\(\sqrt{x-2\sqrt{x-1}}=\sqrt{x-1}\)ĐK : x > = 1 

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

\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}-1\right)^2}=\sqrt{x-1}\Leftrightarrow\left|\sqrt{x-1}-1\right|=\sqrt{x-1}\)

TH1 : \(\sqrt{x-1}-1=\sqrt{x-1}\Leftrightarrow0=1\)( vô lí )

TH2 : \(\sqrt{x-1}-1=-\sqrt{x-1}\Leftrightarrow2\sqrt{x-1}=1\Leftrightarrow x-1=\frac{1}{4}\Leftrightarrow x=\frac{1}{4}+1=\frac{5}{4}\)( tm ) 

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

=> \(\sqrt{x-3}=5\)

=>  \(x-3=25\)

=>  \(x=28\)

\(\frac{x-3}{\sqrt{x-3}}=5\Rightarrow x-3=5\sqrt{x-3}\)ĐK : x >= 3 

\(\Leftrightarrow x^2-6x+9=25\left(x-3\right)\Leftrightarrow x^2-6x+9=25x-75\)

\(\Leftrightarrow x^2-31x+84=0\Leftrightarrow x=28;x=3\)

đkxđ  :    \(x\ge0\)      và \(x\ne1\)

ĐKXĐ : \(x\ge0;\sqrt{x}-1\ne0\Leftrightarrow x\ge0;x\ne1\)

4 lớn hơn 2/3

-5 nhỏ hơn -2

\(a,\sqrt{4\left(1+6x+9x^2\right)^2}=\sqrt{4\left(3x+1\right)^4}\)

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

\(=2\left(3.-\sqrt{2}+1\right)^2=2.\left(18-6\sqrt{2}+1\right)\)

\(=38-12\sqrt{2}\)

\(b,\sqrt{9a^2\left(b^2-4b+4\right)}=\left|3a\right|.\left|b-2\right|=\left|3.\left(-2\right)\right|.\left|-\sqrt{3}-2\right|\)

\(=6.\left(\sqrt{3}+2\right)=12+6\sqrt{3}\)

Xét tam giác ABC vuông tại A, đường cao AH

* Áp dụng hệ thức : \(\frac{1}{AH^2}=\frac{1}{AC^2}+\frac{1}{AB^2}\Rightarrow\frac{1}{576}=\frac{1}{1600}+\frac{1}{AB^2}\Rightarrow AB=30\)cm 

Theo Pytago tam giác ABC vuông tại A

\(BC=\sqrt{AC^2+AB^2}=50\)cm 

* Áp dụng hệ thức : \(AB^2=BH.BC\Rightarrow BH=\frac{AB^2}{BC}=\frac{900}{50}=18\)cm 

=> CH = BC - BH = 50 - 18 = 32 cm