bình phương 2 vế lên ta được
\(x+2\sqrt{x-1}+x-2\sqrt{x-1}+2\sqrt{x^2-4\left(x-1\right)}=\frac{\left(x+3\right)^2}{4}\)
\(< =>2x+2\sqrt{x^2-4x+1}=\frac{x^2+6x+9}{4}\)
\(< =>2\sqrt{x^2-4x+1}=\frac{x^2-2x+9}{4}\)
\(< =>\sqrt{x^2-4x+1}=\frac{x^2-2x+9}{8}\)
tiếp tục mình phương 2 vế thì sẽ ra
\(b,(\sqrt{6}+\sqrt{2})\left(\sqrt{3}-2\right)\sqrt{\sqrt{3}+2}\)
\(=(\sqrt{2}.\sqrt{3}+\sqrt{2})\left(\sqrt{3}-2\right)\sqrt{\sqrt{3}+2}\)
\(=\sqrt{2}.\left(\sqrt{3}+1\right)\left(\sqrt{3}-2\right)\sqrt{\sqrt{3}+2}\)
\(=\sqrt{2}.\sqrt{\sqrt{3}+2}\left(\sqrt{3}+1\right)\left(\sqrt{3}-2\right)\)
\(=\sqrt{2\sqrt{3}+4}\left(3+\sqrt{3}-2\sqrt{3}-2\right)\)
\(=\sqrt{\sqrt{3}^2+2\sqrt{3}+1^2}\left(1-\sqrt{3}\right)\)
\(=\sqrt{\left(1+\sqrt{3}\right)^2}\left(1-\sqrt{3}\right)\)
\(=\left(1+\sqrt{3}\right)\left(1-\sqrt{3}\right)\)
\(=1^2-\sqrt{3}^2\)
\(=1-3=-2\)
\(a,\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=\frac{x+3}{2}\)
\(\Leftrightarrow\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1+2\sqrt{x-1}+1}=\frac{x+3}{2}\)
\(\Leftrightarrow\left|\sqrt{x-1}+1\right|+\left|\sqrt{x-1}-1\right|=\frac{x+3}{2}\)
\(\Leftrightarrow\orbr{\begin{cases}-\sqrt{x-1}-1-\sqrt{x-1}+1=\frac{x-3}{2}\\\sqrt{x-1}+1+\sqrt{x-1}-1=\frac{x-3}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}-2\sqrt{x-1}=\frac{x-3}{2}\\2\sqrt{x-1}=\frac{x-3}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}-4\sqrt{x-1}=x-3\\4\sqrt{x-1}=x-3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{16x-16}=3-x\\\sqrt{16x-16}=x-3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}16x-16=9-6x+x^2\\16x-16=x^2-6x+9\end{cases}}\)
\(\Leftrightarrow16x-16=x^2-6x+9\)
\(\Leftrightarrow x^2-22x+25=0\)
