\(5;;\sqrt{\left(x+5\right)\left(3x+4\right)}>4\left(x-1\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}4\left(x-1\right)\le0\\\left(x+5\right)\left(3x+4\right)\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}4\left(x-1\right)\ge0\\\left(x+5\right)\left(3x+4\right)\ge0\\\left(x+5\right)\left(3x+4\right)>16\left(x-1\right)^2\end{matrix}\right.\end{matrix}\right.\)
\(TH:\left\{{}\begin{matrix}4\left(x-1\right)\le0\\\left(x+5\right)\left(3x+4\right)\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le1\\\left[{}\begin{matrix}x\le-5\\x\ge-\dfrac{4}{3}\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow x\in(-\infty;-5]\cup\left[-\dfrac{4}{3};1\right]\left(1\right)\)
\(TH:\left\{{}\begin{matrix}4\left(x-1\right)\ge0\\\left(x+5\right)\left(3x+4\right)\ge0\\\left(x+5\right)\left(3x+4\right)>16\left(x-1\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\\left[{}\begin{matrix}x\le-5\\x\ge-\dfrac{4}{3}\end{matrix}\right.\\-\dfrac{1}{13}< x< 4\\\end{matrix}\right.\)\(\Rightarrow x\in[1;4)\left(2\right)\)
\(\left(1\right)\left(2\right)\Rightarrow x\in(-\infty;5]\cup[\dfrac{-4}{3};4)\)
\(6;;;;\sqrt{7x+7}+\sqrt{7x-6}+2\sqrt{49x^2+7x-42}< 181-14x\)
(đoạn 49x^2+7x+42 chắc bạn viết sai đề dấu"-" thành "+")
\(đk:\left\{{}\begin{matrix}7x+7\ge0\\7x-6\ge0\end{matrix}\right.\) \(\Leftrightarrow x\ge\dfrac{6}{7}\)
\(bpt\Leftrightarrow\sqrt{7x+7}+\sqrt{7x-6}+2\sqrt{\left(7x+7\right)\left(7x-6\right)}+14x+1< 182\left(1\right)\)
\(đặt:\sqrt{7x+7}+\sqrt{7x-6}=t>0\)
\(\Rightarrow t^2=14x+1+2\sqrt{\left(7x+7\right)\left(7x-6\right)}\)
\(\Rightarrow\left(1\right)\Leftrightarrow t^2+t< 182\Leftrightarrow-14< t< 13\)
\(\Rightarrow\sqrt{7x+7}+\sqrt{7x-6}< 13\Leftrightarrow14x+1+2\sqrt{\left(7x+7\right)\left(7x-6\right)}< 169\)
\(\Leftrightarrow2\sqrt{\left(7x+7\right)\left(7x-6\right)}< 168-14x\)
\(\Leftrightarrow\left\{{}\begin{matrix}168-14x\ge0\\\left(7x+7\right)\left(7x-6\right)\ge0\\4\left(7x+7\right)\left(7x-6\right)< \left(168-14x\right)^2\\\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le12\\\left[{}\begin{matrix}x\le-1\\x\ge\dfrac{6}{7}\end{matrix}\right.\\x< 6\\\end{matrix}\right.\)\(\Rightarrow\dfrac{6}{7}\le x< 6\)
\(7;\) \(3\sqrt{x}+\dfrac{3}{2\sqrt{x}}< 2x+\dfrac{1}{2x}-1\left(đk:x>0\right)\)
\(\Leftrightarrow3\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)< 2\left(x+\dfrac{1}{4x}\right)-1\left(1\right)\)
\(đặt:\sqrt{x}+\dfrac{1}{2\sqrt{x}}=t>0\)
\(\Leftrightarrow t^2=\sqrt{x}^2+2.\sqrt{x}.\dfrac{1}{2\sqrt{x}}+\left(\dfrac{1}{2\sqrt{x}}\right)^2=x+\dfrac{1}{4x}+1\)
\(\Rightarrow x+\dfrac{1}{4x}=t^2-1\)
\(\left(1\right)\Leftrightarrow3t< 2\left(t^2-1\right)-1\)
\(\Leftrightarrow2t^2-3t-3>0\Leftrightarrow\left[{}\begin{matrix}t< \dfrac{3-\sqrt{33}}{4}\\t>\dfrac{3+\sqrt{33}}{4}\end{matrix}\right.\)
\(\Rightarrow\sqrt{x}+\dfrac{1}{2\sqrt{x}}>\dfrac{3+\sqrt{33}}{4}\)
\(\Leftrightarrow\dfrac{2x+1}{2\sqrt{x}}>\dfrac{3+\sqrt{33}}{4}\)
\(\Leftrightarrow\sqrt{x}< \dfrac{2\left(2x+1\right)}{3+\sqrt{33}}\Leftrightarrow\left\{{}\begin{matrix}x>0\\2\left(2x+1\right)\ge0\\x< \left[\dfrac{2\left(2x+1\right)}{3+\sqrt{33}}\right]^2\\\end{matrix}\right.\)
đến đây dễ dàng rồi như mấy ý trên bạn tự giải quyết để tìm ra x
\(8;x+1+\sqrt{x^2-4x+1}\ge3\sqrt{x}\) \(\left(đk:x\ge2+\sqrt{3}\right)\)
\(bpt\Leftrightarrow x+1-3\sqrt{x}+\sqrt{x^2-4x+1}\ge0\)
\(\Leftrightarrow x+1-5-3\sqrt{x}+6+\sqrt{x^2-4x+1}-1\ge0\)
\(\Leftrightarrow\dfrac{\left(x+1\right)^2-5^2}{x+1+5}-\dfrac{9x-36}{3\sqrt{x}+6}+\dfrac{x^2-4x+1-1}{\sqrt{x^2-4x+1}+1}\ge0\)
\(\Leftrightarrow\dfrac{\left(x-4\right)\left(x+1+5\right)}{x+1+5}-\dfrac{9\left(x-4\right)}{3\sqrt{x}+6}+\dfrac{x\left(x-4\right)}{\sqrt{x^2-4x+1}+1}\ge0\)
\(\Leftrightarrow\left(x-4\right)\left[1-\dfrac{9}{3\sqrt{x}+6}+\dfrac{x}{\sqrt{x^2-4x+1}+1}\right]\ge0\)
\(\Leftrightarrow\left(x-4\right)\left[-\dfrac{9}{3\sqrt{x}+6}+\dfrac{6}{5}+\dfrac{x}{\sqrt{x^2-4x+1}+1}-\dfrac{1}{5}\right]\ge0\)
liên hợp 1 lần nữa trong ngoặc tương tự
\(\Rightarrow\left(x-4\right)\left(x-\dfrac{1}{4}\right)\left[...........\right]\ge0\)
\(\Leftrightarrow\left(x-4\right)\left(x-\dfrac{1}{4}\right)\ge0\Leftrightarrow\left[{}\begin{matrix}x\ge4\\x\le\dfrac{1}{4}\end{matrix}\right.\)
