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16. (12分)计算.
(1) $-1^2 + (π - 3.14)^0 ÷ (-\dfrac{1}{3})^{-2} × (\dfrac{3}{4})^{-1}$；
(2) $(4 × 10^2)^3 ÷ (2 × 10^{-3})^2$；(结果用科学记数法表示)
(3) $(x - y)(x - 3y) - (2x - y)^2$；
(4) $(2x - y - z)(y + z - 2x)$.
$· 3·$
(1) $-1^2 + (π - 3.14)^0 ÷ (-\dfrac{1}{3})^{-2} × (\dfrac{3}{4})^{-1}$；
(2) $(4 × 10^2)^3 ÷ (2 × 10^{-3})^2$；(结果用科学记数法表示)
(3) $(x - y)(x - 3y) - (2x - y)^2$；
(4) $(2x - y - z)(y + z - 2x)$.
$· 3·$
答案
【点拨】本题考查实数的混合运算,整式的混合运算,多项式乘多项式,科学记数法的表示,积的乘方,熟练掌握相关运算方法为解题关键.
【解析】(1)
$-1^2 + (π - 3.14)^0 ÷ (-\dfrac{1}{3})^{-2} × (\dfrac{3}{4})^{-1}$
$= -1 + 1 ÷ 9 × \dfrac{4}{3}$
$= -1 + \dfrac{4}{27}$
$= -\dfrac{23}{27}$.
(2)
$(4×10^2)^3 ÷ (2×10^{-3})^2$
$=64×10^6 ÷ (4×10^{-6})$
$=16×10^{12}$
$=1.6×10^{13}$.
(3)
$(x-y)(x-3y) - (2x-y)^2$
$=x^2 -3xy -xy +3y^2 -4x^2 +4xy -y^2$
$=-3x^2 +2y^2$.
(4)
$(2x-y-z)(y+z-2x)$
$=(2x-y-z)[-(2x-y-z)]$
$=-(2x-y-z)^2$
$=-[2x-(y+z)]^2$
$=-[4x^2 -2×2x×(y+z) + (y+z)^2]$
$=-4x^2 +4xy +4xz -y^2 -2yz -z^2$.
【解析】(1)
$-1^2 + (π - 3.14)^0 ÷ (-\dfrac{1}{3})^{-2} × (\dfrac{3}{4})^{-1}$
$= -1 + 1 ÷ 9 × \dfrac{4}{3}$
$= -1 + \dfrac{4}{27}$
$= -\dfrac{23}{27}$.
(2)
$(4×10^2)^3 ÷ (2×10^{-3})^2$
$=64×10^6 ÷ (4×10^{-6})$
$=16×10^{12}$
$=1.6×10^{13}$.
(3)
$(x-y)(x-3y) - (2x-y)^2$
$=x^2 -3xy -xy +3y^2 -4x^2 +4xy -y^2$
$=-3x^2 +2y^2$.
(4)
$(2x-y-z)(y+z-2x)$
$=(2x-y-z)[-(2x-y-z)]$
$=-(2x-y-z)^2$
$=-[2x-(y+z)]^2$
$=-[4x^2 -2×2x×(y+z) + (y+z)^2]$
$=-4x^2 +4xy +4xz -y^2 -2yz -z^2$.
