x^2+2x+1−y^2

(a+b)^2=a^2+2ab+b^2

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

(x+1)^2−y^2

a^2−b^2=(a−b)(a+b)

(x+1)^2−y^2=(x+1−y)(x+1+y)

=(x−y+1)(x+y+1)

\(x^2\) + 2\(x\) + 1 - y\(^2\)

= (\(x^2\) + 2.\(x\).1 + 1\(^2\)) - y\(^2\)

= (\(x+1\))\(^2\) - y\(^2\)

= (\(x+1-y\))(\(x+1+y\))

Câu 1 : x²-y²+2yz-z²=-(y²-2yz+z²-x²)                    Câu 2: x²-2xy+y²-xz+yz=(x²-2xy+y²)-xz+yz

                                 =-(y-z)² -x²                                                                   =(x-y)²-z(x-y)

                                        =-(y-z-x)(y-z+x)                                                            =(x-y)(x-y-z)

45opkik

a: x^2-2x+y^2-8y+17=0

=>x^2-2x+1+y^2-8y+16=0

=>(x-1)^2+(y-4)^2=0

=>x=1 và y=4

b: Sửa đề: 4x^2-4xy+y^2+y^2+4y+4=0

=>(2x-y)^2+(y+2)^2=0

=>y=-2 và x=-1

x² - y² - x-y

= (x-y)(x+y) - (x+y)

=(x+y)(x-y-1) 

x²-y²-2xy+y²

=x²-2xy

=x(x-2y)

x² - y² - 2xy + y²

= (x² - 2xy +y²) - y²

= (x-y)² - y²

= (x-y-y)(x-y+y)

= x(x-2y)

a)Ta có: \(x^2-6x+9=\left(x-3\right)^2\)

b) Ta có: \(27+27x+9x^2+9x^3=27\left(1+x\right)+9x^2\left(1+x\right)=\left(1+x\right)\cdot\left(27+9x^2\right)=9\left(1+x\right)\left(3+x^2\right)\)

c) Ta có: \(x^2-25-2xy+y^2=\left(x-y\right)^2-5^2=\left(x-y-5\right)\left(x-y+5\right)\)

d) Ta có: \(7y^4-14y^3+7y^2=7y^2\left(y^2-2y+1\right)=7y^2\left(y-1\right)^2\)

e) Ta có: \(1-4x^2=1^2-\left(2x\right)^2=\left(1-2x\right)\left(1+2x\right)\)