ĐKXĐ: \(-1\le x\le1\)

Xét \(\sqrt{\left(1+x\right)^3}-\sqrt{\left(1-x\right)^3}=\left(\sqrt{1+x}-\sqrt{1-x}\right)\left[\left(1+x\right)+\left(1-x\right)+\sqrt{\left(1+x\right)\left(1-x\right)}\right]\)

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

Khi đó phương trình đề trở thành:

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

Vì \(2+\sqrt{1-x^2}>0\)nên ta có thể chia 2 vế cho \(2+\sqrt{1-x^2}\):

\(\Rightarrow\sqrt{1+\sqrt{1-x^2}}\left(\sqrt{1+x}-\sqrt{1-x}\right)=\frac{1}{\sqrt{3}}\),Bình phương 2 vế:

\(\Rightarrow\left(1+\sqrt{1-x^2}\right)\left[\left(1+x\right)+\left(1-x\right)-2\sqrt{\left(1+x\right)\left(1-x\right)}\right]=\frac{1}{3}\)

\(\Leftrightarrow\left(1+\sqrt{1-x^2}\right)\left(2-2\sqrt{1-x^2}\right)=\frac{1}{3}\Leftrightarrow2\left(1+\sqrt{1-x^2}\right)\left(1-\sqrt{1-x^2}\right)=\frac{1}{3}\)\(\Leftrightarrow1-\left(1-x^2\right)=\frac{1}{3}\Leftrightarrow x^2=\frac{1}{6}\Leftrightarrow x=\pm\frac{1}{\sqrt{6}}\)

Ta xét phương trình đề: vế phải luôn không âm vì vậy vế trái phải không âm 

Khi đó \(\sqrt{\left(1+x\right)^3}-\sqrt{\left(1-x\right)^3}\ge0\Leftrightarrow1+x\ge1-x\Leftrightarrow x\ge0\)

Vậy ta chỉ nhận nghiệm duy nhất là \(x=\frac{1}{\sqrt{6}}\)

ĐKXĐ: \(x\ge1\)

Ta có:

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

Ta xét 2 trường hợp sau:

TH1: \(x\ge2\)

Khi đó:

\(\left(1\right)\Leftrightarrow2\sqrt{x-1}-1=\dfrac{x+1}{2}\\ \Leftrightarrow2\sqrt{x-1}=\dfrac{x+3}{2}\\ \Leftrightarrow16\left(x-1\right)=x^2+6x+9\\ \Leftrightarrow x^2-10x+25=0\\ \Leftrightarrow\left(x-5\right)^2=0\\ \Leftrightarrow x=5\left(TMĐK\right)\)

TH2: \(1\le x< 2\)

Khi đó:

\(\left(1\right)\Leftrightarrow1=\dfrac{x+1}{2}\Leftrightarrow x=1\left(TMĐK\right)\)

Vậy x=1 hoặc x=5

\(Áp-dụng-BĐT-\left(a+b\right)\le\sqrt{2\left(a^2+b^2\right)}=>VT=x+\sqrt{2-x^2}\le2\\ VP=4y^2+4y+1\ge2\\ =>1\ge VT=VP\ge1\\ =>2y+1=0vax=\sqrt{2-x^2}.\)

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

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

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

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

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

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

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

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

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

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

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

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!

1/ \(x^3-x^2-x=\frac{1}{3}\Leftrightarrow3x^3-3x^2-3x=1\Leftrightarrow x^3+3x^2+3x+1=4x^3\)

\(\Leftrightarrow\left(x+1\right)^3=\left(\sqrt[3]{4}x\right)^3\Leftrightarrow x+1=\sqrt[3]{4}x\Leftrightarrow x\left(\sqrt[3]{4}-1\right)=1\Leftrightarrow x=\frac{1}{\sqrt[3]{4}-1}\)

2/ ĐKXĐ \(x\ge1\)

 \(3+\sqrt{x-2\sqrt{x-1}}=2\sqrt{x-2\sqrt{x-1}}\Leftrightarrow3=\sqrt{\left(\sqrt{x-1}-1\right)^2}\Leftrightarrow\left|\sqrt{x-1}-1\right|=3\)

Tới đây xét trường hợp rồi giải :)