1/25 - 4/27 - 23/27 - 1/25 - 5/43 + 5/43 - 4/7
G = (1/25 - 1/25) - (4/27 + 23/27) - (5/43 - 5/43) - 4/7
G = -1 - 4/7 = -11/7
\(\frac{3^{11}\div5+3^{11}.3}{3^{10}.2^2}=\frac{3^{11}\cdot\frac{1}{5}+3^{11}\cdot3}{3^{10}.2^2}\)
\(=\frac{3^{11}.\left(\frac{1}{5}+3\right)}{3^{10}.2^2}=\frac{3^{11}.\frac{16}{5}}{3^{10}.2^2}=\frac{\frac{48}{5}}{2^2}=\frac{48}{5}\cdot\frac{1}{4}=\frac{12}{5}\)
\(\frac{1}{3^{10}}\div\frac{1}{9^5}=\frac{1}{3^{10}}\div\frac{1}{3^{10}}=1\)
#)Giải :
Câu 2 :
Ta có : \(\frac{1}{9^5}=\frac{1}{\left(3^2\right)^5}=\frac{1}{3^{10}}\)
Vì \(\frac{1}{3^5}=\frac{1}{3^5}\Rightarrow\frac{1}{3^5}=\frac{1}{9^5}\)
311:5+311 .3/310.22(Câu 1)
\(=\frac{8.3^{11}}{3^{10}.4}=6\)
1/310 : 1/95 (Câu 2)
\(=\frac{1}{3^{10}}.3^{10}=1\)
(-25/36)5:5/62 (câu 3)
\(=\frac{-5^5}{6^{10}}.\frac{6^2}{5}=-\frac{5^4}{6^8}\)
5) \(\left(-2\right)^2+\sqrt{36}-\sqrt{9}+\sqrt{25}\)
=\(4+6-3+5\)
=\(12\)
2) \(\dfrac{11}{25}.\left(-24,8\right)-\dfrac{11}{25}.75,2\)
=\(\dfrac{11}{25}.\left(-24,8-75,2\right)\)
=\(\dfrac{11}{25}.\left(-100\right)\)
=\(-44\)
\(\frac{25}{36}.\frac{-8}{11}+\frac{5}{18}.\frac{14}{11}.\frac{-5}{2}\)
\(=\frac{25}{36}.\frac{-8}{11}+\frac{14}{11}.\frac{-25}{36}\)
\(=\frac{-25}{36}\left(\frac{8}{11}+\frac{14}{11}\right)\)
\(=\frac{-25}{36}.\frac{32}{11}\)
\(=\frac{-200}{99}.\)