ĐKXĐ:x\(\ge-1\)

Đặt \(\sqrt{x+1}=a\ge0\)

\(\Rightarrow\hept{\begin{cases}a^2=x+1\\a^2-1=x\\x^2=a^4-2a^2+1\end{cases}}\)

Khi đó pt trên trở thành : \(4a=a^4-2a^2+1-5\left(a^2-1\right)+14\)

\(\Leftrightarrow a^4-2a^2+1-5a^2+5+14-4a=0\)

\(\Leftrightarrow a^4-7a^2-4a+20=0\)

\(\Leftrightarrow a^4-4a^2-3a^2+6a-10a+20=0\)

\(\Leftrightarrow a^2\left(a-2\right)\left(a+2\right)-3a\left(a-2\right)-10\left(a-2\right)=0\)

\(\Leftrightarrow\left(a-2\right)\left(a^3+2a^2-3a-10\right)=0\)

\(\Leftrightarrow\left(a-2\right)\left(a^3-2a^2+4a^2-8a+5a-10\right)=0\)

\(\Leftrightarrow\left(a-2\right)\left(a-2\right)\left(a^2+4a+5\right)=0\)

\(\Leftrightarrow\left(a-2\right)^2=0\)(vì a²+4a+5=(a+2)²+1\(\ge1>0\))

\(\Leftrightarrow x=2\)(thỏa mãn ĐKXĐ)

\(\frac{1}{\sqrt{x+1}+\sqrt{x+2}}+\frac{1}{\sqrt{x+2}+\sqrt{x+3}}+...+\frac{1}{\sqrt{x+2019}+\sqrt{x+2020}}=11\)

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

\(+...+\frac{\sqrt{x+2020}-\sqrt{x+2019}}{\left(\sqrt{x+2019}+\sqrt{x+2020}\right)\left(\sqrt{x+2020}-\sqrt{x+2019}\right)}=11\)

\(\Leftrightarrow\)\(\frac{\sqrt{x+2}-\sqrt{x+1}}{x+2-x-1}+\frac{\sqrt{x+3}-\sqrt{x+2}}{x+3-x-2}+...+\frac{\sqrt{x+2020}-\sqrt{x+2019}}{x+2020-x-2019}=11\)

\(\Leftrightarrow\)\(\sqrt{x+2}-\sqrt{x+1}+\sqrt{x+3}-\sqrt{x+2}+...+\sqrt{x+2020}-\sqrt{x+2019}=11\)

\(\Leftrightarrow\)\(\sqrt{x+2020}-\sqrt{x+1}=11\)

\(\Leftrightarrow\)\(\sqrt{x+2020}=11+\sqrt{x+1}\)

\(\Leftrightarrow\)\(x+2020=121+22\sqrt{x+1}+x+1\)

\(\Leftrightarrow\)\(22\sqrt{x+1}=1898\)

\(\Leftrightarrow\)\(\sqrt{x+1}=\frac{949}{11}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x+1=\frac{900601}{121}\\x+1=\frac{-900601}{121}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{900480}{121}\\x=\frac{-900722}{121}\end{cases}}\)

Chúc bạn học tốt ~ 

PS : sai thì thui nhá 

Bài của bạn Quân làm đúng ùi nhưng mà căn thì không ra số âm nhé!

ĐKXĐ x\(\ge-2\)

Khi đó pt dã cho \(\Leftrightarrow x^3+8-3\sqrt{x^3+8}=0\)

\(\Leftrightarrow\sqrt{x^3+8}\left(\sqrt{x^3+8}-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x^3+8}=0\\\sqrt{x^3+8}=3\end{cases}\Leftrightarrow\orbr{\begin{cases}x^3+8=0\\x^3+8=9\end{cases}\Leftrightarrow}\orbr{\begin{cases}x^3=-8\\x^3=1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-2\\x=1\end{cases}}}\)(thỏa mãn ĐKXĐ)

Vậy.......................

rảnh quá mak cha nội 

23+5+2004=2032

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