Câu 1:
$B=\sqrt{{{x}^{2}}-2x+5}$
$B=\sqrt{{{x}^{2}}-2x+1+4}$
$B=\sqrt{{{\left( x-1 \right)}^{2}}+4}\ge \sqrt{4}=2$
Dấu “=” xảy ra khi $x=1$
Vậy ${{\min }_{B}}=2$ khi $x=1$
Câu 2:
a)
$AD$ phân giác $\Rightarrow \dfrac{AB}{AC}=\dfrac{BD}{CD}\Rightarrow \dfrac{A{{B}^{2}}}{A{{C}^{2}}}=\dfrac{B{{D}^{2}}}{C{{D}^{2}}}=\dfrac{{{36}^{2}}}{{{60}^{2}}}=\dfrac{9}{25}$
$A{{B}^{2}}=BH.BC$
$A{{C}^{2}}=CH.BC$
$\Rightarrow \dfrac{BH}{CH}=\dfrac{A{{B}^{2}}}{A{{C}^{2}}}=\dfrac{9}{25}$
b)
$BH+CH=BD+CD$
$\Leftrightarrow \dfrac{9}{25}CH+CH=36+60$
$\Leftrightarrow CH=\dfrac{1200}{17}cm$
$\Leftrightarrow BH=\dfrac{432}{17}$
$\Rightarrow AH=\sqrt{CH.BH}=\sqrt{\dfrac{1200}{17}\cdot \dfrac{432}{17}}=\dfrac{720}{17}cm$
