phan tich cac da thuc sau thanh nhan tu a) a^2c-a^2d-b^2d+b^2c b)8x^2+4xy-2ax-ay c) x^2-y^2-2yz-z^2 d) 3a^2-6ab+3b^2-12c^2 e)x^2-2xy+y^2-m^2+2mn-n^2

phan tich cac da thuc sau thanh nhan tu
a) a^2c-a^2d-b^2d+b^2c
b)8x^2+4xy-2ax-ay
c) x^2-y^2-2yz-z^2
d) 3a^2-6ab+3b^2-12c^2
e)x^2-2xy+y^2-m^2+2mn-n^2
g) a^2-10a+25-y^2-4yz-4z^2
h) x^2-2x-3
i) a^4+5a^3+15a-9
k) bc ( b+c)+ac (c-a)-ab (a+b)
l) 4x-4y+x^2-2xy+y^2
co ai la mcau nay ko minh cho 50 chuc ne

0 bình luận về “phan tich cac da thuc sau thanh nhan tu a) a^2c-a^2d-b^2d+b^2c b)8x^2+4xy-2ax-ay c) x^2-y^2-2yz-z^2 d) 3a^2-6ab+3b^2-12c^2 e)x^2-2xy+y^2-m^2+2mn-n^2”

  1. a) a²c – a²d – b²d + b²c

    = ( a²c – a²d ) + ( b²c – b²d )

    = a²( c – d ) + b²( c – d )

    = ( c – d )( a² + b² )

    b. 8x² + 4xy – 2ax – ay

    = ( 8x² + 4xy ) – ( 2ax + ay )

    = 4x( 2x + y ) – a( 2x + y )

    = ( 2x + y )( 4x – a )

    c. x² – y² – 2yz – z²

    = x² – ( y² + 2yz + z² )

    = x² – ( y + z )²

    = ( x – y – z )( x + y + z )

    d. 3a² – 6ab + 3b² – 12c²

    = 3( a² – 2ab + b² – 4c² )

    = 3[( a² – 2ab + b² ) – ( 2c )² ]

    = 3 ( a – b – 2c )( a – b + 2c )

    e. x² – 2xy + y² – m² + 2mn – n²

    = ( x² – 2xy + y² ) – ( m² – 2mn + n² )

    = ( x – y )² – ( m – n )²

    = ( x – y – m + n )( x – y + m – n )

    g. a² – 10a + 25 – y² – 4yz – 4z²

    = ( a² – 10a + 5² ) – [ y² + 4yz +( 2z )²]

    = ( a – 5 )² – ( y + 2z )²

    = ( a – 5 – y – 2z )( a – 5 + y + 2z )

    h. x² – 2x – 3

    = x² – 3x + x – 3

    = ( x² – 3x ) + ( x – 3 )

    = x( x – 3 ) + ( x – 3 )

    = ( x – 3 )( x + 1 )

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  2. Đáp án:

    a. a²c – a²d – b²d + b²c

    = ( a²c – a²d ) – ( b²d – b²c )

    = a²( c – d ) – b²( d – c )

    = a²( c – d ) + b²( c – d )

    = ( c – d )( a² + b² )

    b. 8x² + 4xy – 2ax – ay

    = ( 8x² + 4xy ) – ( 2ax + ay )

    = 4x( 2x + y ) – a( 2x + y )

    = ( 2x + y )( 4x – a )

    c. x² – y² – 2yz – z²

    = x² – ( y² + 2yz + z² )

    = x² – ( y + z )²

    = ( x – y – z )( x + y + z )

    d. 3a² – 6ab + 3b² – 12c²

    = 3( a² – 2ab + b² – 4c² )

    = 3[( a² – 2ab + b² ) – 4c² ]

    = 3[( a – b )² – ( 2c )²]

    = 3( a – b – 2c )( a – b + 2c )

    e. x² – 2xy + y² – m² + 2mn – n²

    = ( x² – 2xy + y² ) – ( m² – 2mn + n² )

    = ( x – y )² – ( m – n )²

    = ( x – y – m + n )( x – y + m – n )

    g. a² – 10a + 25 – y² – 4yz – 4z²

    = ( a² – 10a + 25 ) – ( y² + 4yz + 4z² )

    = ( a – 5 )² – ( y + 2z )²

    = ( a – 5 – y – 2z )( a – 5 + y + 2z )

    h. x² – 2x – 3

    = x² – 3x + x – 3

    = ( x² – 3x ) + ( x – 3 )

    = x( x – 3 ) + ( x – 3 )

    = ( x – 3 )( x + 1 )

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