\(x^2-2xy+5y^2-4y+1=0\)

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

=> \(\left(x-y\right)^2+\left(2y-1\right)^2=0\)

Ta có: \(\left(x-y\right)^2\ge0\forall x;y\)

      \(\left(2y-1\right)^2\ge0\forall y\)

=> \(\left(x-y\right)^2+\left(2y-1\right)^2\ge0\forall x;y\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-y=0\\2y-1=0\end{cases}}\) <=> \(\hept{\begin{cases}x=y\\2y=1\end{cases}}\) <=> \(x=y=\frac{1}{2}\)

Vậy x = y = 1/2 (tm)

\(x^2-2xy+5y^2-4y+1=0\)

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

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

Mà (x-y)²và (2y-1)² > 0

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

a)\(x^2+5y^2-2xy+4y+1=0\)

\(x^2+2xy+y^2+4y^2+4y+1=0\)

\(\left(x+y\right)^2+\left(2y+1\right)^2=0\)

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

      Từ (1) ta đc: x = 1/2

b)\(5x^2+5y^2+8xy-2x+2y+2=0\)

\(4x^2+8xy+4y^2+x^2-2x+1+y^2+2y+1=0\)

\(\left(2x+2y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)

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

CÂU B Sao bạn làm được vậy

Bài làm:

a) \(x^2+5y^2-2xy+4y+1=0\)

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

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

Vì \(\hept{\begin{cases}\left(x-y\right)^2\ge0\\\left(2y+1\right)^2\ge0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}\left(x-y\right)^2=0\\\left(2y+1\right)^2=0\end{cases}\Rightarrow}x=y=-\frac{1}{2}\)

Vậy \(x=y=-\frac{1}{2}\)

b) \(5x^2+5y^2+8xy-2x+2y+2=0\)

\(\Leftrightarrow\left(4x^2+8xy+4y^2\right)+\left(x^2-2x+1\right)+\left(y^2+2y+1\right)=0\)

\(\Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(x+1\right)^2=0\)

Vì \(\hept{\begin{cases}4\left(x+y\right)^2\ge0\\\left(x-1\right)^2\ge0\\\left(y+1\right)^2\ge0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}4\left(x+y\right)^2=0\\\left(x-1\right)^2=0\\\left(y+1\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=1\\y=-1\end{cases}}}\)

Ta có: x^2+2y^2-2xy+2x+2-4y=0

=> x^2 -2xy+y^2+ 2x-2y+1+y^2-2y+1=0

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

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

mà (x-y+1)^2> hoặc=0 với mọi x;y

(y-1)^2> hoặc=0 với mọi x;y

nên x-y+1=0;y-1=0

=> y=1; x=0

a )x²+2y²-2xy+2x-4y+2=0 
<=>x²-2x(y-1)+y²-2y+1+y²-2y+1=0 
<=>x²-2x(y-1)+(y-1)²+(y-1)²=0 
<=>(x-y+1)²+(y-1)²=0 
<=>x-y+1=0 va y-1=0 
<=>x=y-1 y=1 
<=>x=1-1=0 y=1

                       XONG RỒI ĐẤY BẠN

a) \(x^2-2x+2xy=3+4y\)

\(x^2-2x+2xy-4y=3\)

\(x\left(x-2\right)+2y\left(x-2\right)=3\)

\(\left(x-2\right)\left(x+2y\right)=3\)

\(\Rightarrow x-2;x+2y\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

\(\Rightarrow\)Ta có bảng giá trị:

               Vậy, \(\left(x;y\right)\in\left\{\left(3;0\right);\left(1;-2\right);\left(5;-2\right)\left(-1;0\right)\right\}\)

b) \(\left|2x-3y\right|+\left|5y-7z\right|+\left|x^2-y^2-2z^2-45\right|=0\)

             Ta có: \(\left|2x-3y\right|\ge0\)

                        \(\left|5y-7z\right|\ge0\)

                        \(\left|x^2-y^2-2z^2-45\right|\ge0\)

                  \(\Rightarrow\left|2x-3y\right|+\left|5y-7z\right|+\left|x^2-y^2-2z^2-45\right|\ge0\)

            Mà đề cho \(\left|2x-3y\right|+\left|5y-7z\right|+\left|x^2-y^2-2z^2-45\right|=0\)

               \(\Rightarrow\hept{\begin{cases}\left|2x-3y\right|=0\\\left|5y-7z\right|=0\\\left|x^2-y^2-2z^2-45\right|=0\end{cases}\Rightarrow\hept{\begin{cases}2x-3y=0\\5y-7z=0\\x^2-y^2-2z^2-45=0\end{cases}}}\)

               \(\Rightarrow\hept{\begin{cases}2x=3y\\5y=7z\\x^2-y^2-2z^2=45\end{cases}\Rightarrow\hept{\begin{cases}10x=15y\\15y=21z\\x^2-y^2-2z^2=45\end{cases}}}\)

               \(\Rightarrow10x=15y=21z\Rightarrow\frac{x}{21}=\frac{y}{14}=\frac{z}{10}\Rightarrow\frac{x^2}{21^2}=\frac{y^2}{14^2}=\frac{z^2}{10^2}\)và \(x^2-y^2-2z^2=45\)

                             Áp dụng tính chất dãy tỉ số bằng nhau, ta được:

                           \(\frac{x^2}{21^2}=\frac{y^2}{14^2}=\frac{z^2}{10^2}=\frac{2z^2}{2\cdot10^2}=\frac{x^2-y^2-2z^2}{21^2-14^2-2\cdot10^2}\)

                                                                                        \(=\frac{45}{441-196-200}=1\)(vì \(x^2-y^2-2z^2=45\))

                 \(\Rightarrow\hept{\begin{cases}x^2=21^2\\y^2=14^2\\z^2=10^2\end{cases}}\Rightarrow\hept{\begin{cases}x=21\\y=14\\z=10\end{cases}}\)

                           Vậy, \(\left(x;y;z\right)=\left(21;14;10\right)\)

cảm ơn bạn nha Huỳnh Phước Mạnh

a) \(3x^2-3xy-5x+5y\)

\(=\left(3x^2-3xy\right)-\left(5x-5y\right)\)

\(=3x\left(x-y\right)-5\left(x-y\right)\)

\(=\left(x-y\right)\left(3x-5\right)\)

b) \(2x^3y-2xy^3-4xy^2-2xy\)

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

\(=2xy\left[x^2-\left(y^2+2y+1\right)\right]\)

\(=2xy\left[x^2-\left(y+1\right)^2\right]\)

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

c) \(x^2+1+2x-y^2\)

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

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

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

d) \(x^2+4x-2xy-4y+y^2\)

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

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

\(=\left(x-y\right)\left(x-y+4\right)\)

e) \(x^3-2x^2+x\)

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

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

f) \(2x^2+4x+2-2y^2\)

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

\(=2\left[\left(x^2+2x+1\right)+y^2\right]\)

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

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

a: =3x(x-y)-5(x-y)

=(x-y)(3x-5)

b: \(=2xy\left(x^2-y^2-2y-1\right)\)

\(=2xy\left[x^2-\left(y^2+2y+1\right)\right]\)

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

d:

Sửa đề: x^2+4x-2xy-4y+y^2

=x^2-2xy+y^2+4x-4y

=(x-y)^2+4(x-y)

=(x-y)(x-y+4)

e: =x(x^2-2x+1)

=x(x-1)^2

f: =2(x^2+2x+1-y^2)

=2[(x+1)^2-y^2]

=2(x+1+y)(x+1-y)