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

b) \(9-x^2-2xy-y^2=9-\left(x^2+2xy+y^2\right)=9-\left(x+y\right)^2=\left(3-x-y\right)\left(3+x+y\right)\)

c) \(25-x^2+4xy-4y^2=25-\left(x^2-4xy+4y^2\right)=25-\left(x-2y\right)^2=\left(5-x+2y\right)\left(5+x-2y\right)\)

a. x² - 2xy + y² - 1

= (x - y)² - 1²

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

b. 9 - x² - 2xy - y²

= 3² - (x + y)²

= (3 - x - y)(3 + x + y)

c. 25 - x² + 4xy - 4y²

= 5² - \(\left[x^2-4xy+\left(2y\right)^2\right]\)

= 5² - (x - 2y)²

= (5 - x + 2y)(5 + x - 2y)

a) \(2x^2+y^2+2xy+10x+25=0\)

\(\Leftrightarrow x^2+x^2+y^2+2xy+10x+25=0\)

\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(x^2+10x+25\right)=0\)

\(\Leftrightarrow\left(x+y\right)^2+\left(x+5\right)^2=0\)

Vì \(\hept{\begin{cases}\left(x+y\right)^2\ge0\forall x\\\left(x+5\right)^2\ge0\forall x\end{cases}}\)

\(\Rightarrow\left(x+y\right)^2+\left(x+5\right)^2\ge0\forall x\)

Vậy đẳng thức xảy ra\(\Leftrightarrow\hept{\begin{cases}x+y=0\\x+5=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-5\\y=5\end{cases}}\)

b)\(x^2+3y^2+2xy-2y+1=0\)

\(\Leftrightarrow x^2+y^2+2y^2+2xy-2y+\frac{1}{2}+\frac{1}{2}=0\)

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

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

Vì \(\left(x+y\right)^2+\left(\sqrt{2}y-\frac{1}{\sqrt{2}}\right)^2\ge0\)

nên \(\left(x+y\right)^2+\left(\sqrt{2}y-\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}>0\)

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

nên pt vô nghiệm

a) 2x² + y² + 2xy + 10x + 25 = 0

=> (x² + 2xy + y²) + (x² + 10x + 25) = 0

=> (x + y)² + (x + 5)² = 0 

    <=> \(\hept{\begin{cases}x+y=0\\x+5=0\end{cases}}\) <=> \(\hept{\begin{cases}y=-x\\x=-5\end{cases}}\) <=> \(\hept{\begin{cases}y=5\\x=-5\end{cases}}\)

b)c) xem lại đề

Chọn B

\(2xy-x^2-y^2+25=25-\left(x^2-2xy+y^2\right)=25-\left(x-y\right)^2=\left(5-x+y\right)\left(5+x-y\right)\)

=(x+y)^2-25

=(x+y+5)(x+y-5)

\(x^2-25+y^2+2xy=\left(x^2+2xy+y^2\right)-25=\left(x+y\right)^2-5^2=\left(x+y-5\right)\left(x+y+5\right)\)

=(x^2+2xy+y^2)+25=(x+y)^2-5^2=(x+y+5)(x+y-5)

Bài 1: 

\(=3x^3y-6x^2y^2+15xy\)

Bài 2: 

\(=\left(x+y\right)^2-25=\left(x+y+5\right)\left(x+y-5\right)\)

\(x^2+2xy-25+y^2\\ =\left(x^2+2xy+y^2\right)-5^2\\ =\left(x+y\right)^2-5^2\\ =\left(x+y-5\right)\left(x+y+5\right)\)

DÀI THẾ AI LÀM NỔI

\(x^2-25-2xy+y^2=\left(x-y\right)^2-5^2=\left(x-y+5\right)\left(x-y-5\right)\)

\(x^2-25-2xy+y^2\)

\(=\left(x^2-2xy+y^2\right)-25\)

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

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

\(=\left(x-y\right)^2+x=\left(25-5\right)^2+25=20^2+25=400+25=425\)

x²-2xy+y²+x
= (x-y)²+ x
thay số:
= (25-5)²+ 25
= 20²+25
= 400+25
= 425
Vậy biểu thức có giá trị là 425 khi x=25 và y=5.