\(\sqrt{96}-6\sqrt{\frac{2}{3}}+\frac{3}{3+\sqrt{6}}-\sqrt{10-4\sqrt{6}}\)

\(=4\sqrt{6}-\frac{6\sqrt{2}}{\sqrt{3}}+\frac{3\left(3-\sqrt{6}\right)}{3}-\sqrt{6-2.2\sqrt{6}+4}\)

\(=4\sqrt{6}-2\sqrt{6}+3-\sqrt{6}-\sqrt{\left(\sqrt{6}-2\right)^2}\)

\(=\sqrt{6}+3-\sqrt{6}+2=5\)

a, Với \(x\ge0;x\ne\frac{16}{9};4\)

\(P=\frac{2\sqrt{x}-4}{3\sqrt{x}-4}-\frac{4+2\sqrt{x}}{\sqrt{x}-2}+\frac{x+13\sqrt{x}-20}{3x-10\sqrt{x}+8}\)

\(=\frac{2x-8\sqrt{x}+8-4\sqrt{x}-6x+16+x+13\sqrt{x}-20}{\left(3\sqrt{x}-4\right)\left(\sqrt{x}-2\right)}\)

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

b, \(P\ge-\frac{3}{4}\Rightarrow\frac{\sqrt{x}+1}{2-\sqrt{x}}+\frac{3}{4}\ge0\Leftrightarrow\frac{4\sqrt{x}+4+6-3\sqrt{x}}{8-4\sqrt{x}}\ge0\Leftrightarrow\frac{\sqrt{x}+10}{8-4\sqrt{x}}\ge0\)

\(\Rightarrow2-\sqrt{x}\ge0\Leftrightarrow x\le4\)Kết hợp với đk vậy \(0\le x< 4\)

\(\left(3+\frac{3-\sqrt{3}}{1-\sqrt{3}}\right)\left(\frac{\sqrt{21}+\sqrt{7}}{\sqrt{7}}+2\right)\)

\(=\left(3-\frac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\right)\left(\frac{\sqrt{7}\left(\sqrt{3}+1\right)}{\sqrt{7}}+2\right)\)

\(=\left(3-\sqrt{3}\right)\left(\sqrt{3}+3\right)=9-3=6\)

\(3+\frac{3-\sqrt{3}}{1-\sqrt{3}}=3+\frac{1-\sqrt{3}+2}{1-\sqrt{3}}=3+1+\frac{2}{1-\sqrt{3}}=4+\frac{2}{1-\sqrt{3}}\)

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

\(\frac{\sqrt{21}+\sqrt{7}}{\sqrt{7}}+2=\frac{\sqrt{7}\left(\sqrt{3}+1\right)}{\sqrt{7}}+2=\left(\sqrt{3}+3\right)\)

Khi đó \(\left(3+\frac{3-\sqrt{3}}{1-\sqrt{3}}\right)\left(\frac{\sqrt{21}+\sqrt{7}}{\sqrt{7}}+2\right)=-\left(\sqrt{3}+3\right)^2=-12-6\sqrt{3}\)

sửa đề: \(B=2x+4\sqrt{x}+9\)ĐK : x >= 0 

\(=2\left(x+2\sqrt{x}+1-1\right)+9=2\left(\sqrt{x}+1\right)^2+7\ge9\)

Dấu ''='' xảy ra khi x = 0 

Vậy GTNN của B bằng 9 tại x = 0 

\(5,4x^2-12x+3\sqrt{x^2-3x+4}=6\)

\(\Leftrightarrow4x^2-12x+16+3\sqrt{x^2-3x+4}=22\)

đặt \(\sqrt{x^2-3x+4}=a\left(a\ge0\right)\)

ta có : \(4x^2+3a=22\)

\(\Leftrightarrow4a^2+3a-22=0\)

\(\Leftrightarrow4a^2-8a+11a-22=0\)

\(\Leftrightarrow4a\left(a-2\right)+11\left(a-2\right)=0\)

\(\Leftrightarrow\left(4a+11\right)\left(a-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=-\frac{11}{4}\left(loai\right)\\a=2\left(tm\right)\end{cases}}\)

\(a=2\Rightarrow\sqrt{x^2-3x+4}=2\) \(\Leftrightarrow a^2-3a=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)

9, \(ĐK:x\ge\frac{2}{3}\)

\(\sqrt{x+2}+\sqrt{3x-2}+5x=14\)

\(\Leftrightarrow\sqrt{x+2}+\sqrt{3x-2}+5x-14=0\)

\(\Leftrightarrow\frac{\left(\sqrt{x+2}-2\right)\left(\sqrt{x+2}+2\right)}{\sqrt{x+2}+2}+\frac{\left(\sqrt{3x-2}-2\right)\left(\sqrt{3x-2}+2\right)}{\sqrt{3x-2}+2}+5x-10=0\)

\(\Leftrightarrow\frac{x+2-4}{\sqrt{x+2}+2}+\frac{3x-2-4}{\sqrt{3x-2}+2}+5\left(x-2\right)=0\)

\(\Leftrightarrow\frac{x-2}{\sqrt{x+2}+2}+\frac{3x-6}{\sqrt{3x-2}+2}+5\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(\frac{1}{\sqrt{x+2}+2}+\frac{3}{\sqrt{3x-2}+2}+5\right)=0\)

với \(x\ge\frac{2}{3}\Rightarrow\frac{1}{\sqrt{x+2}+2}+\frac{3}{\sqrt{3x-2}+2}+5>0\)

\(\Leftrightarrow x=2\left(tm\right)\)

giới thiệu cho cách 2 câu 3 này :

ĐK : \(\orbr{\begin{cases}x\ge6\\x\le1\end{cases}}\)

\(\sqrt{x^2-7x+6}+1=\sqrt{x-1}+\sqrt{x-6}\)

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

đặt \(\hept{\begin{cases}\sqrt{x-1}=a\\\sqrt{x-6}=b\end{cases}\left(a;b\ge0\right)}\)

pt trở thành : \(ab+1=a+b\)

\(\Leftrightarrow ab-a+1-b=0\)

\(\Leftrightarrow a\left(b-1\right)-\left(b-1\right)=0\)

\(\Leftrightarrow\left(a-1\right)\left(b-1\right)=0\Leftrightarrow\orbr{\begin{cases}a=1\\b=1\end{cases}}\)

thay vào tính x

\(a,ĐK:9x^2-1\ne0\Leftrightarrow x^2\ne\frac{1}{9}\Leftrightarrow x\ne\pm\frac{1}{3}\)

\(b,M=\frac{\sqrt{9x^2-6x+1}}{9x^2-1}=\frac{\sqrt{\left(3x-1\right)^2}}{\left(3x-1\right)\left(3x+1\right)}=\frac{\left|3x-1\right|}{\left(3x-1\right)\left(3x+1\right)}\)

với \(3x-1>0\) ta có \(M=\frac{3x-1}{\left(3x-1\right)\left(3x+1\right)}=\frac{1}{3x+1}\)

với \(3x-1< 0\) ta có \(M=\frac{-\left(3x-1\right)}{\left(3x-1\right)\left(3x+1\right)}=-\frac{1}{3x+1}\)

\(c,\) th1 : \(M=\frac{1}{3x+1}\)  khi \(x>\frac{1}{3}\) mà \(M=\frac{1}{4}\)

\(\Leftrightarrow\frac{1}{3x+1}=\frac{1}{4}\Leftrightarrow x=1\left(thoaman\right)\) 

th2 : \(M=-\frac{1}{3x+1}\) khi \(x< \frac{1}{3}\) mà \(M=\frac{1}{4}\)

\(\Leftrightarrow\frac{-1}{3x+1}=\frac{1}{4}\Leftrightarrow3x+1=-4\Leftrightarrow x=-\frac{5}{3}\left(thoaman\right)\)

\(d,M=\frac{\left|3x-1\right|}{\left(3x-1\right)\left(3x+1\right)}< 0\) có \(\left|3x-1\right|>0\)

\(\Rightarrow\left(3x-1\right)\left(3x+1\right)< 0\)

th1 : \(\hept{\begin{cases}3x-1>0\\3x+1< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x>\frac{1}{3}\\x< -\frac{1}{3}\end{cases}\left(voli\right)}}\)

th2 : \(\hept{\begin{cases}3x-1< 0\\3x+1>0\end{cases}}\Leftrightarrow\hept{\begin{cases}x< \frac{1}{3}\\x>-\frac{1}{3}\end{cases}\Leftrightarrow-\frac{1}{3}< x< \frac{1}{3}}\)