1)(x+2)(x+4)(x+6)(x+8)+16

2)(x+2)(x+3)(x+4)(x+5)-24
nhân caisngaocj 1 với ngoạc cuối –  ๖ۣۜTQT☾♋☽ 28-10-17 07:51 AM
1) Câu hỏi của Phạm Bích Ngọc - Toán lớp 8 | Học trực tuyến
2) 
( x + 2 )( x + 3 )( x + 4 )( x + 5 ) - 24
=  $( x^{2} + 7x + 10 ) ( x^{2} + 7x + 12 ) - 24$
Đặt t = $ x^{2} + 7x + 10$ ta có :
= t( t + 2 ) - 24 = $t^2$ + 2t - 24
= $t^2$ - 4t + 6t - 24 = t(t-4) + 6(t-4)
= ( t-4 )(t+6)
= ( $x^2$ + 7x + 10 - 4)( $x^2$ + 7x + 10 + 6 )
= ( $x^2$ + 7x + 6 ) ( $x^2$ + 7x + 16 )
= ( x + 1 ) ( x + 6 ) ( $x^2$ + 7x + 16 )

Câu 1:
phân tích thành$:(x^2+10x+20)^2$
muốn biết cách làm nhân tung ra rồi làm ngược lại,hoặc nhân cái đầu với cuối,2 cái giữa với nhau rồi đặt x^2+10x=t
Câu 2:
tương tự v:kết quả:$(x+1)(x+6)(x^2+7x+16)$
=[(x+2)(x+8)][(x+4)(x+6)]+16
=($x^2$+10x+16)($x^2$+10x+24)+16
=a(a+8)+16
=$a^2$+8a+16
=$(a+4)^2$
=$($x^2$+10x+20)^2$


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