Điều kiện để phương trình có nghĩa \(x\ge1\) hoặc \(x\le-1\).
Khi \(x\le-1\) thì vế phải âm, vế trái dương, mâu thuẫn.
Khi \(x\ge1\), thì vế phải bé hơn \(x
GPT : \(\sqrt{\sqrt{x}+1-2\sqrt[4]{x}}+\sqrt{\sqrt{x}+9-6\sqrt[4]{x}}=2\)
1.Gpt: \(\dfrac{6}{x-3\sqrt{x-2}+7}=\dfrac{1}{\sqrt{x-2}}+\dfrac{\sqrt{3}}{3\sqrt{2\sqrt{x-2}}-3}\)
2.Ghpt: \(\left\{{}\begin{matrix}x^2-y-z=0\\x^3-y^2-z^2+2=0\end{matrix}\right.\)
gpt \(\sqrt{x^2+3x}+2\sqrt{x+2}=2x+\sqrt{x+\frac{6}{x}}+5\)
gpt ( đặt ẩn phụ không hoàn toàn)
9, \(2\sqrt{1-x}+\sqrt{x+1}+3\sqrt{1-x^2}=3-x\)
10, \(4\sqrt{1-x}+3\sqrt{1-x^2}=x+6+5\sqrt{1+x}\).
GPT
a) \(\sqrt{x}+\sqrt{x+1}=\frac{1}{\sqrt{x}}\)
b) \(\frac{x+3+2\sqrt{x^2-9}}{2x-6+\sqrt{x^2-9}}=\sqrt{2}\)
GPT
\(\sqrt{x+2\sqrt{x-1}}-\sqrt{x-2\sqrt{x-1}}=2.\)
Gpt: \(x^2+2\sqrt{x-1}-2x\sqrt{2-x}+1=0\)
GPT:
a .\(x+\sqrt{5+\sqrt{x-1}}=6.\)
b.\(x=\left(2004+\sqrt{x}\right)\left(1-\sqrt{1-\sqrt{x}}\right)^2\)
c. \(3x^2+21x+18+2\sqrt{x^2+7x+7}=2\)
Gpt \(1-\sqrt{2\left(x^2-x+1\right)}=x-\sqrt{x}\)
