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???

a)(x+1)^2

b)3x+y)^2

c)(5a-2b)^2

d)(x-1/2)^2

e)(3x-1)^2

a) x² + 2x +1 

= (x + 1)²

b) 9x² + y² + 6xy

= (3x + y)²

c) 25a² + 4b² - 20ab

= (5a - 2b)²

d) x² - x + 1/4

= (x - 1/2)²

a) \(x^2+2x+1=\left(x+1\right)^2\)

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

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

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

dùng hằng đẳng thức

a) x²+2x+1=(x+1)²

b)(3x+y)²

c)(5a-2b)²

d)(x-1/2)²

a/ (y+1)²

b/ (3x-y)²

c/ (5a+2b)²

d/ (x-1/2)²

a. (3x+y)²

b. (5a+2b)²

a) =(3x+y)²

câu b) sao kì quá mik k hiểu bạn xem lại đề đi có viết thiếu dấu k

mik giải rõ ràng cho

a)9x^2+y^2+6xy

=9x^2+6xy+y^2

=(3x)^2+2.3x.1+y^2

=(3x+y)2

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

=25a^2-20ab+4b^2

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

=(5a-2b)^2

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

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

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