Toán Lớp 8: $(2x-y)(4x^2 – 2xy + y^2)$
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$\big(3x-5\big)\big(2x+11\big)-\big(2x+3\big)\big(3x+7\big)$
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$\dfrac{1}{xy-x^2}-\dfrac{1}{y^2-xy}$
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Toán Lớp 8: $(2x-y)(4x^2 – 2xy + y^2)$
$\\$
$\big(3x-5\big)\big(2x+11\big)-\big(2x+3\big)\big(3x+7\big)$
$\\$
$\dfrac{1}{xy-x^2}-\dfrac{1}{y^2-xy}$
Comments ( 1 )
Giải đáp:
Lời giải và giải thích chi tiết:
(2x-y)(4x^2 -2xy+y^2)
=(8x^3 -4x^2y + 2xy^2) – (4x^2y-2xy^2 – y^3)
= 8x^3 – 8x^2y +4xy^2 -y^3
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(3x-5)(2x+11)-(2x+3)(3x+7)
= (6x^2+33x – 10x-55)- (6x^2+14x +9x +21)
= (6x^2 -6x^2)+(33x-10x – 14x -9x)+(-55-21)
= – 76
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1/(xy-x^2) -1/(y^2 -xy)
= 1/(x(y-x)) – 1/(y(y-x))
= (y)/(xy(y-x)) – (x)/(xy(y-x))
= (y-x)/(xy(y-x))
= 1/xy