đây mà là toán lớp 1 à.đồ dở hơi.

ban a toán lớp mấy thì  toán vậy bạn có trả lời được ko mà còn ra vẻ hả bạn 'doraemi'

\(\frac{1}{5}-\frac{3}{7}+\frac{5}{9}-\frac{2}{11}+\frac{7}{13}-\frac{9}{16}-\frac{7}{13}+\frac{2}{11}-\frac{5}{9}+\frac{3}{7}-\frac{1}{5}\)

\(=\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{2}{11}-\frac{2}{11}\right)+\left(\frac{5}{9}-\frac{5}{9}\right)+\left(\frac{3}{7}-\frac{3}{7}\right)+\left(\frac{7}{13}-\frac{7}{13}\right)-\frac{9}{16}\)

\(=\frac{-9}{16}\)

1. a

\(\dfrac{8}{5}-\dfrac{5}{6}\cdot\dfrac{3}{4}\)

\(=\dfrac{8}{5}-\dfrac{5\cdot3}{3\cdot2\cdot4}\)

\(=\dfrac{8}{5}-\dfrac{5}{8}=\dfrac{39}{40}\)

1.b

\(=\dfrac{7}{8}+\dfrac{5}{6}\cdot\dfrac{3}{2}\)

\(=\dfrac{7}{8}+\dfrac{5\cdot3}{3\cdot2\cdot2}\)

\(=\dfrac{7}{8}+\dfrac{5}{4}=\dfrac{17}{8}\)

2.a

\(\dfrac{4}{5}+x=\dfrac{11}{10}\)

\(x=\dfrac{11}{10}-\dfrac{4}{5}=\dfrac{3}{10}\)

2.b

\(x-\dfrac{3}{4}=\dfrac{5}{7}\)

\(x=\dfrac{5}{7}+\dfrac{3}{4}=\dfrac{41}{28}\)

\(3x\left(2x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x+1=0\\3x=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x=-1\\x=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=0\end{cases}}\)

\(\frac{\frac{6}{5}+\frac{6}{35}-\frac{6}{125}-\frac{6}{2009}-\frac{6}{2011}}{\frac{7}{5}+\frac{7}{35}-\frac{7}{125}-\frac{7}{2009}-\frac{7}{2011}}\)

\(=\frac{6.(\frac{1}{5}+\frac{1}{35}-\frac{1}{125}-\frac{1}{2009}-\frac{1}{2011})}{7.(\frac{1}{5}+\frac{1}{35}-\frac{1}{125}-\frac{1}{2009}-\frac{1}{2011})}\)

\(=\frac{6}{7}\)

Tìm x

\(a,3x(2x+1)=0\)

\(\Rightarrow\hept{\begin{cases}3x=0\\2x+1=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=0\\x=\frac{-1}{2}\end{cases}}\)

Vậy \(x=0\)hoặc \(x=\frac{-1}{2}\)

\(b.\frac{2}{3}-\frac{1}{3}(x-\frac{3}{2})-\frac{1}{2}(2x+1)=5\)

\(\frac{2}{3}-\frac{1}{3}x+\frac{1}{2}-x-\frac{1}{2}=5\)

\(\frac{2}{3}+\frac{1}{2}-\frac{1}{2}-x(\frac{1}{3}+1)=5\)

\(\frac{4}{3}x=\frac{2}{3}-5\)

\(\frac{4}{3}x=\frac{-13}{3}\)

\(x=\frac{-13}{3}\div\frac{4}{3}\)

\(x=\frac{-13}{4}\)

Chúc ban học tốt