\(\left(2x+y\right)^2=4x^2+4xy+y^2\)

\(4x^2+4xy+y^2\)

a. Ta có: x²+y²-2x+4y+5=0

⇌(x-1)²+(y-2)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y-2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

b. Ta có: 4x²+y²-4x-6y+10=0

⇌ (2x-1)²+(y-3)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}2x-1=0\\y-3=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=3\end{matrix}\right.\)

c.Ta có: 5x²-4xy+y²-4x+4=0

⇌(2x-y)²+(x-2)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}2x-y=0\\x-2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=4\\x=2\end{matrix}\right.\)

d.Ta có: 2x²-4xy+4y²-10x+25=0

⇌ (x-2y)²+(x-5)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x-2y=0\\x-5=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{5}{2}\\x=5\end{matrix}\right.\)

\(\left\{{}\begin{matrix}4x^2+y^2\left(1-4xy\right)=0\\4x^2+2y^2-4xy-1=0\end{matrix}\right.\)

\(\Rightarrow y^2\left(1-4xy\right)-2y^2+4xy+1=0\)

\(\Leftrightarrow-y^2\left(4xy+1\right)+4xy+1=0\)

\(\Leftrightarrow\left(4xy+1\right)\left(1-y^2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4xy=-1\\y^2=1\end{matrix}\right.\)

Bạn tự giải nốt

lai hoi bo kien thuc rong ak

D

D

\(=\left(2x-y\right)^2-1\)

\(=\left(2x-y-1\right)\left(2x-y+1\right)\)