cho làm tắt nha , chỗ nèo không hiểu thì hỏi
a,
\(2\left(4x+6\right)< \left(x+3\right)^2+\left(5-x\right)\left(x+5\right)\)
\(\Leftrightarrow4x+6=2\left(2x+3\right)\)
\(\Leftrightarrow4\left(2x+3\right)-\left(6x+34\right)< 0\)
\(\Leftrightarrow2x-22=2\left(x-11\right)\)
\(\Leftrightarrow2\left(x-11\right)< 0\)
\(\Leftrightarrow x< 11\)
S \(\in\left\{-\infty,11\right\}\)
b, \(5\left(x+2\right)^2< \left(2x+3\right)\left(2x-3\right)+\left(x-5\right)^2+30x\)
\(\Leftrightarrow5\left(x+2\right)^2-\left(5x^2+20x+16\right)< 0\)
mà \(\left(x+2\right)^2=x^2+4x+4\)
\(\Leftrightarrow4< 0\) ( vô lý)
\(S\in\)∅
c,
\(\dfrac{\left(x-3\right)^2}{3}-\dfrac{\left(2x-1\right)^2}{12}\le x\)
\(\Leftrightarrow\left(\dfrac{\left(x-3\right)^2}{3}-\dfrac{\left(2x-1\right)^2}{12}\right)-x\le0\)
\(\Leftrightarrow\dfrac{4\left(x-3\right)^2-\left(\left(2x-1\right)^2\right)}{12}=\dfrac{35-20x}{2}\)
\(\Leftrightarrow\dfrac{35-20x}{12}-x\le0\)
\(\Leftrightarrow35-20x=-5\left(4x-7\right)\)
\(\Leftrightarrow\dfrac{-5\left(4x-7\right)-\left(x.12\right)}{12}=\dfrac{35-32x}{12}\)
\(\Leftrightarrow\dfrac{35-32x}{12}\le0\)
\(\Leftrightarrow32x-35\ge0\)
\(\Leftrightarrow x\ge\dfrac{35}{32}\)
\(S\in[\dfrac{35}{32},\infty)\)
