Luyện tập các phép tính về phân số và số thập phân

\(\dfrac{29}{30}-\left(\dfrac{13}{30}+x\right)=\dfrac{7}{69}\)

\(\Leftrightarrow\dfrac{13}{23}+x=\dfrac{29}{30}-\dfrac{7}{69}\)

\(\Leftrightarrow\dfrac{13}{23}+x=\dfrac{199}{230}\)

\(\Leftrightarrow x=\dfrac{199}{230}-\dfrac{13}{23}\)

\(\Leftrightarrow x=\dfrac{69}{230}\)

\(\frac{-3}{5}+\frac{28}{5}.(\frac{43}{56}+\frac{5}{24}-\frac{21}{63})\)

=\(\frac{-3}{5}+\frac{28}{5}.(\frac{43}{56}+\frac{5}{24}-\frac{1}{3})\)

=\(\frac{-3}{5}+\frac{28}{5}.(\frac{129+35-56}{168})\)

=\(\frac{-3}{5}+\frac{28}{5}.\frac{9}{14}\)

=\(\frac{-3}{5}+\frac{28.9}{5.14}\)

=\(\frac{-3}{5}+\frac{2.9}{5.1}\)

=\(\frac{-3}{5}+\frac{18}{5}\)

=3

1: \(\widehat{xOm}=\dfrac{40^0}{2}=20^0\)

\(\widehat{xOn}=\dfrac{120^0}{2}=60^0\)

=>góc mOn=40 độ

2: Oy là phân giác của góc mOn vì \(\widehat{mOy}=\dfrac{40^0}{2}=20^0=\dfrac{1}{2}\cdot\widehat{mOn}\)

3: \(\widehat{tOz}=180^0-60^0=120^0\)

4a-b=6 nên b=4a-6

\(\dfrac{6a-b}{3a+5}-\dfrac{4a-4b}{3b-5}\)

\(=\dfrac{6a-\left(4a-6\right)}{3a+5}-\dfrac{4a-4\left(4a-6\right)}{3\left(4a-6\right)-5}\)

\(=\dfrac{6a-4a+6}{3a+5}-\dfrac{4a-16a+24}{12a-18-5}\)

\(=\dfrac{2a+6}{3a+5}-\dfrac{-12a+24}{12a-23}\)

\(=\dfrac{2a+6}{3a+5}+\dfrac{12a-24}{12a-23}\)

\(=\dfrac{\left(2a+6\right)\left(12a-23\right)+\left(12a-24\right)\left(3a+5\right)}{\left(3a+5\right)\left(12a-23\right)}\)

\(=\dfrac{24a^2-46a+72a-138+36a^2+60a-72a-120}{\left(3a+5\right)\left(12a-23\right)}\)

\(=\dfrac{60a^2+14a-258}{\left(3a+5\right)\left(12a-23\right)}\)

\(D=(\dfrac{157}{8}:\dfrac{7}{12}-\dfrac{53}{4}:\dfrac{7}{12}).\dfrac{4}{5}\)

\(D=(\dfrac{157}{8}.\dfrac{12}{7}-\dfrac{53}{4}.\dfrac{12}{7}).\dfrac{4}{5}\)

\(D=\dfrac{12}{7}.\left(\dfrac{157}{8}-\dfrac{53}{4}\right).\dfrac{4}{5}\)

\(D=\dfrac{12}{7}.\dfrac{51}{8}.\dfrac{4}{5}=\dfrac{6.51}{7.5}=\dfrac{306}{35}\)

Số học sinh nhóm múa là 30x1/3=10(bạn)

SỐ học sinh nhóm hát là 10x2/3=20/3(vô lý)

=>Đề sai rồi bạn

a)

\(\dfrac{1}{2\cdot3}x+\dfrac{1}{3\cdot4}x+...+\dfrac{1}{49\cdot50}x=1\\ x\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=1\\ x\left(\dfrac{1}{2}-\dfrac{1}{50}\right)=1\\ x\cdot\dfrac{12}{25}=1\\ x=1:\dfrac{12}{25}=1\cdot\dfrac{25}{12}=\dfrac{25}{12}\)

\(\dfrac{1\cdot3\cdot5+2\cdot6\cdot10+4\cdot12\cdot20+7\cdot21\cdot35}{1\cdot5\cdot7+2\cdot10\cdot14+4\cdot20\cdot28+7\cdot35\cdot45}\)

=\(\dfrac{3+6+12+21\cdot35}{14+28+7\cdot45}\)

=\(\dfrac{450}{119}\)

Vì \(\dfrac{450}{119}>1\) mà \(1>\dfrac{303}{708}\)

\(\Rightarrow\)\(\dfrac{450}{119}>\dfrac{303}{708}\)