Biểu thức này chỉ có max khi a;b là số thực dương, đề bài thiếu

Bunhiacopxki:

\(\left(a^3+b\right)\left(\dfrac{1}{a}+b\right)\ge\left(a+b\right)^2\)

\(\Rightarrow\dfrac{1}{a^3+b}\le\dfrac{\dfrac{1}{a}+b}{\left(a+b\right)^2}=\dfrac{ab+1}{a\left(a+b\right)^2}\)

Tương tự: \(\dfrac{1}{b^3+a}\le\dfrac{ab+1}{b\left(a+b\right)^2}\)

\(\Rightarrow P\le\left(a+b\right)\left(\dfrac{ab+1}{a\left(a+b\right)^2}+\dfrac{ab+1}{b\left(a+b\right)^2}\right)-\dfrac{1}{ab}\)

\(P\le\left(a+b\right).\dfrac{ab+1}{\left(a+b\right)^2}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)-\dfrac{1}{ab}=\dfrac{ab+1}{a+b}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)-\dfrac{1}{ab}\)

\(P\le\dfrac{ab+1}{a+b}\left(\dfrac{a+b}{ab}\right)-\dfrac{1}{ab}=\dfrac{ab+1}{ab}-\dfrac{1}{ab}=1+\dfrac{1}{ab}-\dfrac{1}{ab}=1\)

Dấu "=" xảy ra khi \(a=b=1\)

\(A=-\dfrac{1}{5}+\dfrac{1}{5^2}-\dfrac{1}{5^3}+...+\dfrac{1}{5^{100}}\)

\(\Rightarrow5A=-1+\dfrac{1}{5}-\dfrac{1}{5^2}+...+\dfrac{1}{5^{99}}\)

\(\Rightarrow5A+A=-1+\dfrac{1}{5^{100}}\)

\(\Rightarrow6A=-\dfrac{5^{100}+1}{5^{100}}\)

\(\Rightarrow A=-\dfrac{5^{100}+1}{5^{100}\times6}\)

sorry nha em không biết

 em mới lớp 4 mà

\(\left\{{}\begin{matrix}\left|x\right|\ge2\\\left|y\right|\ge2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x^2\ge4\\y^2\ge4\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x^2}\le\dfrac{1}{4}\\\dfrac{1}{y^2}\le\dfrac{1}{4}\end{matrix}\right.\)

\(\left(\dfrac{x+y}{xy}\right)^2=\dfrac{\left(x+y\right)^2}{x^2y^2}\le\dfrac{2\left(x^2+y^2\right)}{x^2y^2}=\dfrac{2}{x^2}+\dfrac{2}{y^2}\le2.\dfrac{1}{4}+2.\dfrac{1}{4}=1\)

\(\Rightarrow\dfrac{x+y}{xy}\le1\)

Dấu "=" xảy ra khi \(x=y=\pm2\)

\(4x^2+8x+2=0\)\(\Leftrightarrow4x^2+8x+4-2=0\)\(\Leftrightarrow4\left(x^2+2x+1\right)-2=0\)\(\Leftrightarrow4\left(x+1\right)^2-\left(\sqrt{2}\right)^2=0\)\(\Leftrightarrow\left[2\left(x+1\right)\right]^2-\left(\sqrt{2}\right)^2=0\)\(\Leftrightarrow\left[2\left(x+1\right)+\sqrt{2}\right]\left[2\left(x+1\right)-\sqrt{2}\right]=0\)\(\Leftrightarrow\left(2x+2+\sqrt{2}\right)\left(2x+2-\sqrt{2}\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}2x+2+\sqrt{2}=0\\2x+2-\sqrt{2}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-2-\sqrt{2}}{2}\\x=\frac{-2+\sqrt{2}}{2}\end{cases}}\)

Vậy tập nghiệm của pt đã cho là \(S=\left\{\frac{-2\pm\sqrt{2}}{2}\right\}\)

ơ ơ ko lm đc 

thay 2 => 4 may ra còn lm đc