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

- x² + 10x - 25

= - ( x² - 10x + 25 )

= - ( x - 5 )²

x² -10x + 25 - y²

= (x-5)² - y²

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

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

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

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

\(x^4+2x^3+10x-25\)

\(=x^4+5x^2+2x^3+10x-5x^2-25\)

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

10x-25-x^2

=-x^2+10x-25

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

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

= -(x-5)^2

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

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

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

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

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

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

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

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

x^2 - 10x + 25

= x^2 - 5x - 5x + 5^2

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

= (x - 5)(x - 5)

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

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

\((x+5)^2+4(x+5)(x-5)+4(x^2-10x+25)=0\\\Rightarrow(x+5)^2+4(x+5)(x-5)+4(x^2-2\cdot x\cdot5+5^2)=0\\\Rightarrow(x+5)^2+2\cdot(x+5)\cdot2(x-5)+4(x-5)^2=0\\\Rightarrow(x+5)^2+2\cdot(x+5)\cdot2(x-5)+[2(x-5)]^2=0\\\Rightarrow[(x+5)+2(x-5)]^2=0\\\Rightarrow(x+5+2x-10)^2=0\\\Rightarrow(3x-5)^2=0\\\Rightarrow3x-5=0\\\Rightarrow3x=5\\\Rightarrow x=\frac53\\\text{#}Toru\)

Sao đề là phân tích mà lại "= 0" vậy bạn?