i, \(\dfrac{1}{7}\) .(- \(\dfrac{3}{8}\)) + \(\dfrac{1}{7}\).(-\(\dfrac{13}{8}\))

=- \(\dfrac{1}{7}\).( \(\dfrac{3}{8}\)+ \(\dfrac{13}{8}\))

= - \(\dfrac{1}{7}\). 2

= -2/7 

k, -4\(\dfrac{1}{3}\). ( \(\dfrac{1}{2}\)- \(\dfrac{1}{6}\))

= - \(\dfrac{13}{3}\). \(\dfrac{1}{3}\)

= - 13/9

l, \(\dfrac{3}{5}\).\(\dfrac{13}{46}\)- \(\dfrac{1}{10}\).\(\dfrac{16}{23}\)

= \(\dfrac{39}{230}\) - \(\dfrac{16}{230}\)

= \(\dfrac{1}{10}\)

m, (- \(\dfrac{3}{4}\)+ \(\dfrac{2}{5}\)): \(\dfrac{3}{7}\) + ( \(\dfrac{3}{5}\)- \(\dfrac{1}{4}\)): \(\dfrac{3}{7}\)

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

= 0 : \(\dfrac{3}{7}\)

= 0

t, \(\dfrac{7}{8}\): ( \(\dfrac{2}{9}\) - \(\dfrac{1}{18}\)) + \(\dfrac{7}{8}\): ( \(\dfrac{1}{36}\) - \(\dfrac{5}{12}\))

= \(\dfrac{7}{8}\): \(\dfrac{1}{6}\) + \(\dfrac{7}{8}\): (- \(\dfrac{7}{18}\))

= \(\dfrac{21}{4}\) - \(\dfrac{9}{4}\)

= 3 

= \(\dfrac{12}{4}\)

                   Giải

diện tích hai mặt đáy là: 6x 8 : 2 x 2 = 48(cm2)

diện tích xung quanh là : 248 - 48 = 200(cm2)

nửa đường chéo bé là 6 : 2 = 3(cm)

nửa đường chéo lớn là: 8 : 2 = 4 (cm)

theo pyta go ta có cạnh hình thoi là :  \(\sqrt{3^2+4^2}\) = 5 (cm)

chu vi đáy là 5 x 4 = 20 (cm)

chiều cao hình lăng trụ là:  200 : 20 = 10 (cm)

thể tích hình lăng trụ là: 6x8:2 x 10 = 240(cm³)

đs...

xin lỗi ban đầu chỗ tính chiều cao nhầm đáy là hình chữ nhật nên mình làm lại bạn thông cảm.

\(\sqrt{ }\)

diện tích hai mặt đáy là: 6x 8 : 2 x 2 = 48(cm²)

diện tích xung quanh là : 248 - 48 = 200(cm²)

chiều cao là: 200 : 2 : (6+8) = \(\dfrac{50}{7}\) (cm)

thể tích là: 6x8: 2 x \(\dfrac{50}{7}\) = \(\dfrac{1200}{7}\) (cm³)

đs....

   - \(\dfrac{3}{4}\) x 31\(\dfrac{11}{23}\) - 0,75 x 8\(\dfrac{12}{23}\)

= - \(\dfrac{3}{4}\) x \(\dfrac{724}{23}\) - 0,75 x \(\dfrac{196}{23}\)

= - \(\dfrac{543}{23}\) - \(\dfrac{147}{23}\)

= - \(\dfrac{690}{23}\)

= -30

gfgfgfg

     A =      1/3 + 1/3² + ...................+ 1/3²⁰¹⁸

3x A = 1 + 1/3  + 1/3² +...+1/3²⁰¹⁷

3A - A = 1 - 1/3²⁰¹⁸

         2A = 1- 1/3²⁰¹⁸ < 1

      ⇔ A < \(\dfrac{1}{2}\)

\(\dfrac{6^7.4^2}{9^2.12^5}\)  = \(\dfrac{3^7.2^7.2^4}{3^4.3^5.4^5}\) = \(\dfrac{2^{11}}{3^2.2^{10}}\) = \(\dfrac{2}{9}\)

\(\dfrac{6^7.4^2}{9^2.12^5}=\dfrac{3^7.2^7.2^4}{3^4.3^5.4^5}=\dfrac{2^{11}}{3^2.2^{10}}=\dfrac{2}{9}\)

Ta có :

\(\left(\dfrac{3}{4}.x-\dfrac{9}{16}\right).\left(1,5+\dfrac{-3}{5}:x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}.x-\dfrac{9}{16}=0\\1,5+\dfrac{-3}{5}:x=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}.x=\dfrac{9}{16}\\\dfrac{3}{5}:x=\dfrac{3}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{16}:\dfrac{3}{4}\\x=\dfrac{3}{5}:\dfrac{3}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{2}{5}\end{matrix}\right.\)

1,8 . x - 30% . x = -4,5

\(\Rightarrow\dfrac{18}{10}.x-\dfrac{3}{10}.x=\dfrac{-9}{2}\)

\(\Rightarrow\left(\dfrac{18}{10}-\dfrac{3}{10}\right).x=\dfrac{-9}{10}\)

\(\Rightarrow5.x=\dfrac{-9}{2}\)

\(\Rightarrow x=\dfrac{-9}{10}\)

\(\dfrac{x+1}{65}+1+\dfrac{x+3}{63}+1=\dfrac{x+5}{61}+1+\dfrac{x+7}{59}+1\)

\(\Leftrightarrow\dfrac{x+66}{65}+\dfrac{x+66}{63}-\dfrac{x+66}{61}-\dfrac{x+66}{59}=0\)

\(\Leftrightarrow\left(x+66\right)\left(\dfrac{1}{65}+\dfrac{1}{63}-\dfrac{1}{61}-\dfrac{1}{59}\ne0\right)=0\Leftrightarrow x=-66\)