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

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)

Đặt \(\left(\frac{1}{x};\frac{1}{y}\right)=\left(a;b\right)\Rightarrow ab+a+b=3\)

\(\Rightarrow ab+2\sqrt{ab}\le3\Rightarrow\left(\sqrt{ab}+3\right)\left(\sqrt{ab}-1\right)\le0\)

\(\Rightarrow\sqrt{ab}\le1\Rightarrow ab\le1\)

\(P=\frac{a}{\sqrt{3+a^2}}+\frac{b}{\sqrt{3+b^2}}=\frac{a}{\sqrt{ab+a+b+a^2}}+\frac{b}{\sqrt{ab+a+b+b^2}}\)

\(=\frac{a}{\sqrt{\left(a+b\right)\left(a+1\right)}}+\frac{b}{\sqrt{\left(a+b\right)\left(b+1\right)}}\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{a}{a+1}+\frac{b}{a+b}+\frac{b}{b+1}\right)\)

\(P\le\frac{1}{2}\left(1+\frac{a}{a+1}+\frac{b}{b+1}\right)=\frac{1}{2}\left(1+\frac{ab+a+ab+b}{ab+a+b+1}\right)=\frac{1}{2}\left(1+\frac{ab+3}{4}\right)\)

\(P\le\frac{1}{2}\left(1+\frac{1+3}{4}\right)=1\)

Dấu " = " xảy ra khi \(a=b=1\) hay \(x=y=1\)

Chúc bạn học tốt !!!

\(\left(x^2-3x+9\right)\left(x^2+5x+9\right)=9x^2\)

\(\Leftrightarrow x^4+5x^3+9x^2-3x^3-15x^2-27x+9x^2+45x+81=9x^2\)

\(\Leftrightarrow x^4+2x^3+3x^2+18x+81=9x^2\)

\(\Leftrightarrow x^4+2x^3+3x^2+18x+81-9x^2=0\)

\(\Leftrightarrow x^4+2x^2-6x^2+18x+81=0\)

\(\Leftrightarrow\left(x^3-x^2-3x+27\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left(x^2-4x+9\right)\left(x+3\right)\left(x+3\right)=0\)

Vì \(x^2-4x+9\ne0\) nên: 

\(\Rightarrow x+3=0\)

     \(x=-3\)

Vậy: nghiệm phương trình là: {-3}