Ta có:

\(a^3+b^3=2\)

\(\Rightarrow\left(a+b\right)\left(a^2-ab+b^2\right)=2\)

\(\Rightarrow a+b=\dfrac{2}{a^2-ab+b^2}\)

Mà: \(2\left(a-b\right)^2\ge0\forall a,b\)

\(\Rightarrow2\left(a^2-2ab+b^2\right)\ge0\)

\(\Rightarrow2a^2-4ab+2b^2\ge0\)

\(\Rightarrow2a^2+2a^2-4ab+2b^2+2b^2\ge2a^2+2b^2\)

\(\Rightarrow4a^2-4ab+4b^2\ge2\left(a+b\right)^2\)

\(\Rightarrow4\left(a^2-ab+b^2\right)\ge2\left(a+b\right)^2\ge\left(a+b\right)^2\)

\(\Rightarrow a^2-ab+b^2\ge\dfrac{\left(a+b\right)^2}{4}\)

\(\Rightarrow\dfrac{2}{a^2-ab+b^2}\le\dfrac{8}{\left(a+b\right)^2}\)

\(\Rightarrow a+b\le\dfrac{8}{\left(a+b\right)^2}\)

\(\Rightarrow\left(a+b\right)^3\le8\)

\(\Rightarrow a+b\le2\)

Vậy: \(A_{max}=2\)

Đặt \(f\left(x\right)=ax^3+bx^2+cx+d\left(a\inℤ^+\right)\)

\(f\left(5\right)=125a+25b+5c+d\)

\(f\left(3\right)=27a+9b+3c+d\)

\(\Rightarrow f\left(5\right)-f\left(3\right)=98a+16b+2c\)

Mà \(f\left(5\right)-f\left(3\right)=2022\) nên \(98a+16b+2c=2022\) 

\(\Leftrightarrow49a+8b+c=1011\)

Lại có \(f\left(7\right)=343a+49b+7c+d\)

\(f\left(1\right)=a+b+c+d\)

\(\Rightarrow f\left(7\right)-f\left(1\right)=342a+48b+6c\) \(=6\left(57a+8b+c\right)\) \(=6\left(8a+1011\right)\) (vì \(49a+8b+c=1011\))

 Mà do \(a\inℤ^+\) nên \(f\left(7\right)-f\left(1\right)\) là hợp số (đpcm)

công thức tổng quát: f(x)=x^{3        sdasdasdadasd}

\(VT\ge\dfrac{4}{x^2+y^2+2xy}=\dfrac{4}{\left(x+y\right)^2}\ge4\) (vì \(x+y\le1\) )

Dấu "=" xảy ra \(\Leftrightarrow x=y=\dfrac{1}{2}\)

Ta có đpcm

\(A=\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}+\dfrac{1}{4xy}+4xy+\dfrac{5}{4xy}\)

\(\ge\dfrac{4}{x^2+y^2+2xy}+2\sqrt{\dfrac{1}{4xy}.4xy}+\dfrac{5}{4.\dfrac{\left(x+y\right)^2}{4}}\)

\(\ge\dfrac{4}{1^2}+2+\dfrac{5}{1^2}\) (do \(x+y\le1\))

\(=11\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=\dfrac{1}{2}\)

Vậy GTNN của A là 11.

f f

ff đi mày

67765

\(\dfrac{2}{3+2\sqrt{2}}-\dfrac{2}{3-2\sqrt{2}}\\ =\dfrac{2\left(3-2\sqrt{2}\right)}{1}-\dfrac{2\left(3+2\sqrt{2}\right)}{1}\\ =6-4\sqrt{2}-6-4\sqrt{2}\\ =-8\sqrt{2}\)

\(\left(\sqrt{14}-3\sqrt{2}\right)^3+6\sqrt{28}\)

\(=\left(\sqrt{14}\right)^2-2\cdot3\sqrt{2}\cdot\sqrt{14}+\left(3\sqrt{2}\right)^2+6\cdot2\sqrt{7}\)

\(=14-6\sqrt{28}+9\cdot2+12\sqrt{7}\)

\(=14-6\cdot2\sqrt{7}+18+12\sqrt{7}\)

\(=\left(14+18\right)+\left(12\sqrt{7}-12\sqrt{7}\right)\)

\(=32\)