câu 1a)\(=-\dfrac{5}{3}+\dfrac{4}{6}=-\dfrac{10}{6}+\dfrac{4}{6}=-\dfrac{6}{6}=-1\)b)\(=\left(\dfrac{-3}{7}+\dfrac{10}{7}\right)+\left(-\dfrac{4}{9}-\dfrac{23}{9}\right)+\dfrac{2}{3}=1-3+\dfrac{2}{3}=-2+\dfrac{2}{3}=-\dfrac{4}{3}\)c)\(=\dfrac{9}{4}\cdot\dfrac{7}{2}+\dfrac{9}{4}\cdot\dfrac{7}{16}=\dfrac{9}{4}\cdot\left(\dfrac{7}{2}+\dfrac{7}{16}\right)=\dfrac{9}{4}\cdot\left(\dfrac{56}{16}+\dfrac{7}{16}\right)=\dfrac{9}{4}\cdot\dfrac{63}{16}=\dfrac{567}{64}\)Câu trả lời: câu 1a)\(=-\dfrac{5}{3}+\dfrac{4}{6}=-\dfrac{10}{6}+\dfrac{4}{6}=-\dfrac{6}{6}=-1\)b)\(=\left(\dfrac{-3}{7}+\dfrac{10}{7}\right)+\left(-\dfrac{4}{9}-\dfrac{23}{9}\right)+\dfrac{2}{3}=1-3+\dfrac{2}{3}=-2+\dfrac{2}{3}=-\dfrac{4}{3}\)c)\(=\dfrac{9}{4}\cdot\dfrac{7}{2}+\dfrac{9}{4}\cdot\dfrac{7}{16}=\dfrac{9}{4}\cdot\left(\dfrac{7}{2}+\dfrac{7}{16}\right)=\dfrac{9}{4}\cdot\left(\dfrac{56}{16}+\dfrac{7}{16}\right)=\dfrac{9}{4}\cdot\dfrac{63}{16}=\dfrac{567}{64}\)

cây 2a)\(=>x\left(\dfrac{3}{5}-1\right)=2-\dfrac{1}{3}=\dfrac{6}{3}-\dfrac{1}{3}=\dfrac{5}{5}\)\(=>x=\dfrac{5}{3}:\left(\dfrac{3}{5}-1\right)=\dfrac{5}{3}:\left(\dfrac{3}{5}-\dfrac{5}{5}\right)=\dfrac{5}{3}:\dfrac{-2}{5}=-\dfrac{25}{6}\)b)\(x=\dfrac{1}{2}-\dfrac{1}{7}=\dfrac{7}{14}-\dfrac{2}{14}=\dfrac{5}{14}\)c)\(=>\left[{}\begin{matrix}4x^2=25\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2=\dfrac{25}{4}\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\\x=-3\end{matrix}\right.\)Câu trả lời: cây 2a)\(=>x\left(\dfrac{3}{5}-1\right)=2-\dfrac{1}{3}=\dfrac{6}{3}-\dfrac{1}{3}=\dfrac{5}{5}\)\(=>x=\dfrac{5}{3}:\left(\dfrac{3}{5}-1\right)=\dfrac{5}{3}:\left(\dfrac{3}{5}-\dfrac{5}{5}\right)=\dfrac{5}{3}:\dfrac{-2}{5}=-\dfrac{25}{6}\)b)\(x=\dfrac{1}{2}-\dfrac{1}{7}=\dfrac{7}{14}-\dfrac{2}{14}=\dfrac{5}{14}\)c)\(=>\left[{}\begin{matrix}4x^2=25\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2=\dfrac{25}{4}\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\\x=-3\end{matrix}\right.\)

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Nhất Chu

17 tháng 5 2022

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#Toán lớp 6

TV Cuber

17 tháng 5 2022

câu 1

a)\(=-\dfrac{5}{3}+\dfrac{4}{6}=-\dfrac{10}{6}+\dfrac{4}{6}=-\dfrac{6}{6}=-1\)

b)\(=\left(\dfrac{-3}{7}+\dfrac{10}{7}\right)+\left(-\dfrac{4}{9}-\dfrac{23}{9}\right)+\dfrac{2}{3}=1-3+\dfrac{2}{3}=-2+\dfrac{2}{3}=-\dfrac{4}{3}\)

c)\(=\dfrac{9}{4}\cdot\dfrac{7}{2}+\dfrac{9}{4}\cdot\dfrac{7}{16}=\dfrac{9}{4}\cdot\left(\dfrac{7}{2}+\dfrac{7}{16}\right)=\dfrac{9}{4}\cdot\left(\dfrac{56}{16}+\dfrac{7}{16}\right)=\dfrac{9}{4}\cdot\dfrac{63}{16}=\dfrac{567}{64}\)

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TV Cuber

17 tháng 5 2022

cây 2

a)\(=>x\left(\dfrac{3}{5}-1\right)=2-\dfrac{1}{3}=\dfrac{6}{3}-\dfrac{1}{3}=\dfrac{5}{5}\)

\(=>x=\dfrac{5}{3}:\left(\dfrac{3}{5}-1\right)=\dfrac{5}{3}:\left(\dfrac{3}{5}-\dfrac{5}{5}\right)=\dfrac{5}{3}:\dfrac{-2}{5}=-\dfrac{25}{6}\)

b)\(x=\dfrac{1}{2}-\dfrac{1}{7}=\dfrac{7}{14}-\dfrac{2}{14}=\dfrac{5}{14}\)

c)\(=>\left[{}\begin{matrix}4x^2=25\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2=\dfrac{25}{4}\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\\x=-3\end{matrix}\right.\)

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