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

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

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

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

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

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

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

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

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

\(\Rightarrow x^2=16\)

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

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

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

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

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

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

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

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

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

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

Đề yêu cầu gì thế em?

Tính tổng hay tìm chữ số tận cùng của tổng em nhỉ?

a) \(\left(2x-3\right)^2=16\)

\(\left(2x-3\right)^2=4^2\)

\(2x-3=4\)

\(2x=7\)

\(x=\dfrac{7}{2}=3,5\)

b) \(\left(3x-2\right)^5=-243\)

\(\left(3x-2\right)^5=-3^5\)

\(3x-2=-3\)

\(3x=-1\)

\(3x=-\dfrac{1}{3}\)

c) \(\left(x-7\right)^{x+1}=\left(x-7\right)^{x+11}\)

\(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)

\(\left(x-7\right)^{x+1}\times\left[1-\left(x-7\right)^{10}\right]=0\)

\(\left(x-7\right)^{x+1}=0\) ; \(1-\left(x-7\right)^{10}=0\)

\(x-7=0;\left(x-7\right)^{10}=1\)

\(x=7;\left(x-7=1;x-7=-1\right)\)

\(x=7;x=8;x=6\)

a, (2\(x\) - 3)² = 16

     \(\left[{}\begin{matrix}2x-3=-4\\2x-3=4\end{matrix}\right.\)

     \(\left[{}\begin{matrix}2x=-1\\2x=7\end{matrix}\right.\)

     \(\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{7}{2}\end{matrix}\right.\)

Vậy \(x\in\){ - \(\dfrac{1}{2}\); \(\dfrac{7}{2}\)}

b, (3\(x\) - 2)⁵   = -243 

    ( 3\(x\) - 2)⁵ =  (-3)⁵

 3\(x\)    -  2     = -3

     3 \(x\)           = -1

        \(x\)          = - \(\dfrac{1}{3}\)

Vậy \(x\) = -\(\dfrac{1}{3}\)

c, \(\left(x-7\right)\)^\(x+1\)  = (\(x-7\))^\(x+11\)

    (\(x-7\))^{\(^{x+1}\)}.( \(\left(x-7\right)^{10}\) -  1 ) = 0

      \(\left[{}\begin{matrix}\left(x-7\right)^{x+1}=0\\\left(x-7\right)^{10}=1\end{matrix}\right.\)

         ^{\(\left[{}\begin{matrix}x=7\\x-7=-1\\x-7=1\end{matrix}\right.\)}

         \(\left[{}\begin{matrix}x=7\\x=6\\x=8\end{matrix}\right.\)

Vậy \(x\in\){ 6; 7; 8}

Đề có cho điểm nào cắt điểm nào không ạ hay chỉ có AC//xy ạ?

A C x B y