Bài 2:
3) ĐKXĐ: \(x\ge1\)Ta có: \(\sqrt{49x-49}-\sqrt{25x-25}=3\)
\(\Leftrightarrow7\sqrt{x-1}-5\sqrt{x-1}=3\)
\(\Leftrightarrow2\sqrt{x-1}=3\)
\(\Leftrightarrow\sqrt{x-1}=\dfrac{3}{2}\)
\(\Leftrightarrow x-1=\dfrac{9}{4}\)
hay \(x=\dfrac{13}{4}\)(thỏa ĐK)
Vậy: \(S=\left\{\dfrac{13}{4}\right\}\)
4) Ta có: \(1+\dfrac{3\left(x-5\right)}{4}>\dfrac{2x-1}{6}-2\)
\(\Leftrightarrow\dfrac{12}{12}+\dfrac{9\left(x-5\right)}{12}-\dfrac{2\left(2x-1\right)}{12}-\dfrac{24}{12}>0\)
\(\Leftrightarrow12+9x-45-4x+2-24>0\)
\(\Leftrightarrow5x-55>0\)
\(\Leftrightarrow5x>55\)
hay x>11
Vậy: S={x|x>11}
5) Ta có: \(\dfrac{2x+3}{x^2+1}< 0\)
mà \(x^2+1>0\forall x\)
nên 2x+3<0
\(\Leftrightarrow2x< -3\)
hay \(x< -\dfrac{3}{2}\)
Vậy: S={x|\(x< -\dfrac{3}{2}\)}