Có lẽ x,y nguyên:v

\(x^2+2xy=24y^2+20\)

\(\Leftrightarrow\left(x+y\right)^2=25y^2+20\)

\(\Leftrightarrow\left(x+y\right)^2-25y^2=20\)

\(\Leftrightarrow\left(x-4y\right)\left(x+6y\right)=20\)

Đến đây bạn làm nốt

X bằng 2

\(5x^2-17x-18=0\)

\(\Leftrightarrow5x^2-30x+3x-18=0\)

\(\Leftrightarrow5x\left(x-6\right)+3\left(x-6\right)=0\)

\(\Leftrightarrow\left(5x+3\right)\left(x-6\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}5x+3=0\\x-6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}5x=-3\\x=6\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{-3}{5}\\x=6\end{cases}}}\)

Vậy \(x=\frac{-3}{5};x=6\)

<3 

Cần CM: \(\frac{a}{\left(1-a\right)^3}\ge\frac{135}{16}a-\frac{27}{16}\)\(\left(0< a< 1\right)\)

thaajt vậy, bđt \(\Leftrightarrow\)\(\left(a-\frac{1}{3}\right)^2\left(15a^2-38a+27\right)\ge0\) đúng 

\(\Sigma\frac{a}{\left(b+c\right)^3}=\Sigma\frac{a}{\left(1-a\right)^3}\ge\frac{135}{16}\left(a+b+c\right)-\frac{81}{16}=\frac{27}{8}\)

dấu "=" xảy ra khi a=b=c=1 

à nhầm, \(a=b=c=\frac{1}{3}\)

Áp dụng liên tiếp AM - GM và Cauchy - Schwarz ta có :

\(\frac{a^2+ab+1}{\sqrt{a^2+3ab+c^2}}\ge\frac{a^2+ab+1}{\sqrt{a^2+ab+c^2+\left(a^2+b^2\right)}}\)

   \(=\frac{a^2+ab+1}{\sqrt{a^2+ab+1}}\)

\(=\sqrt{a^2+ab+1}=\sqrt{a^2+ab+a^2+b^2+c^2}\)

\(=\frac{1}{\sqrt{5}}\sqrt{\left(\frac{9}{4}+\frac{3}{4}+1+1\right)\left[\left(a+\frac{b}{2}\right)^2+\frac{3b^2}{4}+a^2+c^2\right]}\)

\(\ge\frac{1}{\sqrt{5}}\left[\frac{3}{2}\left(a+\frac{b}{2}\right)+\frac{3}{4}b+a+c\right]\)

\(=\frac{1}{\sqrt{5}}\left(\frac{5}{2}a+\frac{3}{2}b+c\right)\)

Chứng minh tương tự và công lại ta có đpcm 

Dấu " = " xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)