\(\left(3\sqrt{2}+\sqrt{6}\right)\sqrt{6-3\sqrt{3}}=\left(3+\sqrt{3}\right)\sqrt{12-6\sqrt{3}}=\left(3+\sqrt{3}\right)\sqrt{9-2.3.\sqrt{3}+3}\)

\(=\left(3+\sqrt{3}\right)\sqrt{3^2-2.3.\sqrt{3}+\left(\sqrt{3}\right)^2}=\left(3+\sqrt{3}\right)\sqrt{\left(3-\sqrt{3}\right)^2}\)

\(=\left(3+\sqrt{3}\right)\left|3-\sqrt{3}\right|=\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)=6\)

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

\(=\frac{a+2\sqrt{ab}+b-a+2\sqrt{ab}-b}{a-b}=\frac{4\sqrt{ab}}{a-b}\)

2) \(x-4-\sqrt{16-8x^2+x^4}=x-4-\sqrt{\left(x^2-4\right)^2}=x-4-\left|x^2-4\right|\)

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

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

4) \(\frac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}:\frac{1}{\sqrt{a}+\sqrt{b}}=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}\cdot\left(\sqrt{a}+\sqrt{b}\right)\)

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

5) \(\frac{a-3\sqrt{a}}{\sqrt{a}-3}-\frac{a+4\sqrt{a}+3}{\sqrt{a}+3}=\frac{\sqrt{a}\left(\sqrt{a}-3\right)}{\sqrt{a}-3}-\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+3\right)}{\sqrt{a}+3}\)

\(=\sqrt{a}-\sqrt{a}-1=-1\)

6) \(\frac{9-x}{\sqrt{x}+3}-\frac{9-6\sqrt{x}+x}{\sqrt{x}-3}-6=\frac{\left(3-\sqrt{x}\right)\left(\sqrt{x}+3\right)}{\sqrt{x}+3}-\frac{\left(\sqrt{x}-3\right)^2}{\sqrt{x}-3}-6\)

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

7) \(\frac{x\sqrt{y}+y\sqrt{x}}{\sqrt{xy}}:\frac{\sqrt{x}-\sqrt{y}}{x-y}=\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}:\frac{\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)

\(=\left(\sqrt{x}+\sqrt{y}\right)^2\)

8) \(\frac{\sqrt{a}+3}{\sqrt{a}-2}-\frac{\sqrt{a}-1}{\sqrt{a}+2}+\frac{4\sqrt{a}-4}{4-a}=\frac{\left(\sqrt{a}+3\right)\left(\sqrt{a}+2\right)-\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)-4\sqrt{a}+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)

=\(\frac{a+5\sqrt{a}+6-a+3\sqrt{a}-2-4\sqrt{a}+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}=\frac{4\sqrt{a}+8}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}=\frac{4}{\sqrt{a}-2}\)

9) \(\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}=\frac{x+2+\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

\(=\frac{x-1-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)

10) \(\left(\frac{2+\sqrt{x}}{2-\sqrt{x}}+\frac{2-\sqrt{x}}{2+\sqrt{x}}-\frac{4x}{x-4}\right):\frac{x-6\sqrt{x}+9}{\left(2-\sqrt{x}\right)\left(\sqrt{x}-3\right)}\)

\(=\left(\frac{\left(2+\sqrt{x}\right)^2-\left(2-\sqrt{x}\right)^2+4x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right):\frac{\left(\sqrt{x}-3\right)^2}{\left(2-\sqrt{x}\right)\left(\sqrt{x}-3\right)}\)

\(=\frac{x+4\sqrt{x}+4-x+4\sqrt{x}-4+4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\cdot\frac{2-\sqrt{x}}{\sqrt{x}-3}\)

\(=\frac{4x+8\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}=\frac{4\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}=\frac{4\sqrt{x}}{\sqrt{x}-3}\)

11) \(\left(\frac{2\sqrt{x}+x}{x\sqrt{x}-1}-\frac{1}{\sqrt{x}-1}\right):\left(\frac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)=\frac{2\sqrt{x}+x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\frac{x+\sqrt{x}+1}{\sqrt{x}+2}\)

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

12) \(\left(\frac{x+\sqrt{x}}{\sqrt{x}+1}+1\right)\left(\frac{\sqrt{x}-x}{\sqrt{x}-1}+1\right)=\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}+1\right)\left(\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}+1\right)\)

\(=\left(\sqrt{x}+1\right)^2\)

a) Ta có:

\(\sqrt{\frac{289}{225}}=\sqrt{\frac{\sqrt{289}}{\sqrt{225}}}=\sqrt{\frac{17^2}{15^2}}=\frac{17}{15}\)

b) Ta có:

\(\sqrt{2\frac{14}{25}}=\sqrt{\frac{64}{25}}=\sqrt{\frac{\sqrt{64}}{\sqrt{25}}}=\sqrt{\frac{8^2}{5^2}}=\frac{8}{5}\)

c) Ta có:

\(\sqrt{\frac{0,25}{9}}=\sqrt{\frac{\sqrt{0,25}}{\sqrt{9}}}=\sqrt{\frac{0,5^2}{3^2}}=\frac{0,5}{3}=\frac{1}{6}\)

d) Ta có:

\(\sqrt{\frac{8,1}{1,6}}=\sqrt{\frac{81.0,1}{16.0,1}}=\sqrt{\frac{81}{16}}=\sqrt{\frac{\sqrt{81}}{\sqrt{16}}}=\sqrt{\frac{9^2}{4^2}}=\frac{9}{4}\)

a)Ta có: \(\sqrt{\frac{289}{225}}=\frac{\sqrt{289}}{\sqrt{225}}=\frac{17}{15}\)

b) Ta có: \(\sqrt{2\frac{14}{25}}=\sqrt{\frac{64}{25}}=\frac{\sqrt{64}}{\sqrt{25}}=\frac{8}{5}\)

c) Ta có: \(\sqrt{\frac{0,25}{9}}=\frac{\sqrt{0,25}}{\sqrt{9}}=\frac{0,5}{3}=\frac{1}{6}\)

d)Ta có : \(\sqrt{\frac{8,1}{1,6}}=\frac{\sqrt{8,1}}{\sqrt{1,6}}=\frac{\sqrt{8,1}.100}{\sqrt{1,6}.100}=\frac{\sqrt{81}}{\sqrt{16}}=\frac{9}{4}\)

ĐK: a \(\ge\)1

\(\sqrt{a+2\sqrt{a-1}}+\sqrt{a-2\sqrt{a-1}}\)

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

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

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

Xét a \(\ge\)2 => \(\sqrt{a+2\sqrt{a-1}}+\sqrt{a-2\sqrt{a-1}}=\sqrt{a-1}+1+\sqrt{a-1}=2\sqrt{a-1}\)

Xét \(0\le a< 2\) => \(\sqrt{a+2\sqrt{a-1}}+\sqrt{a-2\sqrt{a-1}}=\sqrt{a-1}+1+1-\sqrt{a-1}=2\)

A B O C D E M H K

a)Ta có: EA \(\perp\)AB (t/c tiếp tuyến) => \(\widehat{OAE}=90^0\)

       OD \(\perp\)EC (t/c tiếp tuyến) => \(\widehat{ODE}=90^0\)

Xét t/giác AODE có \(\widehat{OAE}+\widehat{ODE}=90^0+90^0=180^0\)

=> t/giác AODE nt đường tròn (vì tổng 2 góc đối diện  = 180⁰)

b) Xét \(\Delta\)EKD và \(\Delta\)EDB

có: \(\widehat{BED}\):chung

 \(\widehat{EDK}=\widehat{EBK}=\frac{1}{2}sđ\widebat{KD}\)

 => \(\Delta\)EKD ∽ \(\Delta\)EDB (g.g)

=> \(\frac{ED}{EB}=\frac{EK}{ED}\)=> ED² = EK.EB (1)

Ta có: AE = ED (t/c 2 tt cắt nhau) => E thuộc đường trung trực của AD

 OA = OD = R => O thuộc đường trung trực của AD
=> EO là đường trung trực của ED => OE \(\perp\)AD

Xét \(\Delta\)EDO vuông tại D có DH là đường cao => ED² = EK.EB (2)

Từ (1) và (2) => EH.EO = DK.EB => \(\frac{EH}{EB}=\frac{EK}{EO}\)

Xét tam giác EHK và tam giác EBO

có: \(\widehat{OEB}\): chung

 \(\frac{EH}{EB}=\frac{EK}{EO}\)(cmt)

=> tam giác EHK ∽ tam giác EBO (c.g.c)

=> \(\widehat{EHK}=\widehat{KBA}\)

c) Ta có: OM // AE (cùng vuông góc với AB) => \(\frac{OM}{AE}=\frac{MC}{EC}\)(hq định lí ta-lét)

=> OM.EC = AE.MC

Ta lại có: \(\frac{EA}{EM}-\frac{MO}{MC}=\frac{EA.MC-MO.EM}{EM.MC}=\frac{MO.EC-MO.EM}{EM.MC}=\frac{OM.MC}{EM.MC}=\frac{OM}{EM}\)

Mặt khác: OM // AE => \(\widehat{MOE}=\widehat{OEA}\)(slt)

mà \(\widehat{AEO}=\widehat{OEM}\)(t/c 2 tt cắt nhau)

=> \(\widehat{MOE}=\widehat{MEO}\) => tam giác OME cân tại M => OM = ME

=> \(\frac{OM}{EM}=1\)

=> \(\frac{EA}{EM}-\frac{OM}{MC}=1\)

a, \(\sqrt{x^2+12x+40}\)

Biểu thức trên xác định \(\Leftrightarrow x^2+12x+40\ge0\)

                                       \(\Leftrightarrow\left(x+6\right)^2+4\ge0\)

Ta có: \(\left(x+6\right)^2\ge0\forall x\)=>\(\left(x+6\right)^2+4\ge4\forall x\)

Vậy biểu thức trên xác định với mọi x

b,\(\frac{1}{\sqrt{9x^2-6x+1}}\)

Biểu thức trên xác định \(\Leftrightarrow9x^2-6x+1>0\)

                                       \(\Leftrightarrow\left(3x-1\right)^2>0\)

Vậy biểu thức trên xác định với mọi \(x\ne\frac{1}{3}\)

c, \(\sqrt{\left(4x^2+2x+3\right)\left(3-2x\right)}\)

Ta có: \(4x^2+2x+3=\left(2x+\frac{1}{2}\right)^2+\frac{11}{4}\)mà\(\left(2x+\frac{1}{2}\right)^2+\frac{11}{4}\ge\frac{11}{4}\forall x\)nên \(4x^2+2x+3\ge0\forall x\)

Biểu thức trên xác định \(\Leftrightarrow3-2x\ge0\)

                                       \(\Leftrightarrow x\le\frac{3}{2}\)

d,\(\sqrt{\frac{2x^2+3x+16}{5-7x}}\)

Biểu thức trên xác định \(\Leftrightarrow\hept{\begin{cases}2x^2+3x+16\ge0\\5-7x>0\end{cases}}\)

Ta có: \(2x^2+3x+16\)\(=\left(\sqrt{2}x+\frac{3\sqrt{2}}{4}\right)^2+\frac{119}{8}\ge0\)\(\forall x\)

Vậy biểu thức trên xác định \(\Leftrightarrow5-7x>0\)\(\Leftrightarrow x< \frac{5}{7}\)

e,\(\sqrt{16x^2-25}\)

=\(\sqrt{\left(4x-5\right)\left(4x+5\right)}\)

Biểu thức trên xác định \(\Leftrightarrow\hept{\begin{cases}4x-5\ge0\\4x+5\ge0\end{cases}}\)hoặc \(\hept{\begin{cases}4x-5\le0\\4x+5\le0\end{cases}}\)

                                      \(\Leftrightarrow\hept{\begin{cases}x\ge\frac{5}{4}\\x\ge\frac{-5}{4}\end{cases}}\) hoặc \(\hept{\begin{cases}x\le\frac{5}{4}\\x\le\frac{-5}{4}\end{cases}}\)

                                       \(\Leftrightarrow x\ge\frac{5}{4}\)hoặc \(x\le\frac{-5}{4}\)

f, \(\sqrt{16-9x^2}\)

=\(\sqrt{\left(4-3x\right)\left(4+3x\right)}\)

Biểu thức trên xác định \(\Leftrightarrow\hept{\begin{cases}4-3x\ge0\\4+3x\ge0\end{cases}}\)hoặc \(\hept{\begin{cases}4-3x\le0\\4+3x\le0\end{cases}}\)

                                       \(\Leftrightarrow\hept{\begin{cases}x\le\frac{4}{3}\\x\ge\frac{-4}{3}\end{cases}}\)hoặc \(\hept{\begin{cases}x\ge\frac{4}{3}\\x\le\frac{-4}{3}\end{cases}}\)(vô lý)

Vậy biểu thức trên xác định \(\Leftrightarrow\frac{-4}{3}\le x\le\frac{4}{3}\)

g, \(\sqrt{\frac{x-1}{x+2}}\)

Biểu thức trên xác định \(\Leftrightarrow\hept{\begin{cases}x-1\ge0\\x+2>0\end{cases}}\)

                                      \(\Leftrightarrow\hept{\begin{cases}x\ge1\\x>-2\end{cases}}\)

                                       \(\Leftrightarrow x\ge1\)

h, \(\frac{1}{\sqrt{x^2-2x-3}}\)

Biểu thức trên xác định \(\Leftrightarrow x^2-2x-3>0\)

                                       \(\Leftrightarrow x^2-3x+x-3>0\)

                                        \(\Leftrightarrow x\left(x-3\right)+\left(x-3\right)>0\)

                                        \(\Leftrightarrow\left(x-3\right)\left(x+1\right)>0\)

                                         \(\Leftrightarrow\hept{\begin{cases}x+1>0\\x-3>0\end{cases}}\)hoặc \(\hept{\begin{cases}x+1< 0\\x-3< 0\end{cases}}\)

                                          \(\Leftrightarrow\hept{\begin{cases}x>-1\\x>3\end{cases}}\)hoặc \(\hept{\begin{cases}x< -1\\x< 3\end{cases}}\)

                                           \(\Leftrightarrow x>3\)hoặc \(x< -1\)

\(P=\left(1+2a\right)\left(1+2bc\right)\le\left(1+2a\right)\left(1+b^2+c^2\right)=\left(1+2a\right)\left(2-a^2\right)\)

\(=\frac{3}{2}\left(\frac{2}{3}+\frac{4}{3}a\right)\left(2-a^2\right)\le\frac{3}{8}\left(\frac{8}{3}+\frac{4}{3}a-a^2\right)^2=\frac{3}{8}\left[\frac{28}{9}-\left(a-\frac{2}{3}\right)^2\right]^2\)

\(\le\frac{3}{8}.\left(\frac{28}{9}\right)^2=\frac{98}{27}\)

Dấu \(=\)khi \(\hept{\begin{cases}b=c\\\frac{2}{3}+\frac{4}{3}a=2-a^2,a-\frac{2}{3}=0\\a^2+b^2+c^2=1\end{cases}}\Leftrightarrow\hept{\begin{cases}a=\frac{2}{3}\\b=c=\frac{\sqrt{\frac{5}{2}}}{3}\end{cases}}\).

Vậy \(maxP=\frac{98}{27}\).

Ta co : \(P=2a+2bc+2abc+1\)

Ap dung bdt Co-si : \(P\le a^2+b^2+c^2+2abc+2=2abc+3\)

Tiep tuc ap dung Co-si : \(1=a^2+b^2+c^2\ge3\sqrt[3]{a^2b^2c^2}< =>\sqrt[3]{a^2b^2c^2}\le\frac{1}{3}\)

\(< =>a^2b^2c^2\le\frac{1}{27}< =>abc\le\frac{1}{\sqrt{27}}\)

Khi do : \(2abc+3\le2.\frac{1}{\sqrt{27}}+3=\frac{2}{\sqrt{27}}+3\)

Suy ra \(P\le a^2+b^2+c^2+2abc+2\le\frac{2}{\sqrt{27}}+3\)

Dau "=" xay ra khi va chi khi \(a=b=c=\frac{1}{\sqrt{3}}\)

Vay Max P = \(\frac{2}{\sqrt{27}}+3\)khi a = b = c = \(\frac{1}{\sqrt{3}}\) 

p/s : khong biet dau = co dung k nua , minh lam bay do

\(\sqrt{\frac{2a}{9}}.\sqrt{\frac{25a}{32}}=\sqrt{\frac{2a}{9}.\frac{25a}{32}}=\sqrt{\frac{2.\left(5a\right)^2}{2.\left(3.4\right)^2}}=\frac{5a}{12}\)

\(\sqrt{\frac{2a}{9}}\). \(\sqrt{\frac{25a}{32}}\)

\(=\)\(\sqrt{\frac{2a.25a}{9.32}}\)

\(=\)\(\sqrt{\frac{2.\left(5a\right)^2}{2.\left(3.4\right)^2}}\)

\(=\)\(\frac{5a}{12}\)

\(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}+\frac{12}{\sqrt{6}-3}-\sqrt{6}\)

\(\frac{15\left(\sqrt{6}-1\right)}{6-1}+\frac{4\left(\sqrt{6}+2\right)}{6-2^2}+\frac{12\left(\sqrt{6}+3\right)}{6-3^2}-\sqrt{6}\)

\(\frac{15\sqrt{6}-15}{5}+\frac{4\sqrt{6}+8}{2}+\frac{12\sqrt{6}+36}{-3}-\sqrt{6}\)

\(\frac{5\left(3\sqrt{6}-3\right)}{5}+\frac{2\left(2\sqrt{6}+4\right)}{2}-\frac{3\left(4\sqrt{6}+12\right)}{3}-\sqrt{6}\)

\(3\sqrt{6}-3+2\sqrt{6}+4-4\sqrt{6}-12-\sqrt{6}\)

\(=-11\)