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

Đặt 2x^2 + x +2013 = a, x^2-5x+2012 = b

Ta có: a^2 + 4b^2 = 4ab

          a^2 - 4ab + 4b^2 = 0

          (a-2b)^2 = 0

Do đó: a = 2b

Hay: 2x^2 + x -2013 = 2(x^2 -5x -2012)     

        2x^2 + x -2013 = 2x^2 -10x -4024

        x-2013 = -10x -4024

        x+10x = -4024+2013

        11x = -2011

         x = -2011/11

Bạn hỏi nhiều câu hay đấy. Chúc bạn học tốt.   

Sửa tí nha kết quả cuối sai dâu phải là \(x=\dfrac{-2011}{11}\)

\(\left(2x^2+x-2013\right)^2+4\left(x^2-5x-2012\right)^2=4\left(2x^2+x-2013\right)\left(x^2-5x-2012\right)\\ \Leftrightarrow\left(2x^2+x-2013\right)^2-4\left(2x^2+x-2013\right)\left(x^2-5x-2012\right)+4\left(x^2-5x-2012\right)^2=0\\ \Leftrightarrow\left[2x^2+x-2013-2\left(x^2-5x-2012\right)\right]^2=0\\ \Leftrightarrow\left(11x+2011\right)^2=0\\ \Leftrightarrow11x+2011=0\\ \Leftrightarrow x=\dfrac{2011}{11}\)

\(\dfrac{ }{ }\)rtttytytrrttr

\(\left(2x^2+x-2013\right)^2+4\left(x^2-5x-2012\right)^2=4\left(2x^2+x-2013\right)\left(x^2-5x-2012\right)\)( * )

Đặt \(a=2x^2+x-2013\)

\(\)Đặt \(b=x^2-5x-2012\)

Khi đó ( * ) trở thành:

\(a^2+4b^2=4ab\)

\(\Leftrightarrow a^2+4b^2-4ab=0\)

\(\Leftrightarrow a^2-4ab+4b^2=0\)

\(\Leftrightarrow\left(a-2b\right)^2=0\)

\(\Leftrightarrow a-2b=0\)

\(\Leftrightarrow a=2b\)

\(\Leftrightarrow2x^2+x-2013=2\left(x^2-5x-2012\right)\)

\(\Leftrightarrow2x^2+x-2013-2x^2+10x+4024=0\)

\(\Leftrightarrow11x+2011=0\)

\(\Leftrightarrow x=\dfrac{-2011}{11}\)

Vậy...

đặt: \(x=2x^2+x-2013\\ y=x^2-5x-2012\), khi đó:

\(x^2+4y^2=4xy\\ \Leftrightarrow x^2-4xy+y^2=0\\ \Leftrightarrow\left(x-2y\right)^2=0\Rightarrow x-2y=0\\ \Leftrightarrow x=2y\\ \Rightarrow2x^2+x-2013=2x^2-10x-4024\)

\(\Leftrightarrow11x=-2011\\ \Leftrightarrow x=-\dfrac{2011}{11}\)

vậy ........

\(\left(2x^2+x-2013\right)^2+4\left(x^2-5x-2012\right)^2=4\left(2x^2+x-2013\right)\left(x^2-5x-2012\right)\)

Đặt \(\left\{{}\begin{matrix}2x^2+x-2013=a\\x^2-5x-2012=b\end{matrix}\right.\) thì ta có:

\(a^2+4b^2=4ab\)\(\Rightarrow a^2+4b^2-4ab=0\)

\(\Rightarrow\left(a-2b\right)^2=0\Rightarrow a-2b\Rightarrow a=2b\)

Tức là \(2x^2+x-2013=2\left(x^2-5x-2012\right)\)

\(\Leftrightarrow2x^2+x-2013=2x^2-10x-4024\)

\(\Leftrightarrow11x+2011=0\Leftrightarrow11x=-2011\Rightarrow x=-\dfrac{2011}{11}\)