\(\left(2x-3y\right).5xy\\ =5xy.2x+5xy.\left(-3y\right)\\ =10x^2y-15xy^2\)

Hôm nay bác Tâm thu hoạch được số kg nhãn là:

  13 500 - 700 = 12 800 (kg nhãn)

Vậy cả hai ngày bác Tâm thu hoạch được số kg nhãn là:

  13 500 + 12 800 = 26 300 (kg nhãn)

     Đáp số: 26 300 kg nhãn

\(N=\dfrac{3}{25}\times3=\dfrac{9}{25}\)

\(N=\dfrac{6}{8\times10}+\dfrac{6}{10\times12}+\dfrac{6}{12\times14}+...+\dfrac{6}{198\times200}\\ N:3=\dfrac{2}{8\times10}+\dfrac{2}{10\times12}+\dfrac{2}{12\times14}+...+\dfrac{2}{198\times200}\\ N:3=\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{14}+...+\dfrac{1}{198}-\dfrac{1}{200}\\ N:3=\dfrac{1}{8}-\dfrac{1}{200}\\ N:3=\dfrac{3}{25}\\ N=\dfrac{3}{25}.3=\dfrac{9}{25}\)

\(\dfrac{x+2022}{2020}+\dfrac{x-2016}{2018}=\dfrac{x+2021}{2019}+\dfrac{x-2019}{2021}\\ \Rightarrow\left(\dfrac{x+2022}{2020}-1\right)+\left(\dfrac{x-2016}{2018}+1\right)=\left(\dfrac{x+2021}{2019}-1\right)+\left(\dfrac{x-2019}{2021}+1\right)\\ \Rightarrow\dfrac{x+2}{2020}+\dfrac{x+2}{2018}-\dfrac{x+2}{2019}-\dfrac{x+2}{2021}=0\\ \Rightarrow\left(x+2\right)\left(\dfrac{1}{2020}+\dfrac{1}{2018}-\dfrac{1}{2019}-\dfrac{1}{2021}\right)=0\\ \)

\(\Rightarrow x+2=0\) ( Vì: \(\dfrac{1}{2020}+\dfrac{1}{2018}-\dfrac{1}{2019}-\dfrac{1}{2021}>0\) )

\(\Rightarrow x=-2\)

\(\dfrac{\left(-\dfrac{1}{3}\right)^6}{\left(\dfrac{1}{6}\right)^2}=\dfrac{\left(\dfrac{1}{3}\right)^6}{\left(\dfrac{1}{6}\right)^2}\\ =\dfrac{\dfrac{1^6}{3^6}}{\dfrac{1^2}{6^2}}=\dfrac{\dfrac{1}{3^6}}{\dfrac{1}{6^2}}\\ =\dfrac{1}{3^6}:\dfrac{1}{6^2}=\dfrac{1}{3^6}.\dfrac{6^2}{1}\\ =\dfrac{6^2}{3^6}=\dfrac{2^2.3^2}{3^2.3^4}\\ =\dfrac{2^2}{3^4}=\dfrac{4}{81}\)

Gọi chữ số hàng chục là: \(a\left(a\inℕ^∗,a\le9\right)\)

Theo đề, suy ra chữ số hàng đơn vị là: \(10-a\)

Số phải tìm có dạng: \(\overline{a\left(10-a\right)}\)

Nếu đổi chỗ, ta được số: \(\overline{\left(10-a\right)a}\)

Mà: Nếu đổi chỗ hai chữ số ấy ta được số mới hơn số cũ 18

Nên ta có pt:

\(\overline{\left(10-a\right)a}-\overline{a\left(10-a\right)}=18\\ \Leftrightarrow\overline{\left(10-a\right)0}+a-\left(\overline{a0}+10-a\right)=18\\ \Leftrightarrow10\left(10-a\right)+a-10a-10+a=18\\ \Leftrightarrow100-10a+a-10a-10+a-18=0\\ \Leftrightarrow-18a+72=0\\ \Leftrightarrow-18a=-72\\ \Leftrightarrow a=4\left(TMDK\right)\)

Vậy SPT là: 46

\(n^2=36=\left(\pm6\right)^2\)

\(\Rightarrow n=\pm6\)

Mà: n là STN

\(\Rightarrow n=6\)

Vậy: n=6

\(\left(4,5-2x\right).\left(-1\dfrac{4}{7}\right)=\dfrac{11}{14}\\ \Rightarrow4,5-2x=\dfrac{11}{14}:\left(-1\dfrac{4}{7}\right)\\ \Rightarrow4,5-2x=\dfrac{11}{14}:\dfrac{-11}{7}\\ \Rightarrow4,5-2x=\dfrac{11}{14}.\dfrac{-7}{11}\\ \Rightarrow4,5-2x=-\dfrac{1}{2}=-0,5\\ \Rightarrow2x=4,5-\left(-0,5\right)=5\\ \Rightarrow x=\dfrac{5}{2}\)

Cách làm khác:

\(\left(x-2\right)^3-\left(x-2\right)\left(x^2+2x+4\right)=0\\ \Leftrightarrow x^3-3.x^2.2+3.x.2^2-2^3-\left(x^3-2^3\right)=0\\ \Leftrightarrow x^3-6x^2+12x-8-x^3+8=0\\ \Leftrightarrow-6x^2+12x=0\\ \Leftrightarrow-6x\left(x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)