\(ĐK:-4\le x\le\dfrac{1}{2}\\ PT\Leftrightarrow\left(\sqrt{x+4}-2\right)-\left(\sqrt{1-x}-1\right)-\left(\sqrt{1-2x}-1\right)=0\\ \Leftrightarrow\dfrac{x}{\sqrt{x+4}+2}+\dfrac{x}{\sqrt{1-x}+1}+\dfrac{2x}{\sqrt{1-2x}+1}=0\\ \Leftrightarrow x\left(\dfrac{1}{\sqrt{x+4}+2}+\dfrac{1}{\sqrt{1-x}+1}+\dfrac{2}{\sqrt{1-2x}+1}\right)=0\)
Ta thấy ngoặc lớn >0 với mọi x
\(\Rightarrow x=0\left(tm\right)\)
ĐKXĐ : \(-4\le x\le\dfrac{1}{2}\)
PT <=> \(x+4+\left(1-x\right)-2\sqrt{\left(x+4\right)\left(1-x\right)}\)\(=1-2x\)
<=> \(\sqrt{\left(x+4\right)\left(1-x\right)}=x+2\)
<=> \(x^2-3x+4=x^2+4x+4+\)
<=> x = 0 (TM)
Vậy x = 0 là nghiệm phương trình
