\(đkxđ:x>0.\\ A=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+2}.\)

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

\(đkxđ:x>0;x\ne1.\\ B=\left(5-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right)\left(5+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right).\\ B=\left(5-\sqrt{x}\right)\left(5+\sqrt{x}\right).\\ B=25-x.\)

Xét \(\Delta ABH\) vuông tại H:

\(+sinB=\dfrac{AH}{AB}\) (Tỉ số lượng giác).

\(\Rightarrow sin80^o=\dfrac{5}{AB}.\\ \Rightarrow AB=\dfrac{5}{sin80^o}.\\ \Leftrightarrow AB\approx5,08\left(cm\right).\\ \tan B=\dfrac{AH}{BH}\left(TSLG\right).\\ \Rightarrow\tan80^o=\dfrac{5}{BH}.\\ \Rightarrow BH\approx0,89\left(cm\right).\)

Xét \(\Delta AHC\) vuông tại H:

\(+sinC=\dfrac{AH}{AC}\left(TSLG\right).\\ \Rightarrow sin30^o=\dfrac{5}{AC}.\\ \Rightarrow AC=\dfrac{5}{sin30^o}=10\left(cm\right).\\ +\tan C=\dfrac{AH}{HC}\left(TSLG\right).\\ \Rightarrow\tan30^o=\dfrac{5}{HC}.\\ \Rightarrow HC=5\sqrt{3}\left(cm\right).\)

\(BC=BH+HC\approx0,89+5\sqrt{3}\approx9,55\left(cm\right).\)