a)\(\frac{x-2}{16}=\frac{-4}{2-x}\)

\(\Rightarrow\frac{x-2}{16}=\frac{4}{x-2}\)

\(\Rightarrow\left(x-2\right).\left(x-2\right)=16.4\)

\(\Rightarrow\left(x-2\right)^2=64\)

Mà ta có: \(64=\left(\pm8\right)^2\)

Suy ra: \(\orbr{\begin{cases}\left(x-2\right)^2=8^2\\\left(x-2\right)^2=\left(-8\right)^2\end{cases}\Rightarrow\orbr{\begin{cases}x-2=8\\x-2=-8\end{cases}\Rightarrow}\orbr{\begin{cases}x=10\\x=-6\end{cases}}}\)

Vậy x = 10 hoặc x = -6

b) \(\left(2x+7\right)^2-28=64\)

\(\Rightarrow\left(2x+7\right)^2=64+28=92\)

Mà: \(92=\left(\pm2\sqrt{23}\right)^2\)

Nên \(\orbr{\begin{cases}2x+7=2\sqrt{23}\\2x+7=-2\sqrt{23}\end{cases}\Rightarrow\orbr{\begin{cases}2x=2\sqrt{23}-7\\2x=-2\sqrt{23}-7\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{2\sqrt{23}-7}{2}\\x=-\frac{7+2\sqrt{23}}{2}\end{cases}}}\)

Vậy .....

 (Bạn xem lại đề nha, kết quả bài b lẻ quá nhưng cách làm vẫn vậy nha!)

làm bừa thui,ai tích mình mình tích lại

Số số hạng là : 

( 99 - 1 ) : 2 + 1 = 50 ( số )

Có số cặp là :

50 : 2 = 25 ( cặp )

Mỗi cặp có giá trị là :

99 - 97 = 2 

Tổng dãy trên là :

25 x 2 = 50

Đáp số : 50

\(\left(\frac{2}{3}-x\right)^3-\frac{7}{64}=\frac{5}{16}\)

\(\Rightarrow\left(\frac{2}{3}-x\right)^3=\frac{5}{16}+\frac{7}{64}=\frac{20}{64}+\frac{7}{64}=\frac{27}{64}\)

\(\Rightarrow\left(\frac{2}{3}-x\right)^3=\left(\frac{3}{4}\right)^3\)

\(\Rightarrow\frac{2}{3}-x=\frac{3}{4}\)

\(\Rightarrow x=\frac{2}{3}-\frac{3}{4}=\frac{8}{12}-\frac{9}{12}=-\frac{1}{12}\)

\(\left(\frac{2}{3}-x\right)^3-\frac{7}{64}=\frac{5}{16}\)

\(\left(\frac{2}{3}-x\right)^3=\frac{5}{16}+\frac{7}{64}\)

\(\left(\frac{2}{3}-x\right)^3=\frac{27}{64}\Leftrightarrow\left(\frac{2}{3}-x\right)^3=\left(\frac{3}{4}\right)^3\)

\(\Leftrightarrow\frac{2}{3}-x=\frac{3}{4}\Leftrightarrow x=\frac{2}{3}-\frac{3}{4}\)

\(\Leftrightarrow x=-\frac{1}{12}\)

\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=\frac{23}{16}\)

\(4x+\frac{15}{16}=\frac{23}{16}\)

\(4x=\frac{1}{2}\)

\(x=\frac{1}{8}\)

             Vậy \(x=\frac{1}{8}\)

\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=\frac{23}{16}\)

\(\Rightarrow\left(x+x+x+x+x\right)+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=\frac{23}{16}\)

\(\Rightarrow5x+\frac{15}{32}=\frac{23}{16}\)

\(\Rightarrow5x=\frac{23}{16}-\frac{15}{32}\)

\(\Rightarrow5x=\frac{31}{32}\)

\(\Rightarrow x=\frac{31}{32}.\frac{1}{5}=\frac{31}{160}\)

Có 4x thôi Nguyễn Quang Trung

x thuộc Z

\(\left(x-2\right):\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\right)=\frac{16}{9}\)

\(\left(x-2\right):\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)=\frac{16}{9}\)

\(\left(x-2\right):\frac{2}{9}=\frac{16}{9}\)

\(x-2=\frac{32}{91}\)

\(x=\frac{32}{91}+2\)

\(x=\frac{212}{91}\)

64=4^3

ta có 4^3/4^x+1=4

suy ra x=1

\(\left|x\right|=\frac{4}{7}\)

\(\Rightarrow x=\frac{4}{7}\)

b,\(\left(2x-3\right)^2=64\)

\(\Leftrightarrow\left(2x-3\right)^2=\left(\pm8\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}2x-3=8\\2x-3=-8\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{11}{2}\\x=-\frac{5}{2}\end{cases}}}\)

c,\(\left(\frac{1}{2}\right)^x=\frac{1}{16}\)

\(\Rightarrow\left(\frac{1}{2}\right)^x=\left(\pm\frac{1}{2}\right)^4\)

\(\Rightarrow x=\pm\frac{1}{2}\)

d,\(3^{x+1}=27\)

\(\Leftrightarrow3^{x+1}=3^3\)

\(\Leftrightarrow x+1=3\)

\(\Leftrightarrow x=2\)