Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\dfrac{2-\sqrt{5}}{\sqrt{17-4\sqrt{\left(\sqrt{5}+2\right)^2}}}=\dfrac{2-\sqrt{5}}{\sqrt{9-4\sqrt{5}}}=\dfrac{2-\sqrt{5}}{\sqrt{5}-2}=-1\)
\(\dfrac{2-\sqrt{5}}{\sqrt{17-4\sqrt{9+4\sqrt{5}}}}=\dfrac{2-\sqrt{5}}{\sqrt{17-4\sqrt{\left(\sqrt{5}+2\right)^2}}}\\ =\dfrac{2-\sqrt{5}}{\sqrt{17-4\left(\sqrt{5}+2\right)}}=\dfrac{2-\sqrt{5}}{\sqrt{9-4\sqrt{5}}}\\ =\dfrac{2-\sqrt{5}}{\sqrt{\left(\sqrt{5}-2\right)^2}}=\dfrac{2-\sqrt{5}}{\sqrt{5}-2}=-1\)
a, \(\sqrt{4-x}=2\sqrt{2}\)đk x =< 4
\(\Leftrightarrow4-x=8\Leftrightarrow x=-4\left(tmđk\right)\)
b, \(\dfrac{5}{3}\sqrt{6x}+\dfrac{1}{3}\sqrt{6x}+2\sqrt{6x}=2\)đk x >= 0
\(\Leftrightarrow4\sqrt{6x}=2\Leftrightarrow\sqrt{6x}=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{4}:6=\dfrac{1}{24}\left(tmđk\right)\)
3,đk x>= -1 \(4\sqrt{1+x}-3\sqrt{1+x}=5\Leftrightarrow x+1=25\Leftrightarrow x=24\left(tmđk\right)\)
4, \(2\left|x-1\right|=4\Leftrightarrow\left|x-1\right|=2\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
đk x>= 3/2 và x > 1 <=> x >= 3/2
\(2x-3=4\left(x-1\right)\Leftrightarrow2x-3=4x-4\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\left(ktm\right)\)
Vậy pt vô nghiệm
\(\left(\dfrac{\sqrt{3}\left(2-\sqrt{2}\right)}{2\sqrt{2}-2}-\dfrac{6\sqrt{6}}{3}\right)\dfrac{1}{\sqrt{6}}=\left(\dfrac{\sqrt{3}}{\sqrt{2}}-2\sqrt{6}\right)\dfrac{1}{\sqrt{6}}=\dfrac{\sqrt{3}}{\sqrt{12}}-2=\sqrt{\dfrac{1}{4}}-2=\dfrac{1}{2}-2=-\dfrac{3}{2}\)
sửa đề \(\left(\dfrac{\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{2}-\sqrt{3}}+1\right)\left(\dfrac{\sqrt{3}\left(\sqrt{7}+\sqrt{5}\right)}{\sqrt{7}+\sqrt{5}}+1\right)\)
\(=\left(1-\sqrt{3}\right)\left(\sqrt{3}+1\right)=1-3=-2\)
