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 = x² + 2.x.1+ 12 = ( x + 1)²

b) 9x² + y² + 6xy = (3x)² + 2.3.x.y + y² = (3x + y)²

c) 25a² + 4b2 – 20ab = (5a)² – 2.5.a.2b. + (2b)² = (5a – 2b)²

Hoặc 25a² + 4b2 – 20ab = (2b)² – 2.2b.5a. + (5a)² = (2b – 5a)²

d) x² – x + \(\dfrac{1}{4}\) = x² – 2.x. \(\dfrac{1}{2}\) + ( \(\dfrac{1}{2}\))2² = ( x - \(\dfrac{1}{2}\) )²

Hoặc x² – x + \(\dfrac{1}{4}\) = \(\dfrac{1}{4}\) - x + x² = (\(\dfrac{1}{2}\))² – 2. \(\dfrac{1}{2}\).x + x² = (\(\dfrac{1}{2}\) - x)²

a) x² + 2x + 1 = x²+ 2 . x . 1 + 1²

= (x + 1)²

b) 9x² + y²+ 6xy = (3x)² + 2 . 3 . x . y + y² = (3x + y)²

c) 25a² + 4b²– 20ab = (5a)² – 2 . 5a . 2b + (2b)² = (5a – 2b)²

Hoặc 25a² + 4b² – 20ab = (2b)² – 2 . 2b . 5a + (5a)² = (2b – 5a)²

d) x² – x + = x² – 2 . x . + =

Hoặc x² – x + = - x + x² = - 2 . . x + x² =

a) x² + 2x + 1 = x²+ 2 . x . 1 + 1²

= (x + 1)²

b) 9x² + y²+ 6xy = (3x)² + 2 . 3 . x . y + y² = (3x + y)²

c) 25a² + 4b²– 20ab = (5a)² – 2 . 5a . 2b + (2b)² = (5a – 2b)²

Hoặc 25a² + 4b² – 20ab = (2b)² – 2 . 2b . 5a + (5a)² = (2b – 5a)²

d) x² – x + = x² – 2 . x . + =

Hoặc x² – x + = - x + x² = - 2 . . x + x² =

Chúc bạn học tốt

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. (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

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. $x^2+4x+4$

$=x^2+2\cdot x\cdot2+2^2$

$=(x+2)^2$

b. $x^2-6xy+9y^2$

$=x^2-2\cdot x\cdot3y+(3y)^2$

$=(x-3y)^2$

c. $4x^2+12x+9$

$=(2x)^2+2\cdot2x\cdot3+3^2$

$=(2x+3)^2$

d. $x^2-x+\dfrac14$

$=x^2-2\cdot x\cdot \dfrac12+\Bigg(\dfrac12\Bigg)^2$

$=\Bigg(x-\dfrac12\Bigg)^2$

`x^2 +4x+4`

`=x^2+2*x*2+2^2`

`=(x+2)^2`

__

`x^2-6xy+9y^2`

`=x^2 - 2*x*3y+(3y)^2`

`=(x-3y)^2`

__

`4x^2 +12x+9`

`=(2x)^2 +2*2x*3+3^2`

`=(2x+3)^2`

__

`x^2-x+1/4`

`=x^2 - 2*x*1/2 +(1/2)^2`

`=(x+1/2)^2`

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

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

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