Question

Viết dưới dạng bình phương của một tổng hoặc một hiệu:

a.y^2-2y+1

b.x^2+x+1/4

c.x^2-6x+9

d.4y^2+4y+1

e.9a^2+6ab+b^2

g.4x^2+12xy+9y^2

f.y^2-y+1/4

Answer 1

a, \(y^2-2y+1=y^2-y-y+1\\ =y.\left(y-1\right)-\left(y-1\right)=\left(y-1\right)^2\)

b, \(x^2+x+\dfrac{1}{4}=x^2+\dfrac{1}{2}x+\dfrac{1}{2}x+\dfrac{1}{4}\\ =x.\left(x+\dfrac{1}{2}\right)+\dfrac{1}{2}.\left(x+\dfrac{1}{2}\right)=\left(x+\dfrac{1}{2}\right)^2\)

c, \(x^2-6x+9=x^2-3x-3x+9\\ =x.\left(x-3\right)-3.\left(x-3\right)=\left(x-3\right)^2\)

d, \(4y^2+4y+1=4y^2+2y+2y+1\\ =2y.\left(2y+1\right)+\left(2y+1\right)=\left(2y+1\right)^2\)

Mấy câu còn lại làm tương tự!

Answer 2

e, \(9a^2+6ab+b^2=9a^2+3ab+3ab+b^2\\ =3a.\left(3a+b\right)+b.\left(3a+b\right)=\left(3a+b\right)^2\)

g, \(4x^2+12xy+9y^2=4x^2+6xy+6xy+9y^2\\ =2x.\left(2x+3y\right)+3y.\left(2x+3y\right)=\left(2x+3y\right)^2\)

f, \(y^2-y+\dfrac{1}{4}=y^2-\dfrac{1}{2}y-\dfrac{1}{2}y+\dfrac{1}{4}\\ =y.\left(y-\dfrac{1}{2}\right)-\dfrac{1}{2}.\left(y-\dfrac{1}{2}\right)=\left(y-\dfrac{1}{2}\right)^2\)

Chúc bạn học tốt!!!

Answer 3

a/ \(y^2-2y+1=\left(y-1\right)^2\)

b/ \(x^2+x+\dfrac{1}{4}=x^2+2\cdot x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2=\left(x+\dfrac{1}{2}\right)^2\)

c/ \(x^2-6x+9=x^2-2\cdot x\cdot3+3^2=\left(x-3\right)^2\)

d/ \(4y^2+4y+1=\left(2y\right)^2+2\cdot2y\cdot1+1^2=\left(2y+1\right)^2\)

e/ \(9a^2+6ab+b^2=\left(3a\right)^2+2\cdot3a\cdot b+b^2=\left(3a+b\right)^2\)

g/ \(4x^2+12xy+9y^2=\left(2x\right)^2+2\cdot2x\cdot3y+\left(3y\right)^2=\left(2x+3y\right)^2\)

f/ \(y^2-y+\dfrac{1}{4}=y^2-2\cdot y\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2=\left(y^2-\dfrac{1}{2}\right)^2\)