\(a.\left(x-2\right)\sqrt{x+5}=x^2-3x+2\) (đkxđ: \(x\ge-5\) )
\(\left(x-2\right)\sqrt{x+5}=\left(x-1\right)\left(x-2\right)\)
\(\left(x-2\right)\sqrt{x+5}-\left(x-1\right)\left(x-2\right)=0\)
\(\) \(\left(x-2\right)\left\lbrack\left(\sqrt{x+5}\right)-\left(x-1\right)\right\rbrack=0\)
\(\Rightarrow\left[\begin{array}{l}x-2=0\\ \sqrt{x+5}=x-1\end{array}\left(x\ge1\right)\right.\Rightarrow\left[\begin{array}{l}x=0+2\\ x+5=\left(x-1\right)^2\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{l}x=2\\ x+5=x^2-2x+1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\\ x^2-3x-4=0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{l}x=2\left(TM\right)\\ x=4\left(TM\right)\\ x=-1\left(KTM\right)\end{array}\right.\)
vậy phương trình có 2 nghiệm là x=2; x=4
\(b.4x+10=2\sqrt{2x^2+5x+2}+4\sqrt{x+3}\)
đkxđ: \(\begin{cases}2x^2+5x+2\ge0\\ x+3\ge0\end{cases}\Longrightarrow\begin{cases}x\le-2\\ x\ge-\frac12\\ x\ge-3\end{cases}\) \(\Rightarrow\begin{cases}-\frac12\le x\\ -3\le x\le-2\end{cases}\)
\(2\cdot\left(2x+5\right)=2\sqrt{2x^2+5x+2}+4\sqrt{x+3}\)
\(2x+5=\sqrt{2x^2+5x+2}+2\sqrt{x+3}\)
theo bất đẳng thức cosi ta có:
\(\sqrt{2x^2+5x+2}\le\frac{2x+1+x+2}{2}=\frac{3x+3}{2}\)
\(\Rightarrow2x+5\le\frac{3x+3}{2}+2\sqrt{x+3}\)
\(4x+10\le3x+3+4\sqrt{x+3}\)
\(x+7\le4\sqrt{x+3}\)
\(\left(x+7\right)^2\le\left(4\sqrt{x+3}\right)^2\)
\(x^2+14x+49\le16x+48\)
\(x^2-2x+1\le0\)
\(\left(x-1\right)^2\le0\)
mà \(\left(x-1\right)^2\ge0\)
nên dấu = xảy ra khi và chỉ khi
\(\left(x-1\right)^2=0\Rightarrow x=1\) (TM)
vậy nghiệm của phương trình là 1
