\(\frac{a^3}{a^2+ab+b^2}\ge\frac{2a-b}{3}\)

\(\Leftrightarrow\left(a^2+ab+b^2\right)\left(2a+b\right)\le3a^3\)

\(\Leftrightarrow2a^3+a^2b+ab^2-b^3\le3a^3\)

\(\Leftrightarrow-a^3+a^2b+ab^2-b^3\le0\)

\(\Leftrightarrow a^3+b^3\ge a^2b+ab^2\)

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

\(\Leftrightarrow a^2-ab+b^2\ge ab\)

\(\Leftrightarrow a^2-2ab+b^2\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\ge0\) (luôn đúng)

Vậy \(\frac{a^3}{a^2+ab+b^2}\ge\frac{2a-b}{3}\)

\(\sqrt{x^2+48}=4x-3+\sqrt{x^2+35}\Leftrightarrow\sqrt{x^2+48}-7=4x-4+\sqrt{x^2+35}-6\)

\(\Leftrightarrow\frac{x^2+48-49}{\sqrt{x^2+48}+7}=4x-4+\frac{x^2+35-36}{\sqrt{x^2+35}+6}\Leftrightarrow\frac{x^2-1}{\sqrt{x^2+48}+7}=4\left(x-1\right)+\frac{x^2-1}{\sqrt{x^2+35}+6}\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{x+1}{\sqrt{x^2+48}+7}-4-\frac{x+1}{\sqrt{x^2+35}+6}\right)=0\)\(\Leftrightarrow x-1=0\Leftrightarrow x=1\).

Ta có: \(3x+4y=5\)

\(\Leftrightarrow x=\frac{5-4y}{3}\)

Ta cần chứng minh:

\(x^2+y^2\ge1\)

\(\Leftrightarrow\left(\frac{5-4y}{3}\right)^2+y^2-1\ge0\)

\(\Leftrightarrow25y^2-40y+16\ge0\)

\(\Leftrightarrow\left(5y-4\right)^2\ge0\)(đúng)

Ta có : \(3x+4y=5\Rightarrow y=\frac{5-3x}{4}\)

\(\Rightarrow x^2+y^2=x^2+\frac{\left(5-3x\right)^2}{16}=x^2+\frac{9x^2-30x+25}{16}\)

\(=\frac{16x^2+9x^2-30x+25}{16}=\frac{25x^2-30x+25}{16}=\frac{\left(25x^2-30x+9\right)+16}{16}\)

\(=\frac{\left(5x-3\right)^2+16}{16}\ge\frac{16}{16}=1\)(đpcm)

\(\frac{cosa+sina}{cosa-sina}=\frac{1+tana}{1-tana}=\frac{1+3}{1-3}=-2\)

=>1+sina/1-sina=1+tana/1-tana=1+3/1-3=4/-2=-2

Ta có:

\(1-cos^2a=sin^2a\Rightarrow\left(1+cosa\right)\left(1-cosa\right)=sin^2a\Rightarrow\frac{1+cosa}{sina}=\frac{sina}{1-cosa}\)

\(A=\sqrt{5-2\sqrt{5}+1}-\sqrt{5+2\sqrt{5}+1}=\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{\left(\sqrt{5}+1\right)^2}\)

    \(=\sqrt{5}-1-\sqrt{5}-1=-2\)

Vậy \(A\in Z\)

Làm tương tự với B.