Question

ai cíu mình với :<

Answer 1

ĐK : x ≥ -3/2

\(\Leftrightarrow x^2+4x+3-\left(2\sqrt{2x+3}-2\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+3\right)-\frac{8x+12-4}{2\left(\sqrt{2x+3}+1\right)}=0\)

\(\Leftrightarrow\left(x+1\right)\left[\left(x+3\right)-\frac{4}{\sqrt{2x+3}+1}\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\\left(x+3\right)-\frac{4}{\sqrt{2x+3}+1}=0\end{cases}}\)

TH1 : x + 1 = 0 <=> x = -1 (tm) (1)

TH2 : \(\left(x+3\right)-\frac{4}{\sqrt{2x+3}+1}=0\Leftrightarrow\left(x+3\right)\sqrt{2x+3}+x+3=4\)

\(\Leftrightarrow\left(x+3\right)\left(\sqrt{2x+3}-1\right)+2x+2=0\)

\(\Leftrightarrow\left(x+3\right)\frac{2x+2}{\sqrt{2x+3}+1}+2x+2=0\)

\(\Leftrightarrow2\left(x+1\right)\left[\frac{x+3}{\sqrt{2x+3}+1}+1\right]=0\)(*)

Dễ thấy với x ≥ -3/2 thì \(\frac{x+3}{\sqrt{2x+3}+1}+1>0\)

nên (*) <=> x + 1 = 0 <=> x = -1 (tm) (2)

Từ (1) và (2) => pt có nghiệm x = -1