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

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

\(A,=\left(x^2-6x+9\right)-y^2=\left(x-3\right)^2-y^2=\left(x-y-3\right)\left(x+y-3\right)\\ B,=25-\left(4x^2+4xy+y^2\right)=25-\left(2x+y\right)^2=\left(5-2x-y\right)\left(5+2x+y\right)\)

A) = (x^2-6x+9)-y^2

= (x-3)^2-y^2

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

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

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

Mình làm và sửa đề đúng luôn nhé !

1) \(36x^2-a^2+10a-25\)

\(=\left(6x\right)^2-\left(a^2-10a+25\right)\)

\(=\left(6x\right)^2-\left(a-5\right)^2\)

\(=\left(6x-a+5\right)\left(6x+a-5\right)\)

2) \(4x^2-4xy+y^2-25a^2+10a-1\)

\(=\left(2x-y\right)^2-\left(5a-1\right)^2\)

\(=\left(2x-y-5a+1\right)\left(2x-y+5a-1\right)\)

3) \(m^2-6m+9-x^2+4xy-4y^2\)

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

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

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

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

Giải:

a) \(25-4x^2-4xy-y^2\)

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

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

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

Vậy ...

b) \(x^2+2xy+y^2-xz-yz\)

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

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

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

Vậy ...

c) \(x^2-4xy+4y^2-z^2+4zt-4t^2\)

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

\(=\left(x-2y\right)^2-\left(z-2t\right)^2\)

\(=\left(x-2y+z-2t\right)\left(x-2y-z+2t\right)\)

Vậy ...

phân tích các đa thức sau thành nhân tử:

1, \(25-4x^2-4xy-y^2\)

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

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

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

2,\(x^2+2xy+y^2-xz-yz\)

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

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

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

3,\(x^2-4xy+4y^2-z^2+4zt-4t^2\)

\(=\left(x^2-4xy+4y^2\right)-\left(z^2-4zt+4t^2\right)^{ }\)

\(=\left(x-2y\right)^2-\left(z-2t\right)^2\)

\(=\left(x-2y-z+2t\right)\left(x-2y+z-2t\right)\)

\(2,=\left(x-y\right)^2-2\left(x-y\right)=\left(x-y\right)\left(x-y-2\right)\\ 3,=\left(3x-5\right)\left(x+1\right)\\ 4,sai.đề\\ 5,=\left(x-1\right)^2-y^2=\left(x-y-1\right)\left(x+y-1\right)\\ 6,=\left(x+3\right)\left(x+5\right)\)

\(a,x^2+4x-y^2+4\)

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

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

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

\(b,25-4x^2-4xy-y^2\)

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

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

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

\(c,x^3-x+y^3-y\)

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

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