Bài 1:
a) \(\left(-2\right)\cdot4\cdot5\cdot38\cdot\left(-25\right)\)
\(=\left(-2\cdot5\right)\cdot\left(-25\cdot4\right)\cdot38\)
\(=\left(-10\right)\cdot\left(-100\right)\cdot38\)
\(=1000\cdot38\)
\(=38000\)
b) \(\dfrac{1}{3}+\dfrac{3}{8}+\dfrac{7}{12}\)
\(=\dfrac{8}{24}+\dfrac{9}{24}+\dfrac{14}{24}\)
\(=\dfrac{31}{24}\)
c) \(\dfrac{-5}{8}\cdot\dfrac{5}{12}+\dfrac{-5}{8}\cdot\dfrac{7}{12}+2\dfrac{1}{8}\)
\(=\dfrac{-5}{8}\left(\dfrac{5}{12}+\dfrac{7}{12}\right)+\dfrac{17}{8}\)
\(=\dfrac{-5}{8}\cdot1+\dfrac{17}{8}\)
\(=\dfrac{-5}{8}+\dfrac{17}{8}\)
\(=\dfrac{3}{2}\)
d) \(\left(\dfrac{-5}{24}+0.75+\dfrac{7}{12}\right):\left(-2\dfrac{1}{8}\right)\)
\(=\left(\dfrac{-5}{24}+\dfrac{3}{4}+\dfrac{7}{12}\right):\left(\dfrac{-17}{8}\right)\)
\(=\left(\dfrac{-5}{24}+\dfrac{18}{24}+\dfrac{14}{24}\right)\cdot\left(\dfrac{-8}{17}\right)\)
\(=\dfrac{9}{8}\cdot\left(\dfrac{-8}{17}\right)\)
\(=\dfrac{-9}{17}\)
Bài 2:
a) \(x-\dfrac{2}{5}=0.24\)
\(x-\dfrac{2}{5}=\dfrac{6}{25}\)
\(x=\dfrac{6}{25}+\dfrac{2}{5}\)
\(x=\dfrac{6}{25}+\dfrac{10}{25}\)
\(x=\dfrac{16}{25}\)
Vậy \(x=\dfrac{16}{25}\).
b) \(\dfrac{2}{3}x+\dfrac{1}{2}=\dfrac{1}{10}\)
\(\dfrac{2}{3}x=\dfrac{1}{10}-\dfrac{1}{2}\)
\(\dfrac{2}{3}x=\dfrac{1}{10}-\dfrac{5}{10}\)
\(\dfrac{2}{3}x=\dfrac{-2}{5}\)
\(x=\dfrac{-2}{5}:\dfrac{2}{3}\)
\(x=\dfrac{-2}{5}\cdot\dfrac{3}{2}\)
\(x=\dfrac{-3}{5}\)
Vậy \(x=\dfrac{-3}{5}\).
c) \(\left(3\dfrac{1}{2}-2x\right)\cdot1\dfrac{1}{3}=7\dfrac{1}{3}\)
\(\left(\dfrac{7}{2}-2x\right)\cdot\dfrac{4}{3}=\dfrac{22}{3}\)
\(\dfrac{7}{2}-2x=\dfrac{22}{3}:\dfrac{4}{3}\)
\(\dfrac{7}{2}-2x=\dfrac{22}{3}\cdot\dfrac{3}{4}\)
\(\dfrac{7}{2}-2x=\dfrac{11}{2}\)
\(2x=\dfrac{7}{2}-\dfrac{11}{2}\)
\(2x=-2\)
\(x=-2:2\)
\(x=-1\)
Vậy \(x=-1\).
Bài 3:
a) Số học sinh Trung bình của lớp:
\(35\cdot\dfrac{3}{7}=15\) (học sinh)
Số học sinh Khá của lớp
\(\left(35-15\right)\cdot80\%=16\) (học sinh)
Số học sinh Giỏi của lớp:
\(35-15-16=4\) (học sinh)
b) Tỉ số phần trăm của số học sinh Giỏi và số học sinh cả lớp:
\(\dfrac{4\cdot100}{35}\approx11\%\) (số học sinh cả lớp)
Bài 4:
x O t y
a) Trên nửa mặt phẳng bờ Ox, có \(\widehat{xOt}< \widehat{xOy}\left(40^o< 80^o\right)\).
b) Theo câu a.
\(\Rightarrow\widehat{xOt}+\widehat{tOy}=\widehat{xOy}\\ 40^o+\widehat{tOy}=80^o\\ \widehat{tOy}=40^o\)
Vậy \(\widehat{tOy}=40^o\).
c) Vì \(\widehat{xOt}=\widehat{tOy}=\widehat{\dfrac{xOy}{2}}\left(=40^o\right)\).
\(\Rightarrow\) Tia Ot là tia phân giác của \(\widehat{xOy}\).
Bài 5:
\(A=\dfrac{6}{3\cdot5}+\dfrac{6}{5\cdot7}+\dfrac{6}{7\cdot9}+\dfrac{6}{9\cdot11}+...+\dfrac{6}{97\cdot99}\)
\(=3\left(\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+\dfrac{1}{7\cdot9}+\dfrac{1}{9\cdot11}+...+\dfrac{1}{97\cdot99}\right)\)
\(=3\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)
\(=3\left(\dfrac{1}{3}-\dfrac{1}{99}\right)\)
\(=3\left(\dfrac{33}{99}-\dfrac{1}{99}\right)\)
\(=3\cdot\dfrac{32}{99}\)
\(=\dfrac{32}{33}\)
