a) \(A=\frac{3}{\sqrt{7}-2}+\frac{7}{\sqrt{7}-\sqrt{28}}=\frac{3}{\sqrt{7}-2}-\sqrt{7}\)

\(=\frac{3-7+2\sqrt{7}}{\sqrt{7}-2}=\frac{-4+2\sqrt{7}}{\sqrt{7}-2}=\frac{2\left(\sqrt{7}-2\right)}{\sqrt{7}-2}=2\)

đk: \(\hept{\begin{cases}x>0\\x\ne4\end{cases}}\)

\(B=\left(\frac{1}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\right)\cdot\frac{\sqrt{x}-2}{\sqrt{x}}\)

\(B=\frac{2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\frac{\sqrt{x}-2}{\sqrt{x}}\)

\(B=\frac{2}{\sqrt{x}+2}\)

b) \(6B-A>0\Leftrightarrow\frac{12}{\sqrt{x}+2}-2>0\)

\(\Leftrightarrow\frac{8-2\sqrt{x}}{\sqrt{x}+2}>0\Rightarrow8-2\sqrt{x}>0\left(because:\sqrt{x}+2>0\right)\)

\(\Rightarrow\sqrt{x}< 4\Rightarrow x< 16\)

Vậy \(\hept{\begin{cases}0< x< 16\\x\ne4\end{cases}}\)

\(\sqrt{x}=0\)  đúng 

Vì 

\(\sqrt{x}\ge0\forall x\)

\(\sqrt{x}=0\)  là đúng

=> Vì căn của 0 = 0

a) đk: \(2-m\ne0\Rightarrow m\ne2\)

b) Nếu \(m=4\)

\(\Rightarrow\left(2-4\right)x-4+1=0\)

\(\Leftrightarrow-2x-3=0\Leftrightarrow x=-\frac{3}{2}\)

Ta có: \(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)

\(=\frac{\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}}{\sqrt{2}}\)

\(=\frac{\sqrt{5-2\sqrt{5}+1}+\sqrt{5+2\sqrt{5}+1}}{\sqrt{2}}\)

\(=\frac{\sqrt{\left(\sqrt{5}-1\right)^2}+\sqrt{\left(\sqrt{5}+1\right)^2}}{\sqrt{2}}=\frac{\sqrt{5}-1+\sqrt{5}+1}{\sqrt{2}}=\sqrt{10}\)

Đặt \(A=\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)

\(\Rightarrow A^2=\left(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right)^2\)

\(=\left(\sqrt{3-\sqrt{5}}\right)^2+2\sqrt{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}+\left(\sqrt{3+\sqrt{5}}\right)^2\)

\(=\left|3-\sqrt{5}\right|+2\sqrt{9-5}+\left|3+\sqrt{5}\right|\)

\(=3-\sqrt{5}+2\sqrt{4}+3+\sqrt{5}=6+4=10\)

Vì \(\hept{\begin{cases}\sqrt{3-\sqrt{5}}>0\\\sqrt{3+\sqrt{5}}>0\end{cases}}\Rightarrow\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}>0\)

=> A > 0 mà A² = 10 => A = √10

\(\frac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\)\(\frac{8}{1-\sqrt{5}}\)

\(=\frac{\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}\)\(+\frac{8}{1-\sqrt{5}}\)

\(=2\sqrt{5}+\frac{8}{1-\sqrt{5}}\)

\(=\frac{2\sqrt{5}-10+8}{1-\sqrt{5}}\)

\(=\frac{-2\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\)

\(=-2\)

kb nhaaaaa

chúc bn hok tốt k mk nha

đk: \(x\le\frac{5}{2}\)

Ta có: \(\sqrt{25-20x+4x^2}+2x=5\)

\(\Leftrightarrow\sqrt{\left(5-2x\right)^2}=5-2x\)

\(\Leftrightarrow\left|5-2x\right|=5-2x\)

\(\Leftrightarrow\orbr{\begin{cases}5-2x=5-2x\\5-2x=2x-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}0x=0\\4x=0\end{cases}}\Rightarrow x=0\)

Vậy x = 0

a) \(\sqrt{9x^2}=6\Leftrightarrow\left|x\right|=2\Rightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)

b) \(\sqrt{\left(x-2\right)^2}=5\Leftrightarrow\left|x-2\right|=5\Leftrightarrow\orbr{\begin{cases}x-2=5\\x-2=-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-3\end{cases}}\)

c) \(\sqrt{x^2-6x+9}=3\Leftrightarrow\left|x-3\right|=3\Leftrightarrow\orbr{\begin{cases}x-3=3\\x-3=-3\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=6\\x=0\end{cases}}\)

d) \(\sqrt{x^2+4x+4}-2x=3\Leftrightarrow\left|x+2\right|=2x+3\Leftrightarrow\orbr{\begin{cases}x+2=2x+3\\x+2=-2x-3\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-1\\3x=-5\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=-\frac{5}{3}\end{cases}}\)

\(a,\sqrt{9x^2}=6\)

\(\sqrt{\left(3x\right)^2}=6\)

\(\left|3x\right|=6\)

\(< =>x=\pm2\)

\(\sqrt{\left(x-2\right)^2}=5\)

\(\left|x-2\right|=5\)

\(\orbr{\begin{cases}x-2=5\\x-2=-5\end{cases}\orbr{\begin{cases}x=7\\x=-3\end{cases}}}\)

\(c,\sqrt{x^2-6x+9}=3\)

\(\sqrt{\left(x-3\right)^2}=3\)

\(\left|x-3\right|=3\)

\(\orbr{\begin{cases}x-3=3\\x-3=-3\end{cases}\orbr{\begin{cases}x=6\\x=0\end{cases}}}\)

\(d,\sqrt{x^2+4x+4}-2x=3\)

\(\sqrt{\left(x+2\right)^2}=3+2x\)

\(\left|x+2\right|=3+2x\)

dạng 3 của trị tuyệt đối

xét \(3+2x\ge0\)

\(x\ge-\frac{3}{2}\)

\(\orbr{\begin{cases}x+2=3+2x\\x+2=-3-2x\end{cases}\orbr{\begin{cases}x=-1\\-\frac{5}{3}\end{cases}}}\)