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

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

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

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

\(=x-\sqrt{x}+1\)

\(f,x^2-2\sqrt{5}x+5=0\)

\(\left(x-\sqrt{5}\right)^2=0\)

\(x-\sqrt{5}=0\)

\(x=\sqrt{5}\)

\(g,\sqrt{x^2-4x+4}-\sqrt{x^2+2x+1}=-3\)

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

\(\left|x-2\right|-\left|x+1\right|=-3\)

lập bảng xét dấu:

\(TH1:x\le-1\)

\(2-x+x+1=-3\)

\(3=-3\left(KTM\right)\)

\(TH2:-1< x\le2\)

\(2-x-x-1=-3\)

\(2x=4\)

\(x=2\left(TM\right)\)

\(TH3:x>2\)

\(x-2-x-1=-3\)

\(0x=0\)

pt vô số n⁰ kết hợp với đkxđ

\(x>2\)

Trả lời:

f) \(x^2-2\sqrt{5}x+5=0\)

\(\Leftrightarrow x^2-2.x.\sqrt{5}+\left(\sqrt{5}\right)^2=0\)

\(\Leftrightarrow\left(x-\sqrt{5}\right)^2=0\)

\(\Leftrightarrow x-\sqrt{5}=0\)

\(\Leftrightarrow x=\sqrt{5}\)

câu f mình trả lời sai ạ

\(=\sqrt{7}-\sqrt{4-2.2\sqrt{7}+7}\)

\(=\sqrt{7}-\sqrt{\left(2-\sqrt{7}\right)^2}\)

\(=\sqrt{7}-\sqrt{7}+2=2\)

\(\sqrt{7}-\sqrt{11-4\sqrt{7}}\)

\(=\sqrt{7}-\sqrt{\sqrt{7}^2-2\cdot\sqrt{7}\cdot2+2^2}\)

\(=\sqrt{7}-\sqrt{\left(\sqrt{7}-2\right)^2}\)

\(=\sqrt{7}-\left|\sqrt{7}-2\right|\)

\(=\sqrt{7}-\sqrt{7}+2=2\)

\(5\sqrt{34}+\left|6-\sqrt{34}\right|\)

\(6>\sqrt{34}\)

\(5\sqrt{34}+6-\sqrt{34}\)

\(4\sqrt{34}+6\)

=) ko bt

a) Ta có:

\(\sqrt{\frac{289}{225}}=\sqrt{\frac{\sqrt{289}}{\sqrt{225}}}=\sqrt{\frac{17^2}{15^2}}=\frac{17}{15}\)

b) Ta có:

\(\sqrt{2\frac{14}{25}}=\sqrt{\frac{64}{25}}=\sqrt{\frac{\sqrt{64}}{\sqrt{25}}}=\sqrt{\frac{8^2}{5^2}}=\frac{8}{5}\)

c) Ta có:

\(\sqrt{\frac{0,25}{9}}=\sqrt{\frac{\sqrt{0,25}}{\sqrt{9}}}=\sqrt{\frac{0,5^2}{3^2}}=\frac{0,5}{3}=\frac{1}{6}\)

d) Ta có:

\(\sqrt{\frac{8,1}{1,6}}=\sqrt{\frac{81.0,1}{16.0,1}}=\sqrt{\frac{81}{16}}=\sqrt{\frac{\sqrt{81}}{\sqrt{16}}}=\sqrt{\frac{9^2}{4^2}}=\frac{9}{4}\)

a)Ta có: \(\sqrt{\frac{289}{225}}=\frac{\sqrt{289}}{\sqrt{225}}=\frac{17}{15}\)

b) Ta có: \(\sqrt{2\frac{14}{25}}=\sqrt{\frac{64}{25}}=\frac{\sqrt{64}}{\sqrt{25}}=\frac{8}{5}\)

c) Ta có: \(\sqrt{\frac{0,25}{9}}=\frac{\sqrt{0,25}}{\sqrt{9}}=\frac{0,5}{3}=\frac{1}{6}\)

d)Ta có : \(\sqrt{\frac{8,1}{1,6}}=\frac{\sqrt{8,1}}{\sqrt{1,6}}=\frac{\sqrt{8,1}.100}{\sqrt{1,6}.100}=\frac{\sqrt{81}}{\sqrt{16}}=\frac{9}{4}\)

\(\sqrt{x}=0\)  đúng 

Vì 

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

\(\sqrt{x}=0\)  là đúng

=> Vì căn của 0 = 0