\(25a^2-20ab+4b^2\)

= \(\left(5a\right)^2\) \(-2.5a.2b\) \(+\left(2b\right)^2\)

= \(\left(5a-2b\right)^2\)

\(=\left(5a\right)^2-2\cdot5\cdot2\cdot a\cdot b+\left(2b\right)^2=\left(5a-2b\right)^2\)

\(25a^2+4b^2-20ab\)

\(=25a^2-20ab+4b^2\)

\(=\left(5a\right)^2-2.5a.2b+\left(2b\right)^2\)

\(=\left(5a-2b\right)^2\)

A=9x^2−6x+1

=(3x)^2−2.3x.1+1^2

=(3x−1)^2

B=(2x+3y)^2+(2x+3y)+1(2x+3y)2+(2x+3y)+1

=[(2x+3y)^2+2.(2x+3y).1/2+(1/2)^2]+3/4

=(2x+3y+1/2)^2+3/4

=(2x+3y+1)(2x+3y)+1

\(25a^2+4b^2-20ab\)

\(=25a^2-20ab+4b^2\)

\(=\left(5a\right)^2-2.5a.2b+\left(2b\right)^2\)

\(=\left(5a-2b\right)^2\)

\(25a^2+4b^2-20ab=\left(5a\right)^2-2.5a.2b+\left(2b\right)^2=\left(5a-2b\right)^2.\)

25a^2+4b^2-20ab=(5a)^2-2.5a.2b+(2b)^2

=(5a-2b)^2

25a^2-20ab+4b^2 =(5a-2b)^2

25a^2+4b^2-20ab=(2b)^2-2*2b*5a+(5a)^2=(2b-5a)^2

\(\left(5a-2b\right)^2\)

(5a -2b)²

a)\(x^2+2x+1=x^2+2x1+1^2=\left(x+1\right)^2\)

b)\(9x^2+y^2+6xy=3^2x^2+y^2+2.3x.y=\left(3x\right)^2+2.3x.y+y^2=\left(3x+y\right)^2\)

c)\(25a^2+4b^2-20ab=5^2a^2+2^2b^2-2.5a.2b=\left(5a\right)^2-2.5a.2b+\left(2b\right)^2=\left(5a-2b\right)^2\)

d)\(x^2-x+\frac{1}{4}=x^2-2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2=\left(x-\frac{1}{2}\right)^2\)

a) \(x^2+2x+1=\left(x+1\right)^2\)

b) \(9x^2+y^2+6xy=\left(3x+y\right)^2\)

c) \(25a^2+4b^2-20ab=\left(5a-2b\right)^2\)

Câu d thì biểu thức là \(\frac{x^2-1}{2x+\frac{1}{10}}\) hay là \(\frac{x^2-1}{\frac{2x+1}{10}}\) z bạn???