Ta có: \(\dfrac{-2}{3}=\dfrac{\left(-2\right).4}{3.4}=\dfrac{-8}{12}\)

\(\dfrac{-1}{4}=\dfrac{\left(-1\right).3}{4.3}=\dfrac{-3}{12}\)

Mà \(\dfrac{-8}{12};\dfrac{-7}{12};\dfrac{-6}{12};\dfrac{-5}{12};\dfrac{-4}{12};\dfrac{-3}{12}\)

\(\Rightarrow\dfrac{-8}{12}< \dfrac{-7}{12}< \dfrac{-6}{12}< \dfrac{-5}{12}< \dfrac{-4}{12}< -\dfrac{3}{12}\)

Lời giải:
$\frac{-2}{3}<\frac{....}{12}< \frac{-1}{4}$

$\Rightarrow \frac{-8}{12}<\frac{....}{12}< \frac{-3}{12}$

$\Rightarrow -8<....< -3$

$\Rightarrow ....$ có thể nhận các giá trị thuộc tập $\left\{-7;-6;-5;-4\right\}$

Vậy các phân số cần tìm là $\frac{-7}{12}, \frac{-6}{12}, \frac{-5}{12}, \frac{-4}{12}$

Lời giải:
$\frac{x-2024}{4}=\frac{1}{x-2024}$ (điều kiện: $x\neq 2024$)

$\Rightarrow (x-2024)^2=4.1=4=2^2=(-2)^2$

$\Rightarrow x-2024=2$ hoặc $x-2024=-2$

$\Rightarrow x=2026$ hoặc $x=2022$

Lần sau bạn lưu ý gõ đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người hiểu đề của bạn hơn nhé.

\(\dfrac{2^{10}.3^{10}-2^{10}.3^9}{2^9.3^{10}}=\dfrac{2^{10}.3^9\left(3-1\right)}{2^9.3^{10}}=\dfrac{2^{10}.3^9.2}{2^9.3^{10}}=\dfrac{2^9.2.3^9.2}{2^9.3.3^9}=\dfrac{4}{3}\)

\(\dfrac{2^{10}.3^{10}-2^{10}.3^9}{2^9.3^{10}}=\dfrac{2^{10}.3^9\left(3-1\right)}{2^9.3^{10}}=\dfrac{2^{10}.3^9.2}{2^9.3^{10}}=\dfrac{2^{10}.2.3^9.2}{2^9.3.3^9}=\dfrac{4}{3}\)

\(\dfrac{-11^5.13^7}{11^5.13^8}=\dfrac{-1.11^5.13^7}{11^5.13^8}=-\dfrac{1}{13}\)

\(\dfrac{-11^5\cdot13^8}{11^5\cdot13^7}\)

\(=\dfrac{-11^5\cdot13^7\cdot13}{11^5\cdot13^7}\)

\(=\dfrac{-11^5\cdot13}{11^5}\)

\(=\dfrac{11^5\cdot\left(-13\right)}{11^5}\)

\(=-13\)

\(\dfrac{2}{3}+\dfrac{5}{6}:5-\dfrac{1}{18}.\left(-3\right)^2\)

\(=\dfrac{2}{3}+\dfrac{5}{6}.\dfrac{1}{5}-\dfrac{1}{18}.9\)

\(=\dfrac{2}{3}+\dfrac{1}{6}-\dfrac{1}{2}\)

\(=\dfrac{4}{6}+\dfrac{1}{6}-\dfrac{3}{6}=\dfrac{2}{6}=\dfrac{1}{3}\)

\(\dfrac{2}{3}+\dfrac{5}{6}:5-\dfrac{1}{18}\cdot\left(-3\right)^2\)

\(=\dfrac{4}{6}+\dfrac{5}{6}:5-\dfrac{1}{18}\times9\)

\(=\dfrac{4}{6}+\dfrac{5}{6}\times\dfrac{1}{5}-\dfrac{1}{18}\times\dfrac{9}{1}\)

\(=\dfrac{4}{6}+\dfrac{5}{30}-\dfrac{9}{18}\)

\(=\dfrac{20}{30}+\dfrac{5}{30}-\dfrac{1}{2}\)

\(=\dfrac{25}{30}-\dfrac{1}{2}\)

\(=\dfrac{25}{30}-\dfrac{15}{30}\)

\(\dfrac{10}{30}=\dfrac{1}{3}\)

      \(\dfrac{2}{9}\) - \(\dfrac{7}{8}\).\(x\) = \(\dfrac{1}{3}\)

         \(\dfrac{7}{8}\).\(x\)  = \(\dfrac{2}{9}\) - \(\dfrac{1}{3}\)

        \(\dfrac{7}{8}\).\(x\) =  - \(\dfrac{1}{9}\)

           \(x\)  = - \(\dfrac{1}{9}\): \(\dfrac{7}{8}\)

           \(x\) = - \(\dfrac{8}{63}\)

\(\dfrac{2}{9}-\dfrac{7}{8}\cdot x=\dfrac{1}{3}\)

\(\dfrac{7}{8}\cdot x=\dfrac{2}{9}-\dfrac{1}{3}\)

\(\dfrac{7}{8}\cdot x=\dfrac{2}{9}-\dfrac{3}{9}\)

\(\dfrac{7}{8}\cdot x=\dfrac{2-3}{9}\)

\(\dfrac{7}{8}\cdot x=\dfrac{-1}{9}\)

\(x=\dfrac{-1}{9}:\dfrac{7}{8}\)

\(x=\dfrac{-1}{9}\times\dfrac{8}{7}\)

\(x=\dfrac{-8}{63}\)

Lời giải:

$\frac{4}{x}+\frac{x}{3}=\frac{5}{6}$

$\frac{12+x^2}{3x}=\frac{5}{6}$

$12+x^2=\frac{5}{2}x$
$2x^2-5x+24=0$
$x^2+(x-2,5)^2=-17,75<0$ (vô lý) 

Do đó không tồn tại $x$ thỏa mãn.

\(P=2.5+5.8+8.11+...+101.104\)

\(\Rightarrow9P=2.5.9+5.8.9+8.11.9+...+101.104.9\)

\(\Rightarrow9P=2.5.9+5.8.\left(11-2\right)+8.11.\left(14-5\right)+...+101.104.\left(107-98\right)\)

\(\Rightarrow9P=2.5.9-2.5.8+5.8.11-5.8.11+8.11.14+...-98.101.104+101.104.107\)

\(\Rightarrow9P=2.5.9-2.5.8+101.104.107\)

\(\Rightarrow9P=1123938\)

\(\Rightarrow P=124882\)

Sửa đề: \(\dfrac{2}{17}+\dfrac{15}{23}+\dfrac{15}{17}+\dfrac{4}{19}+\dfrac{8}{23}\)

\(=\left(\dfrac{2}{17}+\dfrac{15}{17}\right)+\left(\dfrac{15}{23}+\dfrac{8}{23}\right)+\dfrac{4}{19}\)

\(=1+1+\dfrac{4}{19}\)

\(=2+\dfrac{4}{19}\)

\(=\dfrac{42}{19}\)

\(#Tmiamm\)

a; \(\dfrac{6}{x}\) < \(\dfrac{x}{7}\) < \(\dfrac{8}{x}\)

    vì \(x\) \(\in\) N* ta có: 6.7 < \(x.x\) < 7.8

                             42 < \(x^2\) < 56

                            \(x^2\) = 49 

                           \(x\) = \(\pm\) 7

Vì \(x\) \(\in\) N*; \(x\) = 7

b;  \(\dfrac{x}{11}\) < \(\dfrac{12}{x}\) < \(\dfrac{x}{9}\)

   9.12<   \(x^2\) < 11.12 

    108 < \(x^2\) < 132

            \(x^2\) = 121

            \(\left[{}\begin{matrix}x=-11\\x=11\end{matrix}\right.\)

    Vì \(x\in\) N*

   \(x\)  = 11