\(A=\sqrt{23+3\sqrt{5}}\)

\(\sqrt{2}A=\sqrt{46+6\sqrt{5}}=\sqrt{45+2.3\sqrt{5}+1}=\sqrt{\left(3\sqrt{5}\right)^2+2.3\sqrt{5}+1^2}\)

\(=\sqrt{\left(3\sqrt{5}+1\right)^2}=3\sqrt{5}+1\)

\(\Rightarrow A=\frac{3\sqrt{5}+1}{\sqrt{2}}=\frac{3\sqrt{10}}{2}+\frac{\sqrt{2}}{2}\)

=\(\sqrt{3^2+2.3.\sqrt{5}+\sqrt{5^2}}\)

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

\(=\left[3+\sqrt{5}\right]\)(dấu ngoặc vuông thay = dấu giá trị tuyệt đối nhé.)

\(=3+\sqrt{5}\)

\(A=\sqrt{7-3\sqrt{5}}\)

\(\sqrt{2}A=\sqrt{14-6\sqrt{5}}=\sqrt{9-2.3.\sqrt{5}+5}=\sqrt{3^2-2.3.\sqrt{5}+\left(\sqrt{5}\right)^2}\)

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

\(\Rightarrow A=\frac{3-\sqrt{5}}{\sqrt{2}}=\frac{3\sqrt{2}}{2}-\frac{\sqrt{10}}{2}\)

\(=\sqrt{3^2-2.3\sqrt{5}+\sqrt{5^2}}\)

\(=\sqrt{\left(3-\sqrt{5}\right)^2}\)(phương pháp đưa về hằng đẳng thức)

\(=\left[3-\sqrt{5}\right]\)(thay '[...] bằng dấu g/trị tuyệt đối)

\(=3-\sqrt{5}\)

a, \(\sqrt{3+2\sqrt{2}}=\sqrt{\sqrt{2}^2+2\sqrt{2}+1}=\sqrt{\left(\sqrt{2}+1\right)^2}=\left|\sqrt{2}+1\right|=\sqrt{2}+1\)

b, \(\sqrt{3-2\sqrt{2}}=\sqrt{\sqrt{2}^2-2\sqrt{2}+1}=\sqrt{\left(\sqrt{2}-1\right)^2}=\left|\sqrt{2}-1\right|=\sqrt{2}-1\)

c, \(\sqrt{8-2\sqrt{15}}=\sqrt{\sqrt{5}^2-2\sqrt{5.3}+\sqrt{3}^2}=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)

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

\(\sqrt{2+\sqrt{3}}=\frac{\sqrt{4+2\sqrt{3}}}{\sqrt{2}}=\frac{\sqrt{3+2\sqrt{3}+1}}{\sqrt{2}}=\frac{\sqrt{\left(\sqrt{3}\right)^2+2\sqrt{3}+1^2}}{\sqrt{2}}\)

\(=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{2}}=\frac{\sqrt{3}+1}{\sqrt{2}}=\frac{\sqrt{6}}{2}+\frac{\sqrt{2}}{2}\)

Đặt \(A=\sqrt{6-\sqrt{35}}\)

\(\sqrt{2}A=\sqrt{12-2\sqrt{35}}=\sqrt{\left(\sqrt{7}-\sqrt{5}\right)^2}\)

\(=\left|\sqrt{7}-\sqrt{5}\right|=\sqrt{7}-\sqrt{5}\)

Vậy \(A=\frac{\sqrt{7}-\sqrt{5}}{\sqrt{2}}=\frac{\sqrt{14}-\sqrt{10}}{2}\)

Ta có: \(x^2-4x+4=\left(x-2\right)^2\ge0\)nên 

\(\sqrt{x^2-4x+5}+\sqrt{x^2-4x+8}+\sqrt{x^2-4x+9}\)

\(=\sqrt{x^2-4x+4+1}+\sqrt{x^2-4x+4+4}+\sqrt{x^2-4x+4+5}\)

\(\ge\sqrt{0+1}+\sqrt{0+4}+\sqrt{0+5}=3+\sqrt{5}\)

Dấu \(=\)khi \(x=2\).

Vậy nghiệm phương trình đã cho là \(x=2\).

\(\sqrt{-x+1}\ge6\)   

\(-x+1\ge6^2\)   

\(-x+1\ge36\)   

\(-x\ge35\)   

\(x\le-35\)

\(\sqrt{2x-1}\le2\)   

ĐK \(2x-1\ge0\)   

\(x\ge\frac{1}{2}\)   

\(\sqrt{2x-1}\le2\)   

\(2x-1\le2^2\)   

\(2x-1\le4\)   

\(2x\le5\)   

\(x\le\frac{5}{2}\)

\(\sqrt{2x-1}\le2\)ĐK : \(2x-1\ge0\Leftrightarrow x\ge\frac{1}{2}\)

\(\Leftrightarrow2x-1\le2\Leftrightarrow2x\le3\Leftrightarrow x\le\frac{3}{2}\)

Kết hợp với đk vậy \(\frac{1}{2}\le x\le\frac{3}{2}\)

Bài 1 : Với \(x>0;x\ne1\)

a, \(P=\left(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{\sqrt{x}+1}\right):\left(\frac{1}{\sqrt{x}+1}+\frac{2}{x-1}\right)\)

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

b, Ta có : \(x=4-2\sqrt{3}=\left(\sqrt{3}-1\right)^2\)

Thay vào P ta được : \(\frac{\left(\sqrt{3}-1\right)^2+1}{\sqrt{3}}=\frac{\sqrt{3}\left(\sqrt{3}-1\right)^2+\sqrt{3}}{3}\)

Bài 1.2 

\(\hept{\begin{cases}x+2y=6\\2x+3y=7\end{cases}\Leftrightarrow\hept{\begin{cases}2x+4y=12\\2x+3y=7\end{cases}\Leftrightarrow}\hept{\begin{cases}y=5\\x+2y=6\end{cases}}}\)

Thay (1) vào (2) 

\(\left(2\right)\Rightarrow x+10=6\Leftrightarrow x=-4\)

Vậy hệ phương trình có một nghiệm ( x ; y ) = ( - 4 ; 5 )