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H9
HT.Phong (9A5)
CTVHS
28 tháng 6 2023
\(\left(-12\cdot\dfrac{2}{7}+\dfrac{8}{9}:3\dfrac{1}{2}-\dfrac{2}{7}\cdot\dfrac{5}{18}\right)\cdot3\dfrac{1}{2}\)
\(=\left(-12\cdot\dfrac{2}{7}+\dfrac{8}{9}\cdot\dfrac{2}{7}-\dfrac{2}{7}\cdot\dfrac{5}{18}\right)\cdot\dfrac{7}{2}\)
\(=\left(-12+\dfrac{8}{9}-\dfrac{5}{18}\right)\cdot\dfrac{2}{7}\cdot\dfrac{7}{2}\)
\(=-\dfrac{205}{18}\cdot1\)
\(=-\dfrac{205}{18}\)
TN
0
26 tháng 7 2020
a) \(\frac{-5}{8}\cdot\frac{11}{3}+\frac{-5}{8}\cdot\frac{1}{3}=-\frac{5}{8}\left(\frac{11}{3}+\frac{1}{3}\right)=-\frac{5}{8}\cdot4=-\frac{5}{2}\cdot1=-\frac{5}{2}\)
b) \(\frac{2}{3}+\frac{3}{4}\cdot\frac{9}{5}=\frac{2}{3}+\frac{27}{20}=\frac{121}{60}\)
c) Tương tự câu a
d) \(\frac{1}{7}\cdot\frac{3}{8}+\frac{1}{7}\cdot\frac{5}{8}=\frac{1}{7}\left(\frac{3}{8}+\frac{5}{8}\right)=\frac{1}{7}\cdot1=\frac{1}{7}\)
T
26 tháng 7 2020
\(a,\frac{-5}{8}.\frac{11}{3}+\frac{-5}{8}.\frac{1}{3}\)
\(=\frac{-5}{8}\left(\frac{11}{3}+\frac{1}{3}\right)\)
\(=\frac{-5}{8}.4\)
\(=\frac{-5}{2}\)
\(b,\frac{2}{3}+\frac{3}{4}.\frac{9}{5}\)
\(=\frac{2}{3}+\frac{27}{20}\)
\(=\frac{40}{60}+\frac{81}{60}\)
\(=\frac{121}{60}\)
\(c,\frac{-5}{7}.\frac{4}{9}-\frac{5}{9}.\frac{5}{7}\)
\(=\frac{-5}{7}\left(\frac{4}{9}+\frac{5}{9}\right)\)
\(=\frac{-5}{7}.1\)
\(=\frac{-5}{7}\)
\(d,\frac{1}{7}.\frac{3}{8}+\frac{1}{7}.\frac{5}{8}\)
\(=\frac{1}{7}\left(\frac{3}{8}+\frac{5}{8}\right)\)
\(=\frac{1}{7}.1\)
\(=\frac{1}{7}\)
Học tốt
NT
0
28 tháng 9 2017
Bài 1:
\(\dfrac{\left(\dfrac{2}{5}\right)^7\cdot5^7+\left(\dfrac{9}{4}\right)^3:\left(\dfrac{3}{16}\right)^3}{2^7\cdot5^2+512}\)
\(=\dfrac{\left(\dfrac{2}{5}\cdot5\right)^7+\left(\dfrac{9}{4}:\dfrac{3}{16}\right)^3}{2^7\cdot5^2+512}\)
\(=\dfrac{2^7+12^3}{2^7\cdot5^2+512}\)
\(=\dfrac{1856}{3712}\)
\(=0,5\)
Bài 2:
\(\left(5x+1\right)^2=\dfrac{36}{49}\)
\(\Rightarrow5x+1=\dfrac{6}{7}\)
\(\Rightarrow5x=\dfrac{-1}{7}\)
\(\Rightarrow x=\dfrac{-1}{35}\)
HL
1
3 tháng 7 2023
Sửa đề: (2/7)^7*7^7
\(A=\dfrac{\left(2\right)^7+\left(\dfrac{9}{3}:\dfrac{3}{16}\right)^3}{2^7\left(5^2+2^2\right)}\)
\(=\dfrac{\left(2\right)^7+\left(16\right)^3}{2^7\cdot29}\)
\(=\dfrac{2^7+2^7\cdot2^5}{2^7\cdot29}=\dfrac{1+2^5}{29}=\dfrac{33}{29}\)
(-12.\(\dfrac{2}{7}\) + \(\dfrac{8}{9}\): 3\(\dfrac{1}{2}\) - \(\dfrac{2}{7}\).\(\dfrac{5}{18}\)). 3\(\dfrac{1}{2}\)
(-\(\dfrac{24}{7}\) + \(\dfrac{8}{9}\):\(\dfrac{7}{2}\) - \(\dfrac{5}{63}\)). \(\dfrac{7}{2}\)
= (-\(\dfrac{24}{7}\) + \(\dfrac{16}{63}\) - \(\dfrac{5}{63}\)).\(\dfrac{7}{2}\)
= (\(-\dfrac{216}{63}\) + \(\dfrac{16}{63}\) - \(\dfrac{5}{63}\)). \(\dfrac{7}{2}\)
= - \(\dfrac{205}{63}\). \(\dfrac{7}{2}\)
= - \(\dfrac{205}{18}\)