sửa \(\left(1+\frac{\sqrt{x}}{x+1}\right):\left(\frac{1}{\sqrt{x}-1}-\frac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}\right)\)ĐK : \(x>0;x\ne1\)

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

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

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

\(\frac{1}{3}\sqrt{9+6a+a^2}+\frac{4a}{3}+5\)

\(=\frac{1}{3}\sqrt{\left(a+3\right)^2}+\frac{4a}{3}+5\)

\(=\frac{1}{3}\left|a+3\right|+\frac{4a}{3}+5\)(1)

Với a < 3 \(\left(1\right)=-\frac{1}{3}\left(a+3\right)+\frac{4}{3}a+5=a+4\)

Với a >= 3 \(\left(1\right)=\frac{1}{3}\left(a+3\right)+\frac{4}{3}a+5=\frac{5}{3}a+6\)

Kẻ HK vuông góc BC

 \(\Delta BKM=\Delta BCD\left(g-g\right)\)

 \(\Delta CHM=\Delta CBE\left(g-g\right)\)

\(\frac{\Rightarrow CH}{BC}=\frac{CM}{CE}\)

\(\Rightarrow CH\cdot CE=CM\cdot BC\left(2\right)\)

Từ (1) và (2 ) ta có :

\(BH\cdot BD+CH\cdot CE=BC^2\)

\(\Delta ABD=ACE\left(g-g\right)\)

\(\frac{\Rightarrow AB}{AC}=\frac{AD}{AE}\)

\(\Rightarrow AB\cdot AE=AC\cdot AD\)

Ta có : \(\frac{AB}{AC}=\frac{1}{4}\Rightarrow AB=\frac{1}{4}AC\)

Xét tam giác ABC vuông tại A, đường cao AH

* Áp dụng hệ thức : \(\frac{1}{AH^2}=\frac{1}{AB^2}+\frac{1}{AC^2}\Leftrightarrow\frac{1}{64}=\frac{1}{\left(\frac{1}{4}AC\right)^2}+\frac{1}{AC^2}\Leftrightarrow AC=8\sqrt{17}\)cm

\(\Rightarrow AB=\frac{8\sqrt{17}}{4}=2\sqrt{17}\)cm 

Theo định lí Pytago tam giác ABC vuông tại A

\(BC=\sqrt{AB^2+AC^2}=34\)cm 

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

-> HC = BC - HB = 32 cm 

\(0,1\sqrt{200}+2\sqrt{0,08}+0,4\sqrt{5}\)

\(=\frac{2\sqrt{5}+7\sqrt{2}}{5}\)