第7页 - 通城学典课时作业本八年级数学人教版南通专版

B

D

$10^{-2}$

-11

$\frac{3}{5}$

$-3a-2$

3

2

$解:(1)原式=(x+ \sqrt{10})(x-\sqrt{10})$

$解:(2)原式=(x-\sqrt{6})²$

$解:(3)原式=(x²+2)(x+\sqrt{2})(x-\sqrt{2})$

$解：∵m=\sqrt{2024}-5$

$∴m+5=\sqrt{2024}$

$∴(m+5)^2=\ \mathrm {m^2}+10m+25=(\sqrt{2024})^2=2024$

$∴\ \mathrm {m^2}+10m+1=\ \mathrm {m^2}+10m+25-24=2024-24=2000$

$解：由题意得\begin{cases}x+y-2024≥0\\2024-x-y≥0\end{cases}$

$∴x+y=2024$

$∴\sqrt{3x+y-z-8}+\sqrt{x+y-z}=0$

$又∵\sqrt{3x+y-z-8}≥0，\sqrt{x+y-z}≥0$

$∴\begin{cases}3x+y-z-8=0\\x+y-z=0\\x+y=2024\end{cases} 解得\begin{cases}x=4\\y=2020\\z=2024\end{cases}$

$∴(z-y)^2=(2024-2020)^2=16$