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

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

\(\Rightarrow\left(x^2-2.x.2+2^2\right)+\left(x^2+2.x.2y+\left(2y\right)^2\right)=0\)

\(\Rightarrow\left(x-2\right)^2+\left(x+2y\right)^2=0\)

\(\Rightarrow\hept{\begin{cases}x-2=0\\x+2y=0\end{cases}\Rightarrow}\hept{\begin{cases}x=2\\y=-1\end{cases}}\)

Chúc bạn học tốt.

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

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

a: x^2-2x+y^2-8y+17=0

=>x^2-2x+1+y^2-8y+16=0

=>(x-1)^2+(y-4)^2=0

=>x=1 và y=4

b: Sửa đề: 4x^2-4xy+y^2+y^2+4y+4=0

=>(2x-y)^2+(y+2)^2=0

=>y=-2 và x=-1

bạn viết lại đề đi, có số mũ, xuống dòng chứ thế này ai mà giải được

2x^ mấy vậy bạn?

a) x²+y²-4x+4y+8=0

⇔ (x-2)²+(y+2)²=0

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

b)5x²-4xy+y²=0

⇔ x²+(2x-y)²=0

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

c)x²+2y²+z²-2xy-2y-4z+5=0

⇔ (x-y)²+(y-1)²+(z-2)²=0

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

b: Ta có: \(5x^2-4xy+y^2=0\)

\(\Leftrightarrow x^2-\dfrac{4}{5}xy+y^2=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{2}{5}y+\dfrac{4}{25}y^2+\dfrac{21}{25}y^2=0\)

\(\Leftrightarrow\left(x-\dfrac{2}{5}y\right)^2+\dfrac{21}{25}y^2=0\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)

d)3x²+3y²+3xy-3x+3y+3=0

⇔ 6x²+6y²+6xy-6x+6y+6=0

⇔ 3(x+y)²+3(x-1)²+3(y+1)²=0

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

Bài 1:

\(a,=3x\left(3xy+5y-1\right)\\ b,=\left(z-2\right)\left(3z-5\right)\\ c,=\left(x+2y\right)^2-4z^2=\left(x+2y+2z\right)\left(x+2y-2z\right)\\ d,=x^2-3x+5x-15=\left(x-3\right)\left(x+5\right)\)

Bài 2:

\(a,\Leftrightarrow x\left(x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\\ b,\Leftrightarrow2x+2-4x^2-12x=9\\ \Leftrightarrow4x^2+10x+7=0\\ \Leftrightarrow4\left(x^2+\dfrac{5}{2}x+\dfrac{25}{16}\right)+\dfrac{3}{4}=0\\ \Leftrightarrow4\left(x+\dfrac{5}{6}\right)^2+\dfrac{3}{4}=0\left(vô.lí\right)\\ \Leftrightarrow x\in\varnothing\\ c,\Leftrightarrow x^2-12x+36=0\\ \Leftrightarrow\left(x-6\right)^2=0\\ \Leftrightarrow x=6\)