HOC24
Lớp học
Môn học
Chủ đề / Chương
Bài học
a) 9x2 - 36
= ( 3x)2 - 62
= ( 3x - 6)( 3x + 6)
b) ax - ay - bx + by
= a( x - y) - b( x - y)
= ( x - y)( a - b)
c) y3 - 4y2 + 3y
= y3 - y2 - 3y2 + 3y
= y2( y - 1) - 3y( y - 1)
= ( y - 1)( y2 - 3y)
= y( y - 3)( y - 1)
a) ( x2 - 1)2 - x( x2 - 1) - 2x2
Đặt : x2 - 1 = a , ta có :
a2 - ax - 2x2
= a2 + ax - 2ax - 2x2
= a( a + x) -2x( a + x)
= ( a + x)( a - 2x)
Thay , x2 - 1 = a , ta có :
( x2 - 1 + x)( x2 - 1 - 2x)
b) ( x2 + 4x + 8)2 + 3x( x2 + 4x + 8) + 2x2
Đặt : x2 + 4x + 8 = b, ta có :
b2 + 3bx + 2x2
= b2 + bx + 2bx + 2x2
= b( b + x) + 2x( b + x)
= ( b + x)( b + 2x)
Thay x2 + 4x + 8 = b, ta có :
( x2 + 4x + 8 + x)( x2 + 4x + 8 + 2x)
= ( x2 + 5x + 8)( x2 + 6x + 8)
Chúc bn hok tốt
3x3 - 12x
= 3x(x2 - 22)
= 3x( x - 2)(x+2)
a) x2 + 2x + 1
= ( x + 1)2
b) 8x3 - 64
= 8( x3 - 23)
= 8( x - 2)( x2 + 2x + 4)
c) x2 - y2 - 6y + 9
= x2 - 2.3y + 32 - y2
= ( x - 3)2 - y2
= ( x - 3 -y)( x - 3 + y)
d) x3 - 2x2 + x - xy2 ( sửa đề)
= x( x2 - 2x + 1 - y2)
= x[( x - 1)2 - y2]
= x( x - 1 - y)( x - 1 +y)
e) x2 - 15x + 14
= x2 - x - 14x + 14
= x( x - 1) - 14(x - 1)
= ( x - 1)( x - 14)
f) 3x2 - 17x + 14
= 3x2 - 3x - 14x + 14
= 3x( x - 1) -14( x - 1)
= ( x - 1)( 3x - 14)
Cách 2 .
x7 + x5 + 1
= x7 + x6 + x5 - x6 + 1
= x5( x2 + x + 1) - [ ( x3 )2 - 1]
= x5( x2 + x + 1) - ( x3 - 1)( x3 + 1)
= x5( x2 + x + 1) -( x - 1)( x2 + x + 1)( x3 + 1)
= ( x2 + x + 1)[ x5 -( x - 1)( x3 + 1)]
= (x2 + x + 1)( x5 - x4 + x3 - x + 1)
x2 - 2x - 4y2 - 4y
= x2 - ( 2y)2 - 2( x + 2y)
= ( x - 2y)( x + 2y) - 2( x + 2y)
= ( x + 2y)( x - 2y - 2)
a) 5x( x - 2) + 3x - 6 =0
(=) 5x( x - 2) + 3( x - 2) = 0
(=) ( x - 2)( 5x + 3) =0
Vậy , x = 2 ; x = \(-\dfrac{3}{5}\)
b) x3 - 9x = 0
(=) x( x2 - 32) = 0
(=) x( x - 3)( x + 3) =0
Vậy , x = 0 ; x = 3 ; x = -3
Giả sử : f( x) = ( x2 - 1).g(x) + ax + b
*) Áp dụng định lý Bezout , ta có :
f( 1) = a + b
(=) 1100 - 150 + 2.125 - 4 = a + b
(=) a + b = -2 (*)
f( -1) = -a + b
(=) ( -1)100 - ( -1)50 + 2.(-1)25 - 4
(=) -a + b = -6 (**)
Từ ( *,**) 2b =-8 -> b = -4 -> a = 2
Vậy số dư là : 2x - 4