PT <=> \(2+\sqrt{x-1}+\left|3-\sqrt{x-1}\right|\) = 5
Với \(\hept{\begin{cases}\hept{\begin{cases}3-\sqrt{x-1}\ge0\\2+\sqrt{x-1}+3-\sqrt{x-1}=5\end{cases}}\\\hept{\begin{cases}3-\sqrt{x-1}< 0\\2+\sqrt{x-1}+\sqrt{x-1}-3=5\end{cases}}\\\end{cases}}\)
PT <=> \(2+\sqrt{x-1}+\left|3-\sqrt{x-1}\right|\) = 5
Với \(\hept{\begin{cases}\hept{\begin{cases}3-\sqrt{x-1}\ge0\\2+\sqrt{x-1}+3-\sqrt{x-1}=5\end{cases}}\\\hept{\begin{cases}3-\sqrt{x-1}< 0\\2+\sqrt{x-1}+\sqrt{x-1}-3=5\end{cases}}\\\end{cases}}\)
\(\left(5\right)\sqrt{x+3-4\sqrt{x-1}}\sqrt{x+8+6\sqrt{x-1}}=5\)
\(\left(6\right)2x^2+3x+\sqrt{2x^2+3x+9}=33\)
\(\left(7\right)\sqrt{3x^2+6x+12}+\sqrt{5x^4-10x^2+30}=8\)
\(\left(8\right)x+y+z+8=2\sqrt{x-1}+4\sqrt{y-2}+6\sqrt{z-3}\)
1\(\sqrt{5+2\sqrt{8}}-\sqrt{5-2\sqrt{8}}\) 2)\(\dfrac{\sqrt{x^2+2\sqrt{3x}+3}}{x^2-3}\) 3) \(\dfrac{\sqrt{x^2-5x+6}}{\sqrt{x-2}}\) 4)\(\dfrac{\sqrt{\left(x-4\right)^2}}{x^2-5x+4}\) 5) \(\dfrac{3x+1}{\sqrt{9x^2+6x+1}}\)
\(\left(6\right)\dfrac{3\sqrt{x}}{5\sqrt{x}-1}\le-3\)
\(\left(7\right)\dfrac{8\sqrt{x}+8}{6\sqrt{x}+9}>\dfrac{8}{3}\)
\(\left(8\right)\dfrac{\sqrt{x}-2}{2\sqrt{x}-3}< -4\)
\(\left(9\right)\dfrac{4\sqrt{x}+6}{5\sqrt{x}+7}\le-\dfrac{2}{3}\)
\(\left(10\right)\dfrac{6\sqrt{x}-2}{7\sqrt{x}-1}>-6\)
1,\(\sqrt{x-5}+\sqrt{x+4}=3\)
2,\(\sqrt{x+4-4\sqrt{x}}+\sqrt{x+6-6\sqrt{x}}=1\)
3,\(\sqrt{x+3}-\sqrt{x-4}=1\)
4,\(\sqrt{15-x}+\sqrt{3-x}=6\)
5,\(\sqrt{10-x}+\sqrt{x+3}=5\)
6,\(\sqrt{2x-1}+\sqrt{x-2}=\sqrt{x+1}\)
7,\(\sqrt{x+6-4\sqrt{x+2}}+\sqrt{x+11-6\sqrt{x+2}}=1\)
8,\(\sqrt{x^2-5x+6}+\sqrt{x-2-3\sqrt{x-3}}=3\)
9,\(2x^2-x+4=2\sqrt{2x+3}\)
1. \(\sqrt{x-2\sqrt{x-1}}+\sqrt{x+3-4\sqrt{x-1}}\left(2< x< 5\right)\)
2. \(\frac{6}{1-\sqrt{3}}-\frac{3\sqrt{3}-1}{\sqrt{3}+1}+\sqrt{3}\)
3. \(\sqrt{29-12\sqrt{5}+\sqrt{24-8\sqrt{3}}}\)
4. \(\sqrt{\frac{4}{9-4\sqrt{5}}}-\sqrt{\frac{4}{9+4\sqrt{5}}}\)
5. \(5\sqrt{\frac{1}{5}}+\frac{1}{2}\sqrt{x}-\frac{5}{4}\sqrt{\frac{4}{5}+\sqrt{5}}\)
6. \(\frac{6-\sqrt{6}}{\sqrt{6}-1}-9\sqrt{\frac{2}{3}}-\frac{4}{2-\sqrt{6}}\)
7. \(\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\frac{\left(\sqrt{x}-1\right)^2}{2}\left(x\ge0,x\ne1\right)\)
phương pháp 1 biến đổi về phương trình có chứa dấu giá trị tuyệt đối
1. \(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8+6\sqrt{x-1}}=5\)
2. \(\sqrt{x+2-3\sqrt{2x-5}}+\sqrt{x-2+\sqrt{2x-5}}=2\sqrt{2}\)
3. \(\sqrt{x+\sqrt{6x-9}}+\sqrt{x-\sqrt{6x-9}}=\sqrt{6}\)
4. \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=\dfrac{x+8}{5}\)
Tìm Gtnn của:
A=\(\sqrt{x-2\sqrt{x-1}}+5\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}\)
a)\(\sqrt{x^2-2x+1}+\sqrt{x^2+4x+4}=3\)
b) \(\sqrt{x+3+4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=5\)
Rút gọn
a)\(\sqrt{m+2\sqrt{m-1}}-\sqrt{m-2\sqrt{m-1}}\left(m>2\right)\)
b)\(\sqrt{x-1-2\sqrt{x-2}}+\sqrt{\sqrt{x-1+2\sqrt{x-2}}}\)
c)\(\sqrt{x+3+4\sqrt{x-1}}+\sqrt{x+8-8\sqrt{x-1}}\left(1< x< 10\right)\)
d)\(\sqrt{2m+4+6\sqrt{2m-5}}+\sqrt{2m-4-2\sqrt{2m-5}}\)