a) \(x^2-2xy+2y^2+2y+1\)

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

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

b) \(4x^2-12x-y^2+2y+8\) (đã sửa đề)

\(=4\left(x^2-3x+\frac{9}{4}\right)-\left(y^2-2y+1\right)\)

\(=\left[2\left(x-\frac{3}{2}\right)\right]^2-\left(y-1\right)^2\)

c) \(z^2-6z+5-t^2-4t\)

\(=\left(z^2-6z+9\right)-\left(t^2+4t+4\right)\)

\(=\left(z-3\right)^2-\left(t+2\right)^2\)

a) Xét tam giác BDC có: MB= MC (gt), ED= EC (gt)

=> ME là đường trung bình tam giác BDC (đ/n)

=> ME // BD (t/c)

b) Vì ME// BD (cmt) => ME // IB // ID ( I thuộc BD)

 - Xét tam giác AME có: ME // ID (cmt), DA= DE (gt)

=> IA = IM (t/c)

Hay I là trung điểm của AM (đpcm)

c) +) Vì ME là đường TB tam giác BDC (cmt) => \(ME=\frac{1}{2}BD\)(t/c)     (1)

    +) Xét tam giác AME có IA= IM (cmt), DA= DE (gt)

=> ID là đường TB tam giác AME (đ/n)

=> \(ID=\frac{1}{2}ME\)(t/c)           (2)

   Từ (1) và (2) có:      \(ID=\frac{1}{4}BD\)

                   =>       4. ID  =  BD       

                    =>      4.ID   =    IB + ID

                   =>      IB       =     3ID  (đpcm)

d) Nối FC, FI.  Kẻ MN // FC.(N thuộc AB)

    +) Xét tam giác BFC có MN // FC (cvẽ), MB = MC (gt)

  => NB = NF (t/c)

        Xét tam giác BFC có NB = NF (cmt), MB = MC (gt)

  => MN là đường TB tam giác BFC (đ/n)

  => MN // FC (t/c)              (3)

    +) Vì AF = 1/3.AB (gt) và AB= FA+ FB

  =>  AF = 1/2.FB mà NB + NF = FB, NB = NF (cmt)

  => AF = NF = NB

     +) Xét tam giác AMN có IA = IM (cmt), FA =FN (cmt)

  =>  FI là đường TB tam giác AMN (đ/n)

  => FI // MN (t/c)              (4)

         Từ (3) và (4) có FI và FC trùng nhau (theo tiên đề Ơ-clit)

                             => 3 điểm F, I, C thẳng hàng (đpcm)

**: Bn tự vẽ hình nhaaaaaaa......

1.

a) x⁴ + x³ + x + 1 = x³( x + 1 ) + ( x + 1 ) = ( x + 1 )( x³ + 1 ) = ( x + 1 )( x + 1 )( x² - x + 1 ) = ( x + 1 )²( x² - x + 1 )

b) x²y + xy² - x - y = xy( x + y ) - ( x + y ) = ( x + y )( xy - 1 )

c) x² - 2xy + y² - xz + yz = ( x² - 2xy + y² ) - ( xz - yz ) = ( x - y )² - z( x - y ) = ( x - y )( x - y - z )

d) ax - ab + b - x = ( ax - x ) - ( ab - b ) = x( a - 1 ) - b( a - 1 ) = ( a - 1 )( x - b )

2.

( 2x - 1 )² - 25 = 0

<=> ( 2x - 1 )² - 5² = 0

<=> ( 2x - 1 - 5 )( 2x - 1 + 5 ) = 0

<=> ( 2x - 6 )( 2x + 4 ) = 0

<=> \(\orbr{\begin{cases}2x-6=0\\2x+4=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)

3x( x - 1 ) + x - 1 = 0

<=> 3x( x - 1 ) + ( x - 1 ) = 0

<=> ( x - 1 )( 3x + 1 ) = 0

<=> \(\orbr{\begin{cases}x-1=0\\3x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{1}{3}\end{cases}}\)

B1:

a) \(x^4+x^3+x+1\)

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

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

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

b) \(x^2y+xy^2-x-y\)

\(=xy\left(x+y\right)-\left(x+y\right)\)

\(=\left(xy-1\right)\left(x+y\right)\)

c) \(x^2-2xy+y^2-xz+yz\)

\(=\left(x-y\right)^2-z\left(x-y\right)\)

\(=\left(x-y\right)\left(x-y-z\right)\)

d) \(ax-ab+b-x\)

\(=a\left(x-b\right)-\left(x-b\right)\)

\(=\left(a-1\right)\left(x-b\right)\)

a) 36 - 4a² + 20ab - 25b² = 36 - ( 4a² - 20ab + 25b² ) = 6² - ( 2a - 5b )² = ( 6 - 2a + 5b )( 6 + 2a - 5b )

b) ( xy + 4 )² - 4( x + y )² = ( xy + 4 )² - 2²( x + y )² = ( xy + 4 )² - [ 2( x + y ) ]² 

                                        = ( xy + 4 )² - ( 2x + 2y )² = ( xy + 4 - 2x - 2y )( xy + 4 + 2x + 2y )

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

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

c) x² + y² - x²y² + xy - x - y

= ( x² - x²y² ) + ( y² - y ) + ( xy - x )

= x²( 1 - y² ) + y( y - 1 ) + x( y - 1 )

= x²( 1 - y )( 1 + y ) - y( 1 - y ) - x( 1 - y )

= ( 1 - y )[ x²( 1 + y ) - y - x ) ]

= ( 1 - y )( x² + x²y - y - x )

= ( 1 - y )[ ( x² - x ) + ( x²y - y ) ]

= ( 1 - y )[ x( x - 1 ) + y( x² - 1 ) ]

= ( 1 - y )[ x( x - 1 ) + y( x - 1 )( x + 1 ) ]

= ( 1 - y )( x - 1 )[ x + y( x + 1 ) ]

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

d) 3x + 3y - x² - 2xy - y²

= 3( x + y ) - ( x² + 2xy + y² )

= 3( x + y ) - ( x + y )²

= ( x + y )( 3 - x - y )

e) ( 2xy + 1 )² - ( 2x + y )²

= ( 2xy + 1 - 2x - y )( 2xy + 1 + 2x + y )

= [ ( 2xy - 2x ) - ( y - 1 ) ][ ( 2xy + 2x ) + ( y + 1 ) ]

= [ 2x( y - 1 ) - ( y - 1 ) ][ 2x( y + 1 ) + ( y + 1 ) ]

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

a) \(36-4a^2+20ab-25b^2\)

\(=36-\left(4a^2-20ab+25b^2\right)\)

\(=36-\left(2a-5b\right)^2\)

\(=\left(6-2a+5b\right)\left(6+2a-5b\right)\)

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

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

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

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

c) \(x^2+y^2-x^2y^2+xy-x-y\)

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

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

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

\(=\left(1-y\right)\left[\left(x^2y-y\right)+\left(x^2-x\right)\right]\)

\(=\left(1-y\right)\left(x-1\right)\left(xy+y+x\right)\)

Ta co : \(x^{2x+1}+y^{2x+1}=\left(x+y\right)^{2x+1}⋮x+y\forall x;y\)( dpcm ) 

Vì \(2n+1\)luôn là số lẻ \(\forall x\inℤ\)

\(\Rightarrow\left(x^{2n+1}+y^{2n+1}\right)⋮\left(x+y\right)\)( đpcm )

x² + y² + z² = xy + yz + xz

<=> 2( x² + y² + z² ) = 2( xy + yz + xz )

<=> 2x² + 2y² + 2z² = 2xy + 2yz + 2xz

<=> 2x² + 2y² + 2z² - 2xy - 2yz - 2xz = 0

<=> ( x² - 2xy + y² ) + ( y² - 2yz + z² ) + ( x² - 2xz + z² ) = 0

<=> ( x - y )² + ( y - z )² + ( x - z )² = 0

Ta có : \(\hept{\begin{cases}\left(x-y\right)^2\\\left(y-z\right)^2\\\left(x-z\right)^2\end{cases}}\ge0\forall x,y,z\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2\ge0\forall x,y,z\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-y=0\\y-z=0\\x-z=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=y\\y=z\\x=z\end{cases}}\Rightarrow x=y=z\left(đpcm\right)\)

Ta có: \(x^2+y^2+z^2=xy+yz+zx\)

\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(x-y\right)^2=0\\\left(y-z\right)^2=0\\\left(z-x\right)^2=0\end{cases}}\Rightarrow x=y=z\)

1a) ( x - 2 )( x² + 2x + 4 ) - x( x - 1 )( x + 1 ) + 3

= x³ - 8 - x( x² - 1 ) + 3

= x³ - 8 - x³ + x + 3

= x - 5

b) Với x = -1/2 => Giá trị của biểu thức = -1/2 - 5 = -11/2

2a) 3x( 5x² - 2xy² + y ) = 15x³ - 6x²y² + 3xy

b) ( x + y )( x² - xy + y² ) = x³ + y³

3) 16x² - ( 4x - 5 )² = 15

<=> 16x² - ( 16x² - 40x + 25 ) = 15

<=> 16x² - 16x² + 40x - 25 = 15

<=> 40x - 25 = 15

<=> 40x = 40

<=> x = 1

=40x ở đâu vậy bạn

ơ cái này đơn giản mà (:

7( x + y ) - 2y( x + y ) = ( x + y )( 7 - 2y )

1/ \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)\left(x+1\right)\)

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

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

\(=\left[\left(x+1\right)-\left(x-1\right)\right]^3=2^3=8\)

2/ \(x\left(x-1\right)\left(x+1\right)-\left(x+1\right)\left(x^2-x+1\right)\)

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

a) x²( 1 - x² ) - 4 - 4x²

= x² - x⁴ - 4 - 4x²

= -x⁴ - 3x² - 4

= x² - ( x⁴ + 4x² + 4 )

= x² - ( x² + 2 )²

= ( x - x² - 2 )( x² + x + 2 )

b) 5x² - 45y² - 30y - 5

= 5( x² - 9y² - 6y - 1 )

= 5[ x² - ( 9y² + 6y + 1 ) ]

= 5[ x² - ( 3y + 1 )² ]

= 5( x - 3y - 1 )( x + 3y + 1 )

c) x⁶ - x⁴ - 9x³ + 9x² = x⁴( x² - 1 ) - 9x( x² - 1 ) = ( x² - 1 )( x⁴ - 9x ) = x( x - 1 )( x + 1 )( x³ - 9 )

d) 3x - 3y + x² - y² = 3( x - y ) + ( x - y )( x + y ) = ( x - y )( 3 + x + y )

e) x² - 2x - 4y² - 4y = ( x² - 2x + 1 ) - ( 4y² + 4y + 1 ) = ( x - 1 )² - ( 2y + 1 )² = ( x - 1 - 2y - 1 )( x - 1 + 2y + 1 ) = ( x - 2y - 2 )( x + 2y )