ĐKXĐ \(\orbr{\begin{cases}x\ge-1\\x\le-\frac{3}{2}\end{cases}}\)

PT

<=> \(4x^2+10x+9=5\sqrt{2x^2+5x+3}\)

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

Đặt \(2x^2+5x+3=t\)

=> \(2t^2-5t+3=0\)

<=> \(\orbr{\begin{cases}t=1\\t=\frac{3}{2}\end{cases}}\)

+ t=1

=> \(2x^2+5x+2=0\)=> \(\orbr{\begin{cases}x=-\frac{1}{2}\\x=-2\end{cases}}\)

+ t=3/2

=> \(2x^2+5x+\frac{3}{2}=0\)=> \(x=\frac{-5\pm\sqrt{13}}{4}\)

Kết hợp với ĐKXĐ

\(S=\left\{\frac{-5\pm\sqrt{13}}{4};-2;-\frac{1}{2}\right\}\)

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

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

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

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

=1

#mã mã#

PT 

<=> \(2\left(x^2+4x+7\right)=2\left(x+4\right)\sqrt{x^2+1}\)

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

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

<=> \(\orbr{\begin{cases}x+4-\sqrt{x^2+1}=\sqrt{3}\left(1\right)\\x+4-\sqrt{x^2+1}=-\sqrt{3}\left(2\right)\end{cases}}\)

Giải (1)

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

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

=> \(x+\sqrt{x^2+1}=\frac{1}{4-\sqrt{3}}=\frac{4+\sqrt{3}}{13}\)

Cộng 2 vế của Pt trên với (1)

=> \(x=\frac{14\sqrt{3}-48}{26}\)

Giải (2) tương tự (1)

ta được \(x=\frac{-48-14\sqrt{3}}{26}\)

Vậy \(S=\left\{\frac{14\sqrt{3}-48}{26};\frac{-48-14\sqrt{3}}{26}\right\}\)

ĐKXĐ \(x\ge-3\)

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

<=> \(4x^2-2x-6=2x\sqrt{x+3}\)

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

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

<=> \(\orbr{\begin{cases}2x=-\sqrt{x+3}\\x=\sqrt{x+3}\end{cases}}\)

 \(S=\left\{-\frac{3}{4};\frac{1+\sqrt{13}}{2}\right\}\)

Ta có: \(A=\frac{\left(1+\frac{2017}{1}\right)\left(1+\frac{2017}{2}\right)...\left(1+\frac{2017}{1009}\right)}{\left(1+\frac{1009}{1}\right)\left(1+\frac{1009}{2}\right)...\left(1+\frac{1009}{2017}\right)}=\frac{\frac{2017+1}{1}\frac{2017+2}{2}...\frac{2017+1009}{1009}}{\frac{1009+1}{1}\frac{1009+2}{2}...\frac{1009+2017}{2017}}\)

\(\Leftrightarrow A=\frac{\frac{2018.2019...3026}{1.2...1009}}{\frac{1010.1011...3026}{1.2...2017}}=\frac{2018.2019...3026}{1.2...1009}.\frac{1.2...2017}{1010.1011...3026}\)

\(\Leftrightarrow A=\frac{1.2...2017.2018.2019...3026}{1.2...1009.1010.1011...3026}=\frac{1.2.3...3026}{1.2.3...3026}=1.\)