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

\(\Leftrightarrow2x+2\sqrt{\left(x-\sqrt{2-x}\right)\left(x+\sqrt{x-2}\right)}=9\)

\(\Leftrightarrow2\sqrt{\left(x-\sqrt{x-2}\right)\left(x+\sqrt{x+2}\right)}=9-2x\)

\(\Leftrightarrow4\left(x-\sqrt{x-2}\right)\left(x+\sqrt{x-2}\right)=\left(9-2x\right)^2\)

\(\Leftrightarrow4x^2-4x+8=81-36x+4x^2\)

\(\Leftrightarrow-4x+8=81-36x\)

\(\Leftrightarrow-4x=81-36x-8\)

\(\Leftrightarrow-4x=-36x+73\)

\(\Leftrightarrow-4x+36x=73\)

\(\Leftrightarrow32x=73\)

\(\Leftrightarrow x=\frac{73}{32}\)

Vậy: nghiệm phương trình là: \(\left\{\frac{73}{32}\right\}\)

Lỗi sai ngu người nhất của Chihiro.Quên viết ĐKXĐ ak em

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

\(ĐKXĐ:x\ge2\)

Bình phương 2 vế của pt ta được

\(2x+2\sqrt{\left(x-\sqrt{x-2}\right)\left(x+\sqrt{x-2}\right)}=9\)

\(\Leftrightarrow2\sqrt{x^2-x+2}=9-2x\)

\(\Leftrightarrow\hept{\begin{cases}9-2x\ge0\Leftrightarrow\frac{9}{2}\ge x\\4\left(x^2-x+2\right)=81-36x+4x^2\left(2\right)\end{cases}}\)

\(\left(2\right)\Leftrightarrow32x-73=0\Leftrightarrow x=\frac{73}{32}\left(tmDK\right)\)

Vậy \(S=\left\{\frac{73}{32}\right\}\)

p/s:học hỏi đi con.

ĐK : \(\left(x\ne-4;x\ne-5;x\ne-6;x\ne-7\right)\)

\(\Rightarrow\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Rightarrow\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)

\(\Rightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)

\(\Rightarrow\frac{3}{x^2+11x+28}=\frac{1}{18}\)

\(\Leftrightarrow x^2+11x+28=54\)

\(\Rightarrow x^2+11x-26=0\)

\(\Rightarrow\left(x-2\right)\left(x+13\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=2\\x=-13\end{cases}}\)

Vậy pt có tập nghiệm là \(S=\left\{2;-13\right\}\)

\(H=\frac{x\left(x+1\right)}{2}.\frac{x\left(x+1\right)\left(2x+1\right)}{6}=x^2\left(x+1\right)^2.\frac{2x+1}{12}\)

tồn tại vô số nguyên dương x để \(\frac{2x+1}{12}\) là số chính phương => ... 

\(\sqrt[3]{3x-5}-\sqrt[3]{3x-4}=-1\)

Đặt \(\hept{\begin{cases}\sqrt[3]{3x-5}=a\\\sqrt[3]{3x-4}=b\end{cases}}\)

Ta có hpt  \(\hept{\begin{cases}a-b=-1\\a^3-b^3=-1\end{cases}}\)

SD pp thế hoặc trừ 2 vế của hpt

Ta có :\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2019.2020}=k\left(\frac{1}{1011}+\frac{1}{1012}+\frac{1}{1013}+....+\frac{1}{2020}\right)\)

\(\Rightarrow1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{2019}-\frac{1}{2020}=k\left(\frac{1}{1011}+\frac{1}{1012}+...+\frac{1}{2020}\right)\)

\(\Rightarrow1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2020}-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2020}\right)=k\left(\frac{1}{1011}+\frac{1}{1012}+...+\frac{1}{2020}\right)\)

\(\Rightarrow1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2020}-1-\frac{1}{2}-\frac{1}{4}-...-\frac{1}{1010}=k\left(\frac{1}{1011}+\frac{1}{1012}+...+\frac{1}{2020}\right)\)

\(\Rightarrow\frac{1}{1011}+\frac{1}{1012}+....+\frac{1}{2020}=k\left(\frac{1}{1011}+\frac{1}{1012}+...+\frac{1}{2020}\right)\)

=> k = 1

=> k là số tự nhiên (đpcm)