\(pt\Leftrightarrow20x+20y+50=25xy\)

\(\Leftrightarrow5y\left(5x-4\right)-4\left(5x-4\right)=66\)

\(\Leftrightarrow\left(5x-4\right)\left(5y-4\right)=66\)

đến đây thì dễ rồi

ĐKXĐ: \(x\ge0;x\ne1\)

a)   \(P=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{\left(\sqrt{x}+1\right)^2+\left(\sqrt{x}-1\right)^2}{x-1}\)

\(=\frac{\left(x+2\sqrt{x}+1\right)+\left(x-2\sqrt{x}+1\right)}{x-1}\)

\(=\frac{2x-2}{x-1}=\frac{2\left(x-1\right)}{x-1}=2\)

Vậy \(P=2\)

ĐK: \(x\ne0;y\ne0\)

Ta có hệ phương trình \(\hept{\begin{cases}x^2+xy-y^2=5\left(1\right)\\\frac{y}{x}-\frac{2x}{y}=\frac{-5}{2}-\frac{2}{xy}\left(2\right)\end{cases}}\)

Từ pt (2) ta có \(\frac{2y^2}{2xy}-\frac{4x^2}{2xy}=\frac{-5xy-4}{2xy}\Rightarrow2y^2-4x^2=-5xy-4\)

\(\Rightarrow4x^2-5xy-2y^2=4\)

Ta có hệ mới \(\hept{\begin{cases}x^2+xy-y^2=5\\4x^2-5xy-2y^2=4\end{cases}}\Leftrightarrow\hept{\begin{cases}4x^2+4xy-4y^2=20\\20x^2-25xy-10y^2=20\end{cases}}\)

\(\Rightarrow4x^2+4xy-4y^2=20x^2-25xy-10y^2\)

\(\Rightarrow-16x^2+29x+6y^2=0\Rightarrow\orbr{\begin{cases}x=2y\\x=-\frac{3y}{16}\end{cases}}\)

Với x = 2y, ta có \(\left(2y\right)^2+2y.y-y^2=5\Rightarrow5y^2=5\Rightarrow\orbr{\begin{cases}y=1\Rightarrow x=2\\y=-1\Rightarrow x=-2\end{cases}}\)

Với \(x=-\frac{3}{16}y\), ta có \(\left(-\frac{3y}{16}\right)^2+\left(-\frac{3y}{16}\right).y-y^2=5\Rightarrow-\frac{296}{256}y^2=5\) (Vô nghiệm)

Vậy hệ phương trình có nghiệm (x;y) = (2;1) hoặc (x;y) = (-2;-1).

Đặt \(a=\frac{1}{x};b=\frac{1}{y};c=\frac{1}{z}\)thì \(x,y,z>0\)và ta cần chứng minh \(\frac{x}{\sqrt{3zx+yz}}+\frac{y}{\sqrt{3xy+zx}}+\frac{z}{\sqrt{3yz+xy}}\ge\frac{3}{2}\)\(\Leftrightarrow\frac{x^2}{x\sqrt{3zx+yz}}+\frac{y^2}{y\sqrt{3xy+zx}}+\frac{z^2}{z\sqrt{3yz+xy}}\ge\frac{3}{2}\)

Áp dụng BĐT Cauchy-Schwarz dạng phân thức, ta có: \(\frac{x^2}{x\sqrt{3zx+yz}}+\frac{y^2}{y\sqrt{3xy+zx}}+\frac{z^2}{z\sqrt{3yz+xy}}\ge\)\(\frac{\left(x+y+z\right)^2}{x\sqrt{3zx+yz}+y\sqrt{3xy+zx}+z\sqrt{3yz+xy}}\)

Áp dụng BĐT Cauchy-Schwarz, ta có: \(x\sqrt{3zx+yz}+y\sqrt{3xy+zx}+z\sqrt{3yz+xy}\)\(=\sqrt{x}.\sqrt{3zx^2+xyz}+\sqrt{y}.\sqrt{3xy^2+xyz}+\sqrt{y}.\sqrt{3yz^2+xyz}\)\(\le\sqrt{\left(x+y+z\right)\left[3\left(xy^2+yz^2+zx^2+xyz\right)\right]}\)

Ta cần chứng minh \(\sqrt{\left(x+y+z\right)\left[3\left(xy^2+yz^2+zx^2+xyz\right)\right]}\le\frac{2}{3}\left(x+y+z\right)^2\)

\(\Leftrightarrow\left(x+y+z\right)^4\ge\frac{9}{4}\left(x+y+z\right)\left[3\left(xy^2+yz^2+zx^2+xyz\right)\right]\)

\(\Leftrightarrow\left(x+y+z\right)^3\ge\frac{27}{4}\left(xy^2+yz^2+zx^2+xyz\right)\)(*)

Không mất tính tổng quát, giả sử \(y=mid\left\{x,y,z\right\}\)thì khi đó \(\left(y-x\right)\left(y-z\right)\le0\Leftrightarrow y^2+zx\le xy+yz\)

\(\Leftrightarrow xy^2+zx^2\le x^2y+xyz\Leftrightarrow xy^2+yz^2+zx^2+xyz\le\)\(x^2y+yz^2+2xyz=y\left(z+x\right)^2=4y.\frac{z+x}{2}.\frac{z+x}{2}\)

\(\le\frac{4}{27}\left(y+\frac{z+x}{2}+\frac{z+x}{2}\right)^3=\frac{4\left(x+y+z\right)^3}{27}\)

Như vậy (*) đúng

Đẳng thức xảy ra khi a = b = c

tính  các độ dài ME,MF