Điều kiện: \(x,y\ge0;\sqrt{x}\ne\sqrt{y}-3.\)

\(A=\frac{x-y+3\sqrt{x}+3\sqrt{y}}{\sqrt{x}-\sqrt{y}+3}=\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)+3\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}+3}\)

\(A=\frac{\left(\sqrt{x}-\sqrt{y}+3\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}+3}=\sqrt{x}+\sqrt{y}\)

\(\sqrt{x^2+1}=-3\)

\(\Rightarrow x^2+1=9\)

Suy ra : x^2 = 8

Suy ra : \(x=2\sqrt{2}\)hoặc \(x=-2\sqrt{2}\)

1.392869546

1,392869546 nha

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

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

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

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

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

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

\(=\left[3+\sqrt{5}\right]\)(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 \(x=2\).