\(ĐK:x\ge\dfrac{1}{2}\\ PT\Leftrightarrow x+\sqrt{2x-1}+x-\sqrt{2x-1}+2\sqrt{\left(x+\sqrt{2x-1}\right)\left(x-\sqrt{2x-1}\right)}=2\\ \Leftrightarrow2x+2\sqrt{x^2-2x+1}=2\\ \Leftrightarrow\sqrt{\left(x-1\right)^2}=1-x\\ \Leftrightarrow\left|x-1\right|=1-x=-\left(x-1\right)\\ \Leftrightarrow x-1\le0\\ \Leftrightarrow x\le1\\ \Leftrightarrow\dfrac{1}{2}\le x\le1\)
Vậy PT có vô số nghiệm trong khoảng \(\left[\dfrac{1}{2};1\right]\)
ĐKXĐ: \(x\ge\dfrac{1}{2}\)
Phương trình tương đương:
\(\sqrt{2x+2\sqrt{2x-1}}+\sqrt{2x-2\sqrt{2x-1}}=2\)
\(\Leftrightarrow\sqrt{2x-1+2\sqrt{2x-1}+1}+\sqrt{2x-1-2\sqrt{2x-1}+1}=2\)
\(\Leftrightarrow\sqrt{\left(\sqrt{2x-1}+1\right)^2}+\sqrt{\left(\sqrt{2x-1}-1\right)^2}=2\)
\(\Leftrightarrow\left|\sqrt{2x-1}+1\right|+\left|\sqrt{2x-1}-1\right|=2\)
\(\Leftrightarrow\sqrt{2x-1}+1+\left|\sqrt{2x-1}-1\right|=2\)
\(\Leftrightarrow\left|\sqrt{2x-1}-1\right|=1-\sqrt{2x-1}\)
\(\Leftrightarrow\text{}\sqrt{2x-1}-1\le0\Leftrightarrow2x-1\le1\Leftrightarrow x\le1\)
Kết hợp với ĐKXĐ, ta có: \(\dfrac{1}{2}\le x\le1\)
Vậy...
