a) \(\sqrt{\left(x+1\right)^2}=-4\)

Vì \(VT\ge0\Rightarrow VP\ge0\Rightarrow-4\ge0\) (vô lý) nên pt vô nghiệm

b) \(\sqrt{4x-20}-3\sqrt{\dfrac{x-5}{9}}=2\left(x\ge5\right)\)

\(\Rightarrow\sqrt{4\left(x-5\right)}-\sqrt{9.\dfrac{x-5}{9}}=2\Rightarrow2\sqrt{x-5}-\sqrt{x-5}=2\)

\(\Rightarrow\sqrt{x-5}=2\Rightarrow x-5=4\Rightarrow x=9\)

a) \(\dfrac{6}{5}\sqrt{50}-3\sqrt{45}-2\sqrt{18}+5\sqrt{20}=\dfrac{6}{5}\sqrt{25.2}-3\sqrt{9.5}-2\sqrt{9.2}+5\sqrt{4.5}\)

\(=6\sqrt{2}-9\sqrt{5}-6\sqrt{2}+10\sqrt{5}=\sqrt{5}\)

b) \(\sqrt{\dfrac{\left(\sqrt{2}-\sqrt{3}\right)^2}{2}}-\sqrt{\dfrac{7}{2}+2\sqrt{3}}=\dfrac{\left|\sqrt{2}-\sqrt{3}\right|}{\sqrt{2}}-\sqrt{\dfrac{7+4\sqrt{3}}{2}}\)

\(=\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{2}}-\sqrt{\dfrac{2^2+2.2.\sqrt{3}+\left(\sqrt{3}\right)^2}{2}}=\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{2}}-\sqrt{\dfrac{\left(2+\sqrt{3}\right)^2}{2}}\)

\(=\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{2}}-\dfrac{\left|2+\sqrt{3}\right|}{\sqrt{2}}=\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{2}}-\dfrac{2+\sqrt{3}}{\sqrt{2}}=\dfrac{-2-\sqrt{2}}{\sqrt{2}}=-\sqrt{2}-1\)

a) \(M=\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}-\dfrac{4\sqrt{ab}}{a-b}+\dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)

\(=\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}-\dfrac{4\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)

\(=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2-4\sqrt{ab}+\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

\(=\dfrac{a-\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}=\dfrac{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}=\dfrac{\sqrt{a}}{\sqrt{a}+\sqrt{b}}\)

b) Ta có: \(\sqrt{b}=\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.1+1^2}=\sqrt{\left(\sqrt{3}-1\right)^2}\)

\(=\left|\sqrt{3}-1\right|=\sqrt{3}-1\)

\(\Rightarrow M=\dfrac{\sqrt{3}}{\sqrt{3}+\sqrt{3}-1}=\dfrac{\sqrt{3}}{2\sqrt{3}-1}=\dfrac{\sqrt{3}\left(2\sqrt{3}+1\right)}{\left(2\sqrt{3}-1\right)\left(2\sqrt{3}+1\right)}=\dfrac{6+\sqrt{3}}{11}\)

a) \(M=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\left(a>0\right)\)

\(=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\dfrac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)

\(=\sqrt{a}\left(\sqrt{a}+1\right)-2\sqrt{a}-1+1=a-\sqrt{a}\)

b) \(a-\sqrt{a}=a-2.\sqrt{a}.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\dfrac{1}{4}=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)

\(\Rightarrow A_{min}=-\dfrac{1}{4}\) khi \(a=\dfrac{1}{4}\)

Vì tam giác ABC cân tại A có đường cao AH nên D là trung điểm BC

Từ C kẻ đường thẳng vuông góc với BC cắt AB tại G

\(\Rightarrow CG\parallel AD\) mà D là trung điểm BC \(\Rightarrow A\) là trung điểm BG

nên AD là đường trung bình tam giác BCG \(\Rightarrow AD=\dfrac{CG}{2}\)

\(\Rightarrow2AD=CG\Rightarrow4AD^2=CG^2\)

tam giác BCG vuông tại C có đường cao CF nên áp dụng hệ thức lượng

\(\Rightarrow\dfrac{1}{BC^2}+\dfrac{1}{CG^2}=\dfrac{1}{CF^2}\Rightarrow\dfrac{1}{BC^2}+\dfrac{1}{4AD^2}=\dfrac{1}{CF^2}\)

\(A=\dfrac{\sqrt{x}}{2-\sqrt{x}}+\dfrac{\sqrt{x}}{2+\sqrt{x}}-\dfrac{6+\sqrt{x}}{4-x}\)

\(=\dfrac{\sqrt{x}}{2-\sqrt{x}}+\dfrac{\sqrt{x}}{2+\sqrt{x}}-\dfrac{6+\sqrt{x}}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\)

\(=\dfrac{\sqrt{x}\left(2+\sqrt{x}\right)+\sqrt{x}\left(2-\sqrt{x}\right)-6-\sqrt{x}}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\)

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

b) Ta có: \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+2\ge2\Rightarrow\dfrac{3}{\sqrt{x}+2}\le\dfrac{3}{2}\Rightarrow-\dfrac{3}{\sqrt{x}+2}\ge-\dfrac{3}{2}\)

\(\Rightarrow A_{min}=-\dfrac{3}{2}\) khi \(x=0\)

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

\(=\dfrac{\sqrt{3}-\sqrt{2}-1}{\left(\sqrt{3}\right)^2-\left(\sqrt{2}+1\right)^2}=\dfrac{\sqrt{3}-\sqrt{2}-1}{-2\sqrt{2}}=\dfrac{\sqrt{2}\left(\sqrt{3}-\sqrt{2}-1\right)}{-2\sqrt{2}.\sqrt{2}}\)

\(=\dfrac{\sqrt{6}-2-\sqrt{2}}{-4}=\dfrac{2+\sqrt{2}-\sqrt{6}}{4}\)

a) \(\sqrt{4x}+\sqrt{\dfrac{x}{4}}+\dfrac{1}{2}\sqrt{49x}=6\left(x\ge0\right)\)

\(\Rightarrow2\sqrt{x}+\dfrac{1}{2}\sqrt{x}+\dfrac{7}{2}\sqrt{x}=6\Rightarrow6\sqrt{x}=6\Rightarrow\sqrt{x}=1\Rightarrow x=1\)

b) ĐKXĐ: \(x\ge\dfrac{1}{2}\)

 \(\sqrt{18x-9}-0,5\sqrt{2x-1}+\dfrac{1}{2}\sqrt{25\left(2x-1\right)}+\sqrt{49\left(2x-1\right)}=24\)

\(\Rightarrow\sqrt{9\left(2x-1\right)}-0,5\sqrt{2x-1}+\dfrac{5}{2}\sqrt{2x-1}+7\sqrt{2x-1}=24\)

\(\Rightarrow3\sqrt{2x-1}-0,5\sqrt{2x-1}+\dfrac{5}{2}\sqrt{2x-1}+7\sqrt{2x-1}=24\)

\(\Rightarrow12\sqrt{2x-1}=24\Rightarrow\sqrt{2x-1}=2\Rightarrow2x-1=4\Rightarrow x=\dfrac{5}{2}\)

c) \(\sqrt{x^2-2x+1}-7=0\Rightarrow\sqrt{\left(x-1\right)^2}=7\Rightarrow\left|x-1\right|=7\)

\(\Rightarrow\left[{}\begin{matrix}x-1=7\\x-1=-7\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=8\\x=-6\end{matrix}\right.\)

d) \(\dfrac{1}{2}\sqrt{\dfrac{49x}{x+2}}-3\sqrt{\dfrac{x}{4x+8}}-2\sqrt{\dfrac{x}{x+2}}-\sqrt{5}=0\left(\dfrac{x}{x+2}\ge0,x\ne-2\right)\)

\(\Rightarrow\dfrac{7}{2}\sqrt{\dfrac{x}{x+2}}-3\sqrt{\dfrac{x}{4\left(x+2\right)}}-2\sqrt{\dfrac{x}{x+2}}=\sqrt{5}\)

\(\Rightarrow\dfrac{7}{2}\sqrt{\dfrac{x}{x+2}}-\dfrac{3}{2}\sqrt{\dfrac{x}{x+2}}-2\sqrt{\dfrac{x}{x+2}}=\sqrt{5}\)

\(\Rightarrow0=\sqrt{5}\) (vô lý) \(\Rightarrow\) pt vô nghiệm

\(\dfrac{\sqrt{a}+2}{\sqrt{a}+3}-\dfrac{5}{a+\sqrt{a}-6}+\dfrac{1}{2-\sqrt{a}}\left(a\ge0,a\ne4\right)\)

\(=\dfrac{\sqrt{a}+2}{\sqrt{a}+3}-\dfrac{5}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}-\dfrac{1}{\sqrt{a}-2}\)

\(=\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)-5-\sqrt{a}-3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}=\dfrac{a-\sqrt{a}-12}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}\)

\(=\dfrac{\left(\sqrt{a}+3\right)\left(\sqrt{a}-4\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}=\dfrac{\sqrt{a}-4}{\sqrt{a}-2}\)

Vì CM,CA là tiếp tuyến nên OC là phân giác góc MOA

\(\Rightarrow\angle MOC=\dfrac{1}{2}\angle MOA\)

Vì DM,DB là tiếp tuyến nên OD là phân giác góc MOB

\(\Rightarrow\angle MOD=\dfrac{1}{2}\angle MOB\)

\(\Rightarrow\angle MOC+\angle MOD=\dfrac{1}{2}\left(\angle MOA+\angle MOB\right)\Rightarrow\angle COD=\dfrac{1}{2}\angle AOB=90\)

b) Vì CM,CA là tiếp tuyến \(\Rightarrow CM=CA\)

Vì DM,DB là tiếp tuyến \(\Rightarrow DM=DB\)

tam giác COD vuông tại O có đường cao OM nên áp dụng hệ thức lượng

\(\Rightarrow OM^2=CM.MD=AC.BD\Rightarrow AC.BD=R^2\)