Bài 2: 

c: \(=x^2\left(x-3\right)-4\left(x-3\right)\)

\(=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)

a) \(\Rightarrow\left(x^2+2\times5x+25\right)+\left(y^2+2y+1\right)\)

\(\Rightarrow\left(x+5\right)^2+\left(y+1\right)^2\)

1)a)x²+10x+26+y²+2y

=(x²+10x+25)+(y²+2y+1)

=(x+5)²+(y+1)²

b)x²-2xy+2y²+2y+1

=(x²-2xy+y²)+(y²+2y+1)

=(x-y)²+(y+1)²

c)z²-6z+13+t²+4t

=(z²-6z+9)+(t²+4t+4)

=(z-3)²+(t+2)²

d)4x²+2z²-4xz-2z+1

=(4x²-4xz+z²)+(z²-2z+1)

=(2x-z)²+(z-1)²

2)a)(x-3)²-4=0

<=>(x-3-2)(x-3+2)=0

<=>(x-5)(x-1)=0

<=>x-5=0 hoặc x-1=0

<=>x=5 hoặc x=1

b)x²-2x=24

<=>x²-2x-24=0

<=>(x²-6x)+(4x-24)=0

<=>x(x-6)+4(x-6)=0

<=>(x-6)(x+4)=0

<=>x-6=0 hoặc x+4=0

<=>x=6 hoặc x=-4

a) x^2 + 10x + 26 + y^2 + 2y

=x²+10x+25+y²+2y+1

=x²+2.x.5+5²+y²+2.y.1+1²

=(x+5)²+(y+1)²

b)x^2 - 2xy + 2y^2 + 2y +1

=x²-2xy+y²+y²+2y+1

=(x-y)²+(y+1)²

c)z^2 - 6z + 13 + t^2 + 4t

=z²-6z+9+t²+4z+4

=z²-2.z.3+3²+t²+2.t.2+2²

=(z-3)²+(t+2)²

d)4x^2 + 2z^2 - 4xz - 2z + 1

=4x²-4xz+z²+z²-2z+1

=(2x)²-2.2x.z+z²+z²-2z.1+1²

=(2x-z)²+(z-1)²

a) x^2 + 10x + 26 + y^2 + 2y

=x²+10x+25+y²+2y+1

=x²+2.x.5+5²+y²+2.y.1+1²

=(x+5)²+(y+1)²

b)x^2 - 2xy + 2y^2 + 2y +1

=x²-2xy+y²+y²+2y+1

=(x-y)²+(y+1)²

c)z^2 - 6z + 13 + t^2 + 4t

=z²-6z+9+t²+4z+4

=z²-2.z.3+3²+t²+2.t.2+2²

=(z-3)²+(t+2)²

d)4x^2 + 2z^2 - 4xz - 2z + 1

=4x²-4xz+z²+z²-2z+1

=(2x)²-2.2x.z+z²+z²-2z.1+1²

=(2x-z)²+(z-1)²

a) (x-3)^2 - 4 =0

<=>(x-3)²-2²=0

<=>(x-3-2)(x-3+2)=0

<=>(x-5)(x-1)=0

<=>x-5=0 hoặc x-1=0

<=>x=5 hoặc x=1

b) x^2 - 2x = 24

<=>x²-2x-24=0

<=>x²-2x+1-25=0

<=>(x-1)²-25=0

<=>(x-1)²-5²=0

<=>(x-1-5)(x-1+5)=0

<=>(x-6)(x+4)=0

<=>x-6=0 hoặc x+4=0

<=>x=6 hoặc x=-4

A) \(\left(x-3\right)^2-\left(x+2\right)^2\)

\(=\left(x-3-x-2\right)\left(x-3+x+2\right)\)

\(=-5.\left(2x-1\right)\)

B) \(\left(4x^2+2xy+y^2\right)\left(2x-y\right)-\left(2x+y\right)\left(4x^2-2xy+y^2\right)\)

\(=\left(2x\right)^3-y^3-\left[\left(2x\right)^3+y^3\right]\)

\(=8x^3-y^3-8x^3-y^3\)

\(=-2y^3\)

C) \(x^2+6x+8\)

\(=x^2+6x+9-1\)

\(=\left(x+3\right)^2-1\)

\(=\left(x+3-1\right)\left(x+3+1\right)\)

\(=\left(x+2\right)\left(x+4\right)\)

bài 3 A) \(x^2-16=0\)

\(\left(x-4\right)\left(x+4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-4=0\\x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)

vậy \(\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)

B) \(x^4-2x^3+10x^2-20x=0\)

\(x^3\left(x-2\right)+10x\left(x-2\right)=0\)

\(\left(x^3+10x\right)\left(x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x^3+10x=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x\left(x^2+10\right)=0\\x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

vậy \(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

x=0

x=2

\(x^2+y^2+26+10x+2y=0\)

\(\Leftrightarrow\left(x^2+10x+25\right)+\left(y^2+2y+1\right)=0\)

\(\Leftrightarrow\left(x+5\right)^2+\left(y+1\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}\left(x+5\right)^2=0\\\left(y+1\right)^2=0\end{cases}}\)( do \(\left(x+5\right)^2\ge0;\left(y+1\right)^2\ge0\))

\(\Leftrightarrow\hept{\begin{cases}x+5=0\\y+1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-5\\y=-1\end{cases}}\)

ban kia lam dung roi do

k tui nha

thanks

Bài 1:

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

b) 2x + 2y - x( x + y) = ( 2x + 2y) - x( x + y)

= 2( x + y ) - x( x + y ) = ( x + y )(2 - x )

c) 5x² - 5xy - 10x + 10y = ( 5x² - 5xy ) - ( 10x - 10y)

= 5x( x - y ) - 10( x - y ) = ( x - y )(5x - 10 )

= 5( x - y )( x - 2 )

d) 4x² + 6xy - 3x - 6y = Mình ko làm được!!! bạn chép có sai đề không

Bài 2:

x ( 2x - 7) - 4x + 14 = 0

⇒ 2x² - 7x - 4x + 14 = 0 ⇒ ( 2x² - 4x ) - ( 7x - 14 ) = 0

⇒ 2x( x - 2 ) - 7(x - 2) = 0

⇒ (x - 2)(2x - 7) = 0

⇒ \(\left[{}\begin{matrix}x-2=0\\2x-7=0\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=2\\x=\dfrac{7}{2}\end{matrix}\right.\)

Vậy x = 2; x = \(\dfrac{7}{2}\)

a) x²+10x+26+y²+2y

=x²+10x+25+y²+2y+1

=(x+5)²+(y+1)²

b) z²-6z+5-t²-4t

=z²-6z+9-t²-4t-4

=(z-3)²-(t²+4t+4)

=(z-3)²-(t+2)²

c)x²-2xy+2y²+2y+1

=x²-2xy+y²+y²+2y+1

=(x-y)²+(y+1)²

d) 4x²-12x-y²+2y+8

=4x²-12x+9-y²+2y-1

=(2x-3)²-(y²-2y+1)

=(2x-3)²-(y-1)²

bạn ơi , bạn lấy bài này ở đâu vậy bạn