Thể tích mực nước trong bể:

\(60\cdot40\cdot25=60000\left(cm^2\right)\)

\(60000cm^2=60dm^2\)

Nhầm lẫn một chút về đơn vị phải là \(cm^3\) và \(dm^3\)

Gọi khoảng tử số cần tìm là x, khi đó:

\(\dfrac{2}{7}< x< \dfrac{6}{7}\)hay \(\dfrac{18}{63}< x< \dfrac{54}{63}\)

Vậy các giá trị x thỏa mãn gói trong tập M:

\(M=\left\{x\in\mathbb{N}|18< x< 54\right\}\)

\(a,\dfrac{2}{3}x-\dfrac{2}{5}=\dfrac{1}{2}x-\dfrac{1}{3}\\ \Rightarrow\dfrac{2}{3}x-\dfrac{1}{2}x-\dfrac{2}{5}=-\dfrac{1}{3}\\ \Rightarrow x\left(\dfrac{2}{3}-\dfrac{1}{2}\right)-\dfrac{2}{5}=-\dfrac{1}{3}\\ \Rightarrow x\dfrac{1}{6}=-\dfrac{11}{15}\\ \Rightarrow x=-\dfrac{22}{5}\\ b,\dfrac{1}{3}x+\dfrac{2}{5}.\left(x+1\right)=0\\ \Rightarrow\dfrac{1}{3}x+\left(x+1\right)=-\dfrac{2}{5}\\ \Rightarrow\dfrac{1}{3}x=-\dfrac{2}{5}-\left(x+1\right)\\ \Rightarrow\dfrac{1}{3}x=-\dfrac{7}{5}-x\\ \Rightarrow\dfrac{1}{3}.2x=-\dfrac{7}{5}\\ \Rightarrow2x=-\dfrac{21}{5}\\ \Rightarrow x=-\dfrac{21}{10}.\)

\(a,101^2=101.\left(100+1\right)=10100+101=10201.\\ b,75^2-50.75+25^2\\ =75.\left(75-50\right)+25^2\\ =75.25+25^2\\ =25.\left(75+25\right)\\ =25.100\\ =2500.\)

\(c,103.97\\ =\left(100+3\right).97\\ =9700+291\\ =9991\)

Đặt:

\(A=\dfrac{7}{11\cdot16}+\dfrac{7}{16\cdot21}+\dfrac{7}{21\cdot26}+...+\dfrac{7}{61\cdot66}\)

\(\dfrac{5}{7}A=\dfrac{5}{11\cdot16}+\dfrac{5}{16\cdot21}+...+\dfrac{5}{61\cdot66}\)

\(\dfrac{5}{7}A=\dfrac{1}{11}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{21}+...+\dfrac{1}{61}-\dfrac{1}{66}\)

\(\dfrac{5}{7}A=\dfrac{1}{11}-\dfrac{1}{66}=\dfrac{6}{66}-\dfrac{1}{66}=\dfrac{5}{66}\)

\(A=\dfrac{5}{66}\cdot\dfrac{7}{5}=\dfrac{7}{66}\)

A = \(\dfrac{\dfrac{3}{8}-\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}}{-\dfrac{5}{8}+\dfrac{5}{10}-\dfrac{5}{11}-\dfrac{5}{12}}\)

A = \(\dfrac{3.\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}{-5.\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}\)

A = - \(\dfrac{3}{5}\)