Toán Lớp 8: Thức hiện phép chia:
(x-2y)^2 – 7(x-2y) – 8 cho (x-2y+1)
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Toán Lớp 8: Thức hiện phép chia:
(x-2y)^2 – 7(x-2y) – 8 cho (x-2y+1)
Comments ( 2 )
$(x-2y)^2-7(x-2y)-8$
$=x^2-4xy+4y^2-7x+14y-8$
$=4y^2-2xy+16y-2xy+x^2-8x-2y+x-8$
$=2y(2y-x+8)-x(2y-x+8)-(2y-x+8)$
$=(2y-x-1)(2y-x+8)$
$=-(x-2y+1)(2y-x+8)$
$⇒[(x-2y)^2-7(x-2y)-8]:(x-2y+1)$
$⇔-(x-2y+1)(2y-x+8):(x-2y+1)$
$=-(2y-x+8)$
$=x-2y-8$
Vậy $[(x-2y)^2-7(x-2y)-8]:(x-2y+1)=(x-2y-8)$.
Giải đáp+Lời giải và giải thích chi tiết:
[(x – 2y)^2 – 7(x – 2y) – 8] : (x – 2y + 1)
= [(x – 2y)^2 + (x – 2y) – 8(x – 2y) – 8] : (x – 2y + 1)
= [(x – 2y)(x – 2y + 1) – 8(x – 2y + 1)] : (x – 2y + 1)
= [(x – 2y + 1)(x – 2y – 8)] : (x – 2y + 1)
= x – 2y – 8