电子课本网 第112页

第112页

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$解:∵ \sqrt{2023-m}与 \sqrt{m-2023}有意义,\ $
$∴m=2023,n=-2,$
$∴\sqrt {\ m-n} = \sqrt{2023+2}=45.$
$解:(1)∵b-c≥0且c-b≥0,∴b=c.$
$(2)由(1),得 \sqrt{2a+b-4}+|a+1|=0,\ $
$∴\begin{cases}{2a+b=4,\ }\ \\ {a+1=0,\ } \end{cases}解得\begin{cases}{ a=-1, }\ \\ { b=6. } \end{cases}$
$∴±\sqrt{-4a+b+c}=±\sqrt {4+6+6} =±4.$
$解:由题意,得 \sqrt{1+x}+ \sqrt{(1-y)^{3} }=0,\ $
$∴1+x=0,1-y=0.$
$∴x=-1,y=1.\ $
$∴x^{2023} -y^{2022} =-1-1=-2.$
$解:∵ \sqrt{x-1}+ \sqrt{9-x}有意义,\ $
$∴x-1≥0,9-x≥0,解得1≤x≤9.\ $
$∴|x-1|+ \sqrt{(x-12)²}=x-1+12-x=11.$
$解:依题意,得\begin{cases}{ x-1≥0, }\ \\ { 1-x≥0, } \end{cases}\ $
$∴x-1=0,解得x=1.\ $
$∴y<3.∴y-3<0,y-4<0.\ $
$∴|y-3|- \sqrt{y²-8y+16}=3-y- \sqrt{(y-4)²}$
$=3-y-(4-y)=-1.$
$解:(1)由(a-2)²+ \sqrt{b-2a}=0,\ $
$得\begin{cases}{ a-2=0, }\ \\ { b-2a=0, } \end{cases}\ 解得\begin{cases}{a=2, }\ \\ {b=4.\ } \end{cases}\ $
$∴b-a的算术平方根是 \sqrt{4-2}=\sqrt{2}.$
$(2)由题意,得\begin{cases}{x-1≥0,\ }\ \\ { 1-x≥0, } \end{cases}\ 解得x=1.\ $
$当x=1时,y=4.$
$∴ \sqrt{x²y}= \sqrt{1²×4}=2.$