Bài 2: ĐKXĐ: x>=-3
Ta có: \(\sqrt{x+3}+\sqrt{9x+27}-\frac12\cdot\sqrt{4x+12}=15\)
=>\(\sqrt{x+3}+3\sqrt{x+3}-\frac12\cdot2\cdot\sqrt{x+3}=15\)
=>\(3\cdot\sqrt{x+3}=15\)
=>\(\sqrt{x+3}=5\)
=>x+3=25
=>x=22(nhận)
Bài 3:
a: ĐKXĐ: \(-x^2+2x+4\ge0\)
=>\(x^2-2x-4\le0\)
=>\(x^2-2x+1\le5\)
=>\(\left(x-1\right)^2\le5\)
=>\(-\sqrt5\le x-1\le\sqrt5\)
=>\(-\sqrt5+1\le x\le\sqrt5+1\)
Ta có: \(\sqrt{4+2x-x^2}=x-2\)
=>\(\begin{cases}x-2\ge0\\ 4+2x-x^2=\left(x-2\right)^2\end{cases}\Rightarrow\begin{cases}x\ge2\\ 4+2x-x^2-x^2+4x-4=0\end{cases}\)
=>\(\begin{cases}x\ge2\\ -2x^2+6x=0\end{cases}\Rightarrow\begin{cases}x\ge2\\ -2x\left(x-3\right)=0\end{cases}\Rightarrow x=3\) (nhận)
b: ĐKXĐ: \(25-x^2\ge0\)
=>\(x^2\le25\)
=>-5<=x<=5
\(\sqrt{25-x^2}=x-1\)
=>\(\begin{cases}x-1\ge0\\ 25-x^2=\left(x-1\right)^2\end{cases}\Rightarrow\begin{cases}x\ge1\\ 25-x^2-x^2+2x-1=0\end{cases}\)
=>\(\begin{cases}x\ge1\\ -2x^2+2x+24=0\end{cases}\Rightarrow\begin{cases}1\le x\le5\\ x^2-x-12=0\end{cases}\)
=>\(\begin{cases}1\le x\le5\\ \left(x-4\right)\left(x+3\right)=0\end{cases}\Rightarrow x=4\)
c: \(\sqrt{3x^2-9x+1}+2=x\)
=>\(\sqrt{3x^2-9x+1}=x-2\)
=>\(\begin{cases}x-2\ge0\\ 3x^2-9x+1=\left(x-2\right)^2\end{cases}\Rightarrow\begin{cases}x\ge2\\ 3x^2-9x+1-x^2+4x-4=0\end{cases}\)
=>\(\begin{cases}x\ge2\\ 2x^2-5x-3=0\end{cases}=>\begin{cases}x\ge2\\ \left(2x-3\right)\left(x+1\right)=0\end{cases}\)
=>x∈∅
