Question

Розв'яжіть рівняння: √x+3+√x-2=√4x+1.

Answer 1

$\sqrt{x+3}+\sqrt{x-2}=\sqrt{4x+1}\Leftrightarrow \cfrac{\left ( \sqrt{x+3}+\sqrt{x-2} \right )\left ( \sqrt{x+3}-\sqrt{x-2} \right )}{\sqrt{x+3}-\sqrt{x-2}}=\sqrt{4x+1}\\\cfrac{5}{\sqrt{x+3}-\sqrt{x-2}}=\sqrt{4x+1}\Leftrightarrow \cfrac{5}{\sqrt{4x+1}}=\sqrt{x+3}-\sqrt{x-2}$

$\begin{cases}\sqrt{x+3}+\sqrt{x-2}=\sqrt{4x+1}\\ \sqrt{x+3}-\sqrt{x-2}=\cfrac{5}{\sqrt{4x+1}}\end{cases}\Rightarrow 2\sqrt{x+3}=\cfrac{4x+6}{\sqrt{4x+1}}\Leftrightarrow 2\sqrt{x+3}\sqrt{4x+1}=4x+6$$\begin{cases}4(x+3)(4x+1)=16x^2+48x+36\\ 4x+6\geq 0\end{cases}\Leftrightarrow \begin{cases}4x-24=0\\ x\geq -\cfrac{2}{3}\end{cases}\Rightarrow x=6$