3, ta có:

\(B=\dfrac{\sqrt{x}-3+2\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x-3}\right)}\cdot\dfrac{2\left(\sqrt{x}+3\right)}{\sqrt{x}-1}\\ =\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{2\left(\sqrt{x}+3\right)}{\sqrt{x}-1}\\ =\dfrac{6}{\sqrt{x}-3}\)

để B=3 thì ta có:

\(\dfrac{6}{\sqrt{x}-3}=3\\ \Leftrightarrow\dfrac{6}{\sqrt{x}-3}=\dfrac{3\sqrt{x}-9}{\sqrt{x}-3}\\ \Leftrightarrow6=3\sqrt{x}-9\\ \Leftrightarrow3\sqrt{x}=15\\ \Leftrightarrow\sqrt{x}=5\\ \Leftrightarrow x=25\)

vậy để B=3 thì x=25

1:

a: \(A=\dfrac{\sqrt{x}+1-2}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\)

căn x+1>=1

=>2/căn x+1<=2

=>-2/căn x+1>=-2

=>A>=-2+1=-1

Dấu = xảy ra khi x=0

b: 

a: Xét ΔABC vuông tại A có sin C=AB/BC

=>6/BC=sin30=1/2

=>BC=12cm

=>AC=6*căn 3(cm)

HB=AB^2/BC=3cm

HC=12-3=9cm

b: Xét ΔABH vuông tại H có sin B=AH/AB=1/2

=>góc B=30 độ

=>góc C=60 độ

BH=căn 12^2-6^2=6*căn 3(cm)

CH=AH^2/HB=2*căn 3(cm)

BC=căn 3^2+4^2=5cm

AH=3*4/5=2,4cm

\(AD=\dfrac{2\cdot3\cdot4}{3+4}\cdot cos45=\dfrac{12}{7}\sqrt{2}\left(cm\right)\)

cos HAD=AH/AD=7/10*căn 2

=>góc HAD=8 độ

\(\dfrac{x}{\sqrt{x}-1}\ge0\) (ĐK: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\))

\(\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x< 0\\\sqrt{x}-1< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-1\ge0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x< 0\\x< 1\end{matrix}\right.\\\left\{{}\begin{matrix}x\ge0\\x\ge1\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x< 0\left(ktm\right)\\x\ge1\end{matrix}\right.\) (mà \(x\ne1\))

\(\Leftrightarrow x>1\)

\(\sqrt{7-\sqrt{24}}-\dfrac{\sqrt{50}-5}{\sqrt{10}-\sqrt{5}}+\sqrt{\left(11+\sqrt{120}\right)\left(11+2\sqrt{30}\right)^2}\)

\(=\sqrt{7-2\sqrt{6}}-\dfrac{5\left(\sqrt{2}-1\right)}{\sqrt{5}\left(\sqrt{2}-1\right)}+\left|11+2\sqrt{30}\right|\sqrt{11-2\sqrt{30}}\)

\(=\sqrt{1^2-2\sqrt{6}\cdot1+\left(\sqrt{6}\right)^2}-\dfrac{\sqrt{5}\cdot\sqrt{5}}{\sqrt{5}}+\left(11+2\sqrt{30}\right)\sqrt{\left(\sqrt{6}\right)^2-2\sqrt{5}\cdot\sqrt{6}+\left(\sqrt{5}\right)^2}\)

\(=\sqrt{\left(1-\sqrt{6}\right)^2}-\sqrt{5}+\left(11+2\sqrt{30}\right)\sqrt{\left(\sqrt{6}-\sqrt{5}\right)^2}\)

\(=\left|1-\sqrt{6}\right|-\sqrt{5}+\left(11+2\sqrt{30}\right)\left|\sqrt{6}-\sqrt{5}\right|\)

\(=-1+6-\sqrt{5}+\left(\sqrt{6}+\sqrt{5}\right)^2\left(\sqrt{6}-\sqrt{5}\right)\)

\(=\sqrt{6}-1-\sqrt{5}+\left[\left(\sqrt{6}\right)^2-\left(\sqrt{5}\right)^2\right]\left(\sqrt{6}+\sqrt{5}\right)\)

\(=\sqrt{6}-1-\sqrt{5}+\left(6-5\right)\left(\sqrt{6}+\sqrt{5}\right)\)

\(=\sqrt{6}-1-\sqrt{5}+\sqrt{6}+\sqrt{5}\)

\(=2\sqrt{6}-1\)

\(=\sqrt{6+1-2\sqrt{6}}-\dfrac{\sqrt{5}\left(\sqrt{10}-\sqrt{5}\right)}{\sqrt{10}-\sqrt{5}}+\sqrt{\left(11-\sqrt{120}\right)\left(11+\sqrt{120}\right)^2}\\ =\sqrt{\left(\sqrt{6}-\sqrt{1}\right)^2}-\sqrt{5}+\sqrt{\left(11^2-120\right)\left(11+2\sqrt{30}\right)}\\ =\sqrt{6}-\sqrt{1}-\sqrt{5}+\sqrt{1\left(6+5+2\sqrt{6\cdot5}\right)}\\ =\sqrt{6}-\sqrt{1}-\sqrt{5}+\sqrt{\left(\sqrt{6}+\sqrt{5}\right)^2}\\ =\sqrt{6}-\sqrt{1}-\sqrt{5}+\sqrt{6}+\sqrt{5}=2\sqrt{6}-\sqrt{1}\)

1: ΔOMN cân tại O 

mà OA vuông góc MN

nên OA là trung trực của MN

=>AM=AN

góc AMB=góc ANB=1/2*sđ cung AB=90 độ

Xét ΔAMB vuông tại M và ΔANB vuông tại N có

AB chung

AM=AN

=>ΔAMB=ΔANB

=>BM=BN

=>AM,AN là tiếp tuyến của (B;BM)

2: MH^2=AH*HB

=>4*MH^2=4*AH*HB

=>MN^2=4*AH*HB

3: góc MBA=90-60=30 độ

=>góc MBN=60 độ

=>ΔMBN đều

\(B=\left(\dfrac{1}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{x-9}\right)\cdot\dfrac{2\sqrt{x}+6}{\sqrt{x}-1}\)

\(=\dfrac{\sqrt{x}-3+2\sqrt{x}}{x-9}\cdot\dfrac{2\left(\sqrt{x}+3\right)}{\sqrt{x}-1}\)

\(=\dfrac{3\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\cdot\dfrac{2}{\sqrt{x}-3}=\dfrac{6}{\sqrt{x}-3}\)