Ta có: \(\dfrac{7-4\sqrt{3}}{\sqrt{3}-2}-\dfrac{28-10\sqrt{3}}{5-\sqrt{3}}\)

\(=\dfrac{\left(\sqrt{3}-2\right)^2}{\sqrt{3}-2}-\dfrac{\left(5-\sqrt{3}\right)^2}{5-\sqrt{3}}\)

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

=-7

a)

ĐKXĐ: \(x\notin\left\{3;-3\right\}\)

Ta có: \(\dfrac{2x}{x-3}=\dfrac{x^2+11x-6}{x^2-9}\)

\(\Leftrightarrow\dfrac{2x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{x^2+11x-6}{\left(x-3\right)\left(x+3\right)}\)

Suy ra: \(2x^2+6x=x^2+11x-6\)

\(\Leftrightarrow2x^2+6x-x^2-11x+6=0\)

\(\Leftrightarrow x^2-5x+6=0\)

\(\Leftrightarrow x^2-2x-3x+6=0\)

\(\Leftrightarrow x\left(x-2\right)-3\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\x=3\left(loại\right)\end{matrix}\right.\)

Vậy: S={2}

b) Ta có: \(3x^2+\left(1-\sqrt{3}\right)x+\sqrt{3}-4=0\)

\(\Leftrightarrow3x^2-\left(\sqrt{3}-1\right)x+\sqrt{3}-4=0\)

\(\Leftrightarrow3x^2-\left(\sqrt{3}-1\right)x+\sqrt{3}-1-3=0\)

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

\(\Leftrightarrow3\left(x-1\right)\left(x+1\right)-\left(\sqrt{3}-1\right)\left(x-1\right)=0\)

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

\(\Leftrightarrow\left(x-1\right)\left(3x+4-\sqrt{3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\3x+4-\sqrt{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\3x=\sqrt{3}-4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{\sqrt{3}-4}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{1;\dfrac{\sqrt{3}-4}{3}\right\}\)

a: \(\dfrac{5+2\sqrt{5}}{\sqrt{5}+\sqrt{2}}=\dfrac{\left(5+2\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)}{3}=\dfrac{5\sqrt{5}-5\sqrt{2}+10-2\sqrt{10}}{3}\)

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

1)

\(5x^2-\sqrt{3}x-1=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{23}-\sqrt{3}}{10}\\x=\frac{\sqrt{23}+\sqrt{3}}{10}\end{cases}}\)

a: ĐKXĐ: 2x+5>=0 và 1-x>=0

=>-5/2<=x<=1

PT =>2x+5=1-x

=>3x=-4

=>x=-4/3(nhận)

b: ĐKXĐ: x^2-x>=0 và 3-x>=0

=>x<=3 và (x>=1 hoặc x<=0)

=>x<=0 hoặc (1<=x<=3)

PT =>x^2-x=3-x

=>x^2=3

=>x=căn 3(nhận) hoặc x=-căn 3(nhận)

c: ĐKXĐ: 2x^2-3>=0 và 4x-3>=0

=>x>=3/4 và x^2>=3/2

=>x>=3/4 và \(\left[{}\begin{matrix}x>=\dfrac{\sqrt{6}}{4}\\x< =\dfrac{-\sqrt{6}}{4}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x>=\dfrac{3}{4}\\x< =-\dfrac{\sqrt{6}}{4}\end{matrix}\right.\)

PT =>2x^2-3=4x-3

=>2x^2-4x=0

=>2x(x-2)=0

=>x=0(loại) hoặc x=2(nhận)

\(\sqrt{2x+5}=\sqrt{1-x}\) (ĐK: \(-\dfrac{5}{2}\le x\le1\)) 

\(\Leftrightarrow2x+5=1-x\)

\(\Leftrightarrow2x+x=1-5\)

\(\Leftrightarrow3x=-4\)

\(\Leftrightarrow x=-\dfrac{4}{3}\left(tm\right)\)

b) \(\sqrt{x^2-x}=\sqrt{3-x}\) (ĐK: \(\left[{}\begin{matrix}1\le x\le3\\x\le0\end{matrix}\right.\))

\(\Leftrightarrow x^2-x=3-x\)

\(\Leftrightarrow x^2=3\)  

\(\Leftrightarrow x=\pm\sqrt{3}\left(tm\right)\)  

c) \(\sqrt{2x^2-3}=\sqrt{4x-3}\) (ĐK: \(x\ge\dfrac{\sqrt{6}}{2}\)) 

\(\Leftrightarrow2x^2-3=4x-3\)

\(\Leftrightarrow2x^2=4x\)

\(\Leftrightarrow x^2=2x\)

\(\Leftrightarrow x\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=2\left(tm\right)\end{matrix}\right.\)

\(\sqrt{3}-\frac{5}{2}>\sqrt{3}-4\text{ vì }-\frac{5}{2}>-4\)

\(\Rightarrow2.\left(\sqrt{3}-\frac{5}{2}\right)>\sqrt{3}-4\)

\(\Rightarrow2.\sqrt{3}-5>\sqrt{3}-4\)

b) vì \(\sqrt{5}-\sqrt{12}< 0\), ta có: 

 \(5\sqrt{5}-2\sqrt{3}=4\sqrt{5}+\sqrt{5}-\sqrt{12}< 4\sqrt{5}< 4\sqrt{5}+6\) 

Vậy \(5\sqrt{5}-2\sqrt{3}< 6+4\sqrt{5}\)

c)\(\sqrt{2}-\sqrt{6}=\sqrt{2}.\left(\sqrt{1}-\sqrt{3}\right)>\left(1-\sqrt{3}\right)\)

Vậy \(\sqrt{2}-\sqrt{6}>1-\sqrt{3}\)

\(<=>x^2-\sqrt{3}x-\sqrt{5}x+\sqrt{15}=0<=>x\left(x-\sqrt{3}\right)-\sqrt{5}\left(x-\sqrt{3}\right)=0<=>\left(x-\sqrt{3}\right)\left(x-\sqrt{5}\right)=0\)

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