\(2\left(3-4x\right)=10-\left(2x-5\right)\Rightarrow6-8x=10-2x+5\)

\(\Rightarrow6-8x-10+2x-5=0\)

\(-9-6x=0\Rightarrow-6x=9\Rightarrow x=\frac{-3}{2}\)

Bunhiacopxki:

\(\left(a^2+bc+ca\right)\left(b^2+bc+ca\right)\ge\left(ab+bc+ca\right)^2\)

\(\Rightarrow\dfrac{ab}{a^2+bc+ca}\le\dfrac{ab\left(b^2+bc+ca\right)}{\left(ab+bc+ca\right)^2}\)

Tương tự: \(\dfrac{bc}{b^2+ca+ab}\le\dfrac{bc\left(c^2+ca+ab\right)}{\left(ab+bc+ca\right)^2}\)

\(\dfrac{ca}{c^2+ab+bc}\le\dfrac{ca\left(a^2+ab+bc\right)}{\left(ab+bc+ca\right)^2}\)

\(\Rightarrow VT\le\dfrac{ab\left(b^2+bc+ca\right)+bc\left(c^2+ca+ab\right)+ca\left(a^2+ab+bc\right)}{\left(ab+bc+ca\right)^2}\)

Nên ta chỉ cần chứng minh:

\(\dfrac{ab\left(b^2+bc+ca\right)+bc\left(c^2+ca+ab\right)+ca\left(a^2+ab+bc\right)}{\left(ab+bc+ca\right)^2}\le\dfrac{a^2+c^2+c^2}{ab+bc+ca}\)

\(\Leftrightarrow ab\left(b^2+bc+ca\right)+bc\left(c^2+ca+ab\right)+ca\left(a^2+ab+bc\right)\le\left(a^2+b^2+c^2\right)\left(ab+bc+ca\right)\)

Nhân phá và rút gọn 2 vế:

\(\Leftrightarrow a^3b+b^3c+c^3a\ge abc\left(a+b+c\right)\)

\(\Leftrightarrow\dfrac{a^3b+b^3c+c^3a}{abc}\ge a+b+c\)

\(\Leftrightarrow\dfrac{a^2}{c}+\dfrac{b^2}{a}+\dfrac{c^2}{b}\ge a+b+c\)

Đúng do: \(\dfrac{a^2}{c}+\dfrac{b^2}{a}+\dfrac{c^2}{b}\ge\dfrac{\left(a+b+c\right)^2}{a+b+c}=a+b+c\)

Dấu "=" xảy ra khi \(a=b=c\)

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 \(x=y=\pm2\)