Bài 10 . Câu a) ( a + b +c)3 -a3 - b3 -c3
= ( a +b +c -a).[( a +b + c)2 + ( a +b+c).a + a2] - ( b3 + c3)
= ( b+c).[( a +b + c)2 + ( a +b+c).a + a2] -( b+c).( b2 - bc +c2)
= ( b+c).[( a +b + c)2 + ( a +b+c).a + a2 - b2 + bc - c2]
Đến đây, cậu chỉ cần tính như thường thôi ha
b) x2( y - z) + y2( z-x) + z2( x-y)
= x2y - x2z + y2z - y2x + z2( x -y)
= xy( x -y) - z( x2 - y2) + z2( x -y)
= xy( x -y) - z( x -y).( x+y) + z2( x -y)
= ( x - y).( xy - zx - zy + z2)
= ( x - y).[ x( y -z) - z( y - z)]
= ( x -y).(x -z).( y -z)
c) ( x +2).( x+3).(x+4).(x+5) - 24
= ( x+2).(x+5).(x+3).(x+4) - 24
=( x2 + 7x + 10).( x2 + 7x + 12) - 24
Đặt : x2 + 7x + 11 = y
Ta có : ( y -1).( y +1) - 24
= y2 - 1 - 24
= y2 - 25
= y2 - 52
= ( y - 5 ).( y+5)
Thay : x2 + 7x + 11 = y
Ta có : ( x2 + 7x +6).( x2 + 7x + 16)
d) x8 + x7 + 1
= x8 + x7 - x6 + x6 + 1
= x6( x2 + x +1) - [ (x3)2 - 1]
= x6( x2 + x +1) - ( x3 - 1)( x3 + 1)
= x6( x2 + x +1) - ( x - 1).( x2 + x +1).( x3 +1)
= ( x2 + x +1).[ x6 - ( x-1).( x3 +1)]
= ( x2 + x +1).[x6 - ( x4+ x - x3 -1)]
= ( x2 + x +1).( x6 - x4 - x + x3 +1)