a, x=85/3

b, x=6 hoặc x=-12

a)\(\left(2x-3\right)\left(x+1\right)< 0\)

\(\Leftrightarrow\begin{cases}2x-3>0\\x+1< 0\end{cases}\)  hoặc \(\begin{cases}2x-3< 0\\x+1>0\end{cases}\)

\(\Leftrightarrow\begin{cases}x>\frac{3}{2}\\x< -1\end{cases}\) (loại)  hoặc \(\begin{cases}x< \frac{3}{2}\\x>-1\end{cases}\)

\(\Leftrightarrow-1< x< \frac{3}{2}\)

b) \(\left(x-\frac{1}{2}\right)\left(x+3\right)>0\)

\(\Leftrightarrow\begin{cases}x-\frac{1}{2}>0\\x+3>0\end{cases}\) hoặc \(\begin{cases}x-\frac{1}{2}< 0\\x+3< 0\end{cases}\)

\(\Leftrightarrow\begin{cases}x>\frac{1}{2}\\x>-3\end{cases}\) hoặc \(\begin{cases}x< \frac{1}{2}\\x< -3\end{cases}\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x>\frac{1}{2}\\x< -3\end{array}\right.\)

c) Sai đề phải là \(\frac{x}{\left(x+3\right)\left(x+7\right)}\)

Có: \(\frac{3}{\left(x+3\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+3\right)\left(x+17\right)}\)

\(\Leftrightarrow\)\(\frac{1}{x+3}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+7}=\frac{x}{\left(x+3\right)\left(x+7\right)}\)

\(\Leftrightarrow\)\(\frac{1}{x+3}-\frac{1}{x+7}=\frac{x}{\left(x+3\right)\left(x+7\right)}\)

\(\Leftrightarrow\)\(\frac{4}{\left(x+3\right)\left(x+7\right)}=\frac{x}{\left(x+3\right)\left(x+7\right)}\)

\(\Leftrightarrow x=4\)

đề câu c sai rầu kìa má

Bạn đăng ít một thôi!

mk lỡ đăng rồi bạn ạ 

nên các bạn giải bài nào cũng được

a)\(5^x.\left(5^3\right)^2=625\)

\(5^x.5^6=5^4\)

\(5^x=5^{-2}\)

\(x=-2\)

b)\(27< 81^3:3^x< 243\)

\(3^3< \left(3^4\right)^3:3^x< 3^5\)

\(3^3< 3^{12}:3^x< 3^5\)

\(3^{12}:3^x=3^4\)

\(3^x=3^3\)

\(x=3\)

c)\(\left(5x+1\right)^2=\frac{36}{49}\) 

\(\left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\)

\(5x+1=\frac{6}{7}\)

\(5x=\frac{-1}{7}\)

\(x=\frac{-1}{35}\)

d)\(\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}\right)^6\)

\(\left(x-\frac{2}{9}\right)^3=\left[\left(\frac{2}{3}\right)^2\right]^3\)

\(x-\frac{2}{9}=\frac{4}{9}\)

\(x=\frac{6}{9}=\frac{2}{3}\)

\(5^x.\left(5^3\right)^2=625\)

\(\Rightarrow5^x.5^6=5^4\)

\(\Rightarrow5^{x+6}=5^4\Rightarrow x+6=4\Rightarrow x=-2\)

Đề sai rồi bạn : Phải là :

 \(5^x:\left(5^3\right)^2=625\)

\(\Rightarrow5^x:5^6=5^4\)

\(\Rightarrow5^{x-6}=5^4\)

\(\Rightarrow x-6=4\Rightarrow x=10\)

Nhứng nếu đề đúng thì bạn có thể lấy KQ trên

1. Tìm x, biết :

a. ( x - \(\frac{3}{4}\)) \(^2\)= 0

=> x - \(\frac{3}{4}\)= 0

=> x = 0 + \(\frac{3}{4}\)

=> x = \(\frac{3}{4}\)

b. ( x + \(\frac{1}{2}\)) \(^2\)= \(\frac{9}{64}\)

=> ( x + \(\frac{1}{2}\)) \(^2\)= ( \(\frac{3}{8}\)) \(^2\)

=> x + \(\frac{1}{2}\)= \(\frac{3}{8}\)

=> x = \(\frac{3}{8}\)- \(\frac{1}{2}\)

=> x = \(\frac{-1}{8}\)

c.  \(\frac{\left(-2\right)^x}{16}=-8\)

=> \(\frac{\left(-2\right)^x}{16}=\frac{-8}{1}=\frac{-128}{16}\)

=> ( -2)\(^x\)= -128

=> ( -2 ) \(^x\)= ( -2) \(^7\)

=> x = 7

\(9,5-\frac{3}{4}\left|X-\frac{1}{3}\right|=6\frac{1}{3}-\frac{1}{3}\left|\frac{1}{3}-X\right|\)

\(\frac{19}{2}-\frac{3}{4}\left|X-\frac{1}{3}\right|=\frac{19}{3}-\frac{1}{3}\left|X-\frac{1}{3}\right|\)

\(\frac{19}{2}-\frac{3}{4}\left|X-\frac{1}{3}\right|+\frac{1}{3}\left|X-\frac{1}{3}\right|=\frac{19}{3}\)

\(\frac{19}{2}-\left(\frac{3}{4}\left|X-\frac{1}{3}\right|-\frac{1}{3}\left|X-\frac{1}{3}\right|=\frac{19}{3}\right)\)

\(\left|X-\frac{1}{3}\right|\left(\frac{3}{4}-\frac{1}{3}\right)=\frac{19}{2}-\frac{19}{3}\)

\(\frac{5}{12}\left|X-\frac{1}{3}\right|=\frac{19}{6}\)

\(\left|X-\frac{1}{3}\right|=\frac{19}{6}\div\frac{5}{12}\)

\(\left|X-\frac{1}{3}\right|=\frac{38}{5}\)

\(\Rightarrow\orbr{\begin{cases}X-\frac{1}{3}=\frac{38}{5}\\X-\frac{1}{3}=\frac{-38}{5}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{119}{15}\\x=\frac{-109}{15}\end{cases}}\)

Vậy.....................

P/s: sai thì bỏ qua nha!