\(1.\Leftrightarrow\sqrt{x+3}=x^2-2\Leftrightarrow\left\{{}\begin{matrix}x^2-2\ge0\\x+3=\left(x^2-2\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\le-\sqrt{2}\\x\ge\sqrt{2}\end{matrix}\right.\\x^4-4x^2-x+1=0\left(1\right)\end{matrix}\right.\)
\(giải\left(1\right)\Rightarrow x\)
\(2.pt\Leftrightarrow\sqrt{x+1}+\sqrt{4-x}+\sqrt{\left(4-x\right)\left(x+1\right)}=5\)
\(đặt:\sqrt{x+1}+\sqrt{4-x}=t\)
\(t\le\sqrt{2\left(x+1+4-x\right)}=\sqrt{2.5}=\sqrt{10}\)
\(t\ge\sqrt{x+1+4-x}=\sqrt{5}\)
\(\Rightarrow\sqrt{5}\le t\le\sqrt{10}\)(1)
\(\Rightarrow pt\Leftrightarrow t+\dfrac{t^2-5}{2}=5\left(2\right)\)
\(giải\left(2\right)\Rightarrow t=...\)\(\)(đối chiếu điều kiện (1)) rồi tính ra x
\(3.bn\) \(xem\) \(lại\) \(đề\)
\(5.pt\Leftrightarrow x^2-2x-11+2x^2-2x-5-2\sqrt{2x^2-2x-5}+1=0\)
\(\Leftrightarrow x^2-2x-11+\left(\sqrt{2x^2-2x-5}-1\right)^2=0\left(1\right)\)
\(đặt:\sqrt{2x^2-2x-5}=t\ge0\Leftrightarrow t^2=2x^2-2x-5\Leftrightarrow x^2-2x=\dfrac{t^2}{2}+\dfrac{5}{2}\)
\(pt\Leftrightarrow t^2+\dfrac{5}{2}-11+\left(t-1\right)^2=0\left(2\right)\)
giải (2) \(\Rightarrow t\Rightarrow x=....\)
\(4..x^2+22x+5=16\sqrt{2x+51}\)
\(đặt:\sqrt{2x+51}=t\ge0\Leftrightarrow2x+51=t^2\Leftrightarrow x=\dfrac{t^2-51}{2}\Rightarrow pt\Leftrightarrow\dfrac{\left(t^2-51\right)^2}{4}+11\left(t^2-51\right)+5-16t=0\Leftrightarrow\dfrac{1}{4}\left(t^2-4t-29\right)\left(t^2+4t-13\right)=0\Leftrightarrow\left[{}\begin{matrix}t=....\\t=....\end{matrix}\right.\) đối chiếu \(t\ge0\Rightarrow t\Rightarrow x\)
