\(\left(-\dfrac{4}{3}+\dfrac{5}{13}\right):\dfrac{2}{7}-\left(\dfrac{9}{4}+\dfrac{8}{13}\right):\dfrac{2}{7}\\ =\left(-\dfrac{4}{3}+\dfrac{5}{13}-\dfrac{9}{4}-\dfrac{8}{13}\right):\dfrac{2}{7}\\ =-\dfrac{595}{156}:\dfrac{2}{7}\\ =-\dfrac{595}{156}.\dfrac{7}{2}=-\dfrac{4165}{312}\)

\(\dfrac{-4165}{312}\)

(3\(x\) - 2)(\(x+4\)) - (1- \(x\))(2-\(x\)) =(\(x+1\))(\(x-2\))

3\(x^2\) + 12\(x\) - 2\(x\) - 8  - (\(x+1\))(\(x-2\)) -  [-(\(x-2\))](1- \(x\)) = 0 

                    3\(x^2\) + 10\(x\) - 8  -  (\(x-2\))( \(x\) + 1 - 1 + \(x\)) = 0

                    3\(x^2\) + 10\(x\) - 8 - (\(x-2\)). 2\(x\)  = 0

                    3\(x^2\) + 10\(x\) - 8 - 2\(x^2\) + 4\(x\)     = 0

                      \(x^2\) + 14\(x\) -  8 = 0

                    \(x^2\) + 7\(x\) + 7\(x\) + 49 - 57 = 0 

                     \(x\)( \(x\) + 7) + 7(\(x\) + 7) = 57

                       (\(x+7\))(\(x\) + 7) =57

                       (\(x+7\))² = 57

                        \(\left[{}\begin{matrix}x+7=\sqrt{57}\\x+7=-\sqrt{57}\end{matrix}\right.\)

                         \(\left[{}\begin{matrix}x=-7+\sqrt{57}\\x=-7-\sqrt{57}\end{matrix}\right.\)

Vậy \(x\) \(\in\) { -7 - \(\sqrt{57}\); - 7 + \(\sqrt{57}\)}

Hình thì bn tự vẽ

a, Vì điểm M nằm giữa 2 điểm C và D nên ta có:

  ⇒   CM + MD = CD

Thay số : 1 + MD = 6

              ⇒ MD = 5 cm.

b, Vì I là trung điểm của đoạn MD nên ta có:

   ⇒ MI = ID = MD/2 = 2,5cm

Vậy ID = 2,5 cm.

sai rồi bạn ơi

cảm ơn bạn

\(-\dfrac{1}{8}\)

\(\left(2x+5\right)\left(y-3\right)=22\)

\(\Rightarrow\left(2x+5\right);\left(y-3\right)\in\left\{1;2;11;22\right\}\)

\(\Rightarrow\left(x;y\right)\in\left\{\left(-2;25\right);\left(-\dfrac{3}{2};14\right);\left(3;5\right);\left(\dfrac{17}{2};4\right)\right\}\)

\(\Rightarrow\left(x;y\right)\in\left\{\left(3;5\right)\right\}\left(\left(x;y\inℤ^+\right)\right)\)

\(\dfrac{x}{\dfrac{4}{2}}=\dfrac{4}{\dfrac{x}{2}}\)

\(\Rightarrow\dfrac{x}{2}=\dfrac{8}{x}\)

\(\Rightarrow x^2=8\cdot2\)

\(\Rightarrow x^2=16\)

\(\Rightarrow x^2=4^2\)

\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

\(1+\dfrac{1}{3}+1\dfrac{3}{4}\div2\dfrac{1}{2}+\dfrac{2}{3}\)

\(=1+\dfrac{1}{3}+\dfrac{7}{4}\div\dfrac{5}{2}+\dfrac{2}{3}\)

\(=1+\dfrac{1}{3}+\dfrac{7}{10}+\dfrac{2}{3}\)

\(=\left(1+\dfrac{7}{10}\right)+\left(\dfrac{1}{3}+\dfrac{2}{3}\right)\)

\(=\dfrac{17}{10}+1\)

\(=\dfrac{27}{10}\)

\(a,\left(-\dfrac{1}{3}\right)^2.\dfrac{4}{11}+\dfrac{7}{11}.\left(-\dfrac{1}{3}\right)^2=\left(-\dfrac{1}{3}\right)^2.\left(\dfrac{4}{11}+\dfrac{7}{11}\right)\\ =\dfrac{1}{9}.1=\dfrac{1}{9}\\ b,\left(\dfrac{1}{2}\right)^4.\dfrac{5}{21}+\dfrac{16}{21}.\left(-\dfrac{1}{4}\right)^2\\ =\dfrac{1}{16}.\dfrac{5}{21}+\dfrac{16}{21}.\dfrac{1}{16}\\ =\left(\dfrac{5}{21}+\dfrac{16}{21}\right).\dfrac{1}{16}=1.\dfrac{1}{16}=\dfrac{1}{16}\\ c,\dfrac{12}{7}.\left(-\dfrac{2}{5}\right)^2-\dfrac{5}{7}.\left(\dfrac{2}{5}\right)^2\\ =\left(\dfrac{12}{7}-\dfrac{5}{7}\right).\dfrac{4}{25}\\ =1.\dfrac{4}{25}=\dfrac{4}{25}\)

\(d,\left(-\dfrac{1}{7}\right)^0-2\dfrac{4}{9}.\left(\dfrac{2}{3}\right)^2=1-\dfrac{22}{9}.\dfrac{4}{9}\\ =1-\dfrac{88}{99}=1-\dfrac{8}{9}=\dfrac{1}{9}\\ e,\left[\dfrac{10^5:10^3}{50}-\left(\dfrac{24}{23}\right)^0\right]^{2024}=\left[\dfrac{10^2}{50}-1\right]^{2024}=\left[\dfrac{100}{50}-1\right]^{2024}=\left[2-1\right]^{2024}\\ =1^{2024}=1\\ f,\left(-\dfrac{1}{2}\right)^0.\left(\dfrac{1}{2}\right)^2+\left(-2\right)^2:8=1.\dfrac{1}{4}+-4:8\\ =\dfrac{1}{4}-\dfrac{1}{2}=-\dfrac{1}{4}\)