Vì \(\left|2x-27\right|\ge0\Rightarrow\left|2x-27\right|^{2011}\ge0\);  \(\left(3y+10\right)^{2012}\ge0\)

=>\(\left|2x-27\right|^{2011}+\left(3y+10\right)^{2012}\ge0\)

Dấu "=" xảy ra khi \(\left|2x-27\right|^{2011}=\left(3y+10\right)^{2012}=0\Leftrightarrow\hept{\begin{cases}\left|2x-27\right|=0\\\left(3y+10\right)^{2012}=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}2x-27=0\\3y+10=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{27}{2}\\y=-\frac{10}{3}\end{cases}}\)

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)

\(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{\left(bk+b\right)^2}{\left(dk+d\right)^2}=\frac{b^2\left(k+1\right)^2}{d^2\left(k+1\right)^2}=\frac{b^2}{d^2}\)(1)

\(\frac{a^2+b^2}{c^2+d^2}=\frac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\frac{b^2.k^2+b^2}{d^2.k^2+d^2}=\frac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\frac{b^2}{d^2}\)(2)

\(\frac{a^2-b^2}{c^2-d^2}=\frac{\left(bk\right)^2-b^2}{\left(dk\right)^2-d^2}=\frac{b^2k^2-b^2}{d^2k^2-d^2}=\frac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\frac{b^2}{d^2}\)(3)

Từ (1),(2),(3) =>\(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{a^2+b^2}{c^2+d^2}=\frac{a^2-b^2}{c^2-d^2}\)(đpcm)

bang 210 

k cho minh nhe

\(a)\) Ta có : 

\(A=\left|x-1,35\right|\ge0\)

Dấu "=" xảy ra khi \(\left|x-1,35\right|=0\)

\(\Rightarrow\)\(x-1,35=0\)

\(\Rightarrow\)\(x=1,35\)

Vậy \(A_{min}=0\) khi \(x=1,35\)

\(b)\) Ta có : 

\(\left|2x-8\right|\ge0\)

\(\Rightarrow\)\(\frac{3}{14}-\left|2x-8\right|\le\frac{3}{14}\)

Dấu "=" xảy ra khi \(\left|2x-8\right|=0\)

\(\Rightarrow\)\(2x-8=0\)

\(\Rightarrow\)\(2x=8\)

\(\Rightarrow\)\(x=4\)

Vậy \(B_{max}=\frac{3}{14}\) khi \(x=4\)

Chúc bạn học tốt ~ 

tui cũng đang hỏi câu này

Đặt \(\frac{x}{2}=\frac{y}{5}=k\)

\(\Rightarrow x=2k\)

     \(y=5k\)   

Thay vào : \(x\cdot y=40\)

\(\Rightarrow2k\cdot5k=40\)

\(\Rightarrow10k^2=40\)

\(\Rightarrow k^2=4\Rightarrow\orbr{\begin{cases}k=2\\k=-2\end{cases}}\)

Ta có : \(x=2k=\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)

            \(y=5k=\orbr{\begin{cases}y=10\\y=-10\end{cases}}\)

Đặt \(\frac{x}{2}=\frac{y}{5}=k\)

\(\Rightarrow\hept{\begin{cases}x=2k\\y=5k\end{cases}}\)

Có \(x\times y=40\)

\(\Rightarrow2k\times5k=40\)

\(\Rightarrow10\times k^2=40\)

\(\Rightarrow k^2=40:10\)

\(\Rightarrow k^2=4\)

\(\Rightarrow\orbr{\begin{cases}k=2\\k=-2\end{cases}}\)

có \(k=2\Rightarrow\hept{\begin{cases}x=2\times2=4\\y=5\times2=10\end{cases}}\)

     \(k=-2\Rightarrow\hept{\begin{cases}x=2\times\left(-2\right)=-4\\y=5\times\left(-2\right)=-10\end{cases}}\)

\(33^{52}=3^{52}.11^{52}=81^{13}.\left(11^4\right)13\)

\(44^{39}=4^{39}.11^{39}=64^{13}.\left(11^3\right)^{13}\)

Ta có\(11^4>11^3\)\(\Rightarrow11^{52}>11^{39}\)(1)

\(81^{13}>64^{13}\Rightarrow3^{52}>4^{39}\)(2)

Từ (1) và (2)\(\Rightarrow33^{52}>44^{39}\)

mk nghĩ là  \(33^{52}\)>\(44^{39}\)ngắn gọn nhất