Sửa lại đề : AM = NC

                    

Ta có : AB // CD ( tứ giác ABCD là hình bình hành )

\(\Rightarrow BM//DN\left(1\right)\)

Ta có : AB = AM + MB

             DC = DN + NC

mà AB = DC ( tứ giác ABCD là hình bình hành ) ; AM = NC (gt)

\(\Rightarrow MB=DN\left(2\right)\)

\(\Rightarrow\)Tứ giác BMDN là hình bình hành (đpcm)

a ) xy + 1 - x - y

= x ( y - 1 ) + 1 - y

= x ( y - 1 ) - ( y - 1 )

= ( x - 1 ) ( y - 1 )

b ) x² + ab + ac + bx

= ( b + x )x + a( b + c )

= ( b + x )1x + 1a( b + c )

= ( b + x ) ( x + a ) ( b + c )

c ) ax + bx - cx + a + b - c

= ( a + b - c )x + a + b - c

= ( a + b - c )x + ( a + b - c )1

= ( a + b - c ) ( x + 1 )

1. Ta có: \(ab+bc+ca=3abc\)

\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3\)

Đặt \(\hept{\begin{cases}\frac{1}{a}=m\\\frac{1}{b}=n\\\frac{1}{c}=p\end{cases}}\) khi đó \(\hept{\begin{cases}m+n+p=3\\M=2\left(m^2+n^2+p^2\right)+mnp\end{cases}}\)

Áp dụng Cauchy ta được:

\(\left(m+n-p\right)\left(m-n+p\right)\le\left(\frac{m+n-p+m-n+p}{2}\right)^2=m^2\)

\(\left(n+p-m\right)\left(n+m-p\right)\le n^2\)

\(\left(p-n+m\right)\left(p-m+n\right)\le p^2\)

\(\Rightarrow\left(m+n-p\right)\left(n+p-m\right)\left(p+m-n\right)\le mnp\)

\(\Leftrightarrow m^3+n^3+p^3+3mnp\ge m^2n+mn^2+n^2p+np^2+p^2m+pm^2\)

\(\Leftrightarrow\left(m+n+p\right)\left(m^2+n^2+p^2-mn-np-pm\right)+6mnp\ge mn\left(m-n\right)+np\left(n-p\right)+pm\left(p-m\right)\)

\(=mn\left(3-p\right)+np\left(3-m\right)+pm\left(3-n\right)\)

\(\Leftrightarrow3\left(m^2+n^2+p^2\right)-3\left(mn+np+pm\right)+6mnp\ge3\left(mn+np+pm\right)-3mnp\)

\(\Leftrightarrow3\left(m^2+n^2+p^2\right)+9mnp\ge6\left(mn+np+pm\right)\)

\(\Leftrightarrow xyz\ge\frac{2}{3}\left(mn+np+pm\right)-\frac{1}{3}\left(m^2+n^2+p^2\right)\)

\(\Rightarrow M\ge2\left(m^2+n^2+p^2\right)+\frac{2}{3}\left(mn+np+pm\right)-\frac{1}{3}\left(m^2+n^2+p^2\right)\)

\(=\frac{5}{3}\left(m^2+n^2+p^2\right)+\frac{2}{3}\left(mn+np+pm\right)\)

\(=\frac{4}{3}\left(m^2+n^2+p^2\right)+\frac{1}{3}\left(m^2+n^2+p^2+2mn+2np+2pm\right)\)

\(=\frac{4}{3}\left(m^2+n^2+p^2\right)+\frac{1}{3}\left(m+n+p\right)^2\)

\(\ge\frac{4}{3}\cdot3+\frac{1}{3}\cdot3^2=4+3=7\)

Dấu "=" xảy ra khi: \(m=n=p=1\Leftrightarrow a=b=c=1\)

Sửa đề : \(9\left(x+y-1\right)^2-4\left(2x+3y+1\right)^2\)

\(=\left(9x+9y-9\right)^2-\left(8x+12y+4\right)^2\)

\(=\left(9x+9y-9-8x-12y-4\right)\left(9x+9y-9+8x+12y+4\right)\)

\(=\left(x-3y-13\right)\left(17x+21y-5\right)\)

Đúng là Tú có khác (:

9( x + y - 1 )² - 4( 2x + 3y + 1 )²

= 3²( x + y - 1 )² - 2²( 2x + 3y + 1 )²

= [ 3( x + y - 3 ) ]² - [ 2( 2x + 3y + 1 ) ]²

= ( 3x + 3y - 3 )² - ( 4x + 6y + 2 )²

= [ ( 3x + 3y - 3 ) - ( 4x + 6y + 2 ) ][ ( 3x + 3y - 3 ) + ( 4x + 6y + 2 ) ]

= ( 3x + 3y - 3 - 4x - 6y - 2 )( 3x + 3y - 3 + 4x + 6y + 2 )

= ( -x - 3y - 5 )( 7x + 9y - 1 )

Nỏ có đáp án mô anh

a , \(-q^3+12q^2x-48qx^2+64x^3\)

 \(=-\left(q^3-12q^2x+48qx^2-64x^3\right)\)

\(=\)\(-\left(q-4x\right)^3\)

b , x² + 2xy - y² - 9 

= - ( x² - 2xy + y² ) - 9

= - ( x - y )² - 9

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

3 , 1 - m² + 2mn - n²

= 1 - ( m² - 2mn + n² )

= 1 - ( m - n )²

= ( 1 - m + n ) ( 1 + m - n )

4 , x³ - 8 + 6a² - 12a

  = x³ +  6a² - 12a + 8 

  = x³ + 6a² - 12a + 4 + 4

  = x³ + ( 6a² - 12a + 4 ) + 4

  = x³ + ( 3a - 2 )² + 4

  = ( x + 3a - 2 + 2 ) ( x² + 3a + 2 + 2 )

( Mai làm tiếp mấy ý sau '-' muộn rồi ~ )

5 , x² - 2xy + y² - xz - yz

  = ( x² - 2xy + y² ) - ( xz + yz )

  = (  x - y )² - z ( x + y )

  = ( x - y ) ² - z ( x - y )

  = ( x - y ) ( x - y - z )

6 , x² - 4xy + 4y ² - z² + 4z - 4t²

 =(  x² - 4xy + 4y ² ) - (z² - 4z +4 ) . t²

 = ( x - y )² - ( z - 2  )² . t²

 = ( x - y - z - 2 ) ( x - y + z - 2 ) t²

7 , 25 - 4x² - 4xy - y²

  = 25 + ( - 4x² - 4xy + y² )

  = 25 + ( 2x - y )²

  = ( 5 + 2x - y ) ( 5 + 2x + y )

8 ,

       x³ + y³ + z³ - 3xyz

    = (x+y)³ - 3xy (x  - y ) + z³ - 3xyz 
    = [ ( x + y)³ + z³ ] - 3xy ( x + y + z ) 
    = ( x + y + z )³ - 3z ( x + y )( x + y + z ) - 3xy ( x - y - z ) 
    = ( x + y + z )[( x + y + z )² - 3z ( x + y ) - 3xy ] 
    = ( x + y + z )( x² + y² + z² + 2xy + 2xz + 2yz - 3xz - 3yz - 3xy) 
    = ( x + y + z)(x² + y² + z² - xy - xz - yz)