-x²+10x-25=-(x²-10x+25)=-(x²-2.5+25)=-(x-5)²

x² -10x + 25 - y²

= (x-5)² - y²

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

\(x^2-10x+25-y^2\)

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

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

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- x² + 10x - 25

= - ( x² - 10x + 25 )

= - ( x - 5 )²

a.\(x^2y-xz+z-y=\)\(\left(x^2y-y\right)-\left(xz-z\right)=\)\(y\left(x^2-1\right)-z\left(x-1\right)\)

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

b.\(x^4-x^3+x^2-1=x^3\left(x-1\right)+\left(x+1\right)\left(x-1\right)\)=\(\left(x-1\right)\left(x^3+x+1\right)\)

c.\(x^4-x^2+10x-25=x^4-\left(x^2-10x+25\right)\)=\(\left(x^2\right)^2-\left(x-5\right)^2=\left(x^2+x-5\right)\left(x^2-x+5\right)\)

bạn hoàng tính như nào mà ra đc dòng cuối vậy?

\(x^2+4y^2+4xy-10x-20y+25\)

\(=\left(x^2-10x+25\right)+\left(4xy-20y\right)+4y^2\)

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

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

10x-25-x^2

=-x^2+10x-25

= - (x^2-10x+25)

=-(x^2-2x5+25)

= -(x-5)^2

a)Ta có:  \(x\left(x-y\right)+5x-5y=x\left(x-y\right)+\left(5x-5y\right)\)

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

b) \(x^2-y^2+10x+25=\left(x^2+10x+25\right)-y^2=\left(x+5\right)^2-y^2\)

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

\(x^2-10x+25\)

\(=x^2-2\cdot x\cdot5+5^2\)

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

x^2 - 10x + 25

= x^2 - 5x - 5x + 5^2

= x(x - 5) - 5(x - 5)

= (x - 5)(x - 5)

Cách khác có thể áp dụng cho nhiều bài toán :

\(x^2-10x+25\)

\(=x^2-5x-5x+25\)

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

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