\(DK:\left\{{}\begin{matrix}x\ge0\\\sqrt{x}+1\ne0\end{matrix}\right.\)

\(< =>\left\{{}\begin{matrix}x\ge0\\\sqrt{x}\ne-1\end{matrix}\right.\) 

\(< =>x\ge0\) ( Vì : \(\forall x\ge0=>\sqrt{x}\ge0\) )

\(B=\dfrac{1}{\sqrt{x}+1}\)

Biểu thức B xác định khi:

\(\left\{{}\begin{matrix}\sqrt{x}+1\ne0\\x\ge0\end{matrix}\right.\)

Mà: \(\sqrt{x}+1\ne0\) (luôn đúng) nên:

\(\Leftrightarrow x\ge0\)

Vậy: ... 

\(DK:\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-2\ne0\\3-\sqrt{x}\ne0\\x-5\sqrt{x}+6=\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)\ne0\end{matrix}\right.\)

\(< =>\left\{{}\begin{matrix}x\ge0\\x\ne4\\x\ne9\end{matrix}\right.\\ < =>x\ge0,x\ne\left\{4;9\right\}\)

Lời giải:
$BC=BH+HC=61+84=145$ (cm) 

Áp dụng hệ thức lượng trong tam giác vuông: 

$AH^2=BH.CH=61.84=5124$ 

Áp dụng định lý Pitago cho tam giác vuông $ABH, ACH$:

$AB=\sqrt{AH^2+BH^2}=\sqrt{5124+61^2}\approx 94$ (cm) 

$AC=\sqrt{AH^2+CH^2}=\sqrt{5124+84^2}\approx 110,4$ (cm)

$\cos B =\frac{AB}{BC}=\frac{94}{145}\Rightarrow \widehat{B}\approx 50^0$

$\widehat{C}=90^0-\widehat{B}\approx 90^0-50^0=40^0$

Hình vẽ:

Lời giải:
ĐKXĐ: $x\geq 0 ; x\neq 9$

\(Q=\frac{(\sqrt{x}+1)(\sqrt{x}+3)+2\sqrt{x}(\sqrt{x}-3)}{(\sqrt{x}-3)(\sqrt{x}+3)}-\frac{7\sqrt{x}+3}{(\sqrt{x}-3)(\sqrt{x}+3)}\)

\(=\frac{x+4\sqrt{x}+3+2x-6\sqrt{x}}{(\sqrt{x}-3)(\sqrt{x}+3)}-\frac{7\sqrt{x}+3}{(\sqrt{x}-3)(\sqrt{x}+3)}\)

$=\frac{3x-2\sqrt{x}+3-7\sqrt{x}-3}{(\sqrt{x}-3)(\sqrt{x}+3)}$

$=\frac{3x-9\sqrt{x}}{(\sqrt{x}-3)(\sqrt{x}+3)}=\frac{3\sqrt{x}(\sqrt{x}-3)}{(\sqrt{x}-3)(\sqrt{x}+3)}=\frac{3\sqrt{x}}{\sqrt{x}+3}$

Ta có đpcm.

α=-9,889492973

\(B=\dfrac{2010}{4x+20\sqrt{x}+30}\)

\(B=\dfrac{2010}{\left(2\sqrt{x}\right)^2+2\cdot2\sqrt{x}\cdot5+25+5}\)

\(B=\dfrac{2010}{\left(2\sqrt{x}+5\right)^2+5}\)

Ta có: \(\left(2\sqrt{x}+5\right)^2+5\ge5\)

\(\Rightarrow B=\dfrac{2010}{\left(2\sqrt{x}+5\right)^2+5}\le\dfrac{2010}{5}=402\)

Vậy: \(B_{min}=402\)