\(C=x-2\sqrt{xy}+3y-2\sqrt{x}+1\)à

\(C=x-2\sqrt{xy}+3y-2\sqrt{x}+1\)

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

\(=\left(\sqrt{x}-\sqrt{y}\right)^2-2\left(\sqrt{x}-\sqrt{y}\right)+1+2\left(y-\sqrt{y}+\frac{1}{4}\right)-\frac{1}{2}\)

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

Đến đây dễ rồi

\(\hept{\begin{cases}5x+3y=15\\4x-5y=6\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}20x+12y=60\\20x-25y=30\end{cases}}\)

\(\Leftrightarrow12y+25y=30\)

\(\Leftrightarrow y\left(12+25\right)=30\)

P/s đén đây bạn tứ tính tiếp nhé !!!!!!!!!!!

\(\frac{x^2}{\sqrt{x-2}}=\frac{8}{\sqrt{x-2}}\)

\(\Leftrightarrow x^2=8\)

\(\Leftrightarrow x=\pm\sqrt{8}\)

\(\sqrt{1-x}-\sqrt{x-1}=2-x\)

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

\(1-x+x-1-2.\sqrt{\left(1-x\right)\left(x-1\right)}=4-4x+x^2\)

\(2\sqrt{\left(x-1\right)^2}-4+4x-x^2=0\)

\(2x-2-4+4x-x^2=0\)

\(x^2-6x+6=0\)

Đenta=b'² -ac=(-3)² -1.6=3

=> phương trình có 2 nghiệm phân biệt.

\(\hept{\begin{cases}x_1=6+\sqrt{3}\\x_2=6-\sqrt{3}\end{cases}}\)

Vô nghiệm

Ta có: \(\hept{\begin{cases}x^2+2xy=x\\y^2+2xy=y\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\left(x+2y\right)=x\\y\left(y+2x\right)=y\end{cases}\Leftrightarrow\hept{\begin{cases}x+2y=1\\y+2x=1\end{cases}\Leftrightarrow}\hept{\begin{cases}x+2y=1\\4x+2y=2\end{cases}}}\)

\(\Leftrightarrow\hept{\begin{cases}-3x=-1\\x+2y=1\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{1}{3}\\\frac{1}{3}+2y=1\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{3}\\y=\frac{1}{3}\end{cases}}}\)