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1. (2026·广安中考)已知$a, b$为实数,且$8\sqrt{a-3} + 2\sqrt{12-5b} = a -5b +26$, 则$a+\dfrac{5}{11}b$的平方根是
±2√5
.答案
1. ±2√5 解析:设$\sqrt{a-3}=m$,$\sqrt{12-5b}=n$,则$m≥0,n≥0,a≥3,b≤\dfrac{12}{5}$,$\therefore a=m^2+3,b=\dfrac{-n^2+12}{5}$,代入原方程可得$8m+2n=m^2+3-5×\dfrac{-n^2+12}{5}+26$,整理,得$m^2-8m+n^2-2n+17=0$,变形,得$(m-4)^2+(n-1)^2=0.\because (m-4)^2≥0,(n-1)^2≥0$,$\therefore m-4=0,n-1=0$,$\therefore m=4,n=1$,$\therefore a=m^2+3=19,b=\dfrac{-n^2+12}{5}=\dfrac{11}{5}$,符合题意,$\therefore a+\dfrac{5}{11}b=19+\dfrac{5}{11}×\dfrac{11}{5}=19+1=20$.
$\because 20$的平方根为$±2\sqrt{5}$,$\therefore a+\dfrac{5}{11}b$的平方根为$±2\sqrt{5}$.
$\because 20$的平方根为$±2\sqrt{5}$,$\therefore a+\dfrac{5}{11}b$的平方根为$±2\sqrt{5}$.
2. (2026·扬州模拟)对$x,y$定义一种新运算$\Delta$，规定：$\Delta(x,y)=\dfrac{ax+by}{x+3y}$(其中$a,b$均为非零常数)，这里等式右边是通常的四则运算，例如：$\Delta(0,1)=\dfrac{a×0 + b×1}{0+3×1}=\dfrac{b}{3}$。
(1)已知$\Delta(1,-1)=1$，$\Delta(2,1)=4$。
①求$a,b$的值；
②若关于$m$的不等式组$\begin{cases} \Delta(-3m,m+1)≤6, \\ \Delta(6m,1-2m)＜p \end{cases}$恰好有2个整数解，求实数$p$的取值范围。
(2)若$\Delta(x,y)=\Delta(y,x)$对任意实数$x,y$都成立(这里$\Delta(x,y),\Delta(y,x)$都有意义)，则$a,b$应满足怎样的关系式？
(1)已知$\Delta(1,-1)=1$，$\Delta(2,1)=4$。
①求$a,b$的值；
②若关于$m$的不等式组$\begin{cases} \Delta(-3m,m+1)≤6, \\ \Delta(6m,1-2m)＜p \end{cases}$恰好有2个整数解，求实数$p$的取值范围。
(2)若$\Delta(x,y)=\Delta(y,x)$对任意实数$x,y$都成立(这里$\Delta(x,y),\Delta(y,x)$都有意义)，则$a,b$应满足怎样的关系式？
答案
2. (1)①$\because \Delta(x,y)=\dfrac{ax+by}{x+3y}$,$\Delta(1,-1)=1$,$\Delta(2,1)=4$,$\therefore \dfrac{a-b}{1-3}=1$,$\dfrac{2a+b}{2+3}=4$,解得$a=6,b=8$.
②解$\Delta ( - 3m, m + 1 ) ≤ 6$,即$\dfrac{6(-3m)+8(m+1)}{-3m+3(m+1)} ≤ 6$,解得$m≥-1$;
解$\Delta(6m,1-2m)＜p$,即$\dfrac{6×6m+8(1-2m)}{6m+3(1-2m)}＜p$,解得$m＜\dfrac{3p-8}{20}$,
$\therefore$不等式的解集为$-1≤ m＜\dfrac{3p-8}{20}.\because$关于$m$的不等式组$\begin{cases} \Delta(-3m,m+1)≤6, \\ \Delta(6m,1-2m) ＜p \end{cases}$恰好有2个整数解,即$m=-1,0$,$\therefore 0＜\dfrac{3p-8}{20}≤1$,解得$\dfrac{8}{3}＜p≤\dfrac{28}{3}$.
(2)$\because \Delta(x,y)=\Delta(y,x)$对任意实数$x,y$都成立,$\therefore \dfrac{ax+by}{x+3y}=\dfrac{ay+bx}{y+3x}$,整理得$(ax+by)(y+3x)=(ay+bx)(x+3y)$,展开得$axy+3ax^2+by^2+3bxy=axy+3ay^2+bx^2+3bxy$,化简得$3ax^2+by^2=bx^2+3ay^2$,再整理得$(3a-b)x^2=(3a-b)y^2.\because$上式对于任意实数$x,y$都成立,$\therefore 3a-b=0$,$\therefore b=3a$.
②解$\Delta ( - 3m, m + 1 ) ≤ 6$,即$\dfrac{6(-3m)+8(m+1)}{-3m+3(m+1)} ≤ 6$,解得$m≥-1$;
解$\Delta(6m,1-2m)＜p$,即$\dfrac{6×6m+8(1-2m)}{6m+3(1-2m)}＜p$,解得$m＜\dfrac{3p-8}{20}$,
$\therefore$不等式的解集为$-1≤ m＜\dfrac{3p-8}{20}.\because$关于$m$的不等式组$\begin{cases} \Delta(-3m,m+1)≤6, \\ \Delta(6m,1-2m) ＜p \end{cases}$恰好有2个整数解,即$m=-1,0$,$\therefore 0＜\dfrac{3p-8}{20}≤1$,解得$\dfrac{8}{3}＜p≤\dfrac{28}{3}$.
(2)$\because \Delta(x,y)=\Delta(y,x)$对任意实数$x,y$都成立,$\therefore \dfrac{ax+by}{x+3y}=\dfrac{ay+bx}{y+3x}$,整理得$(ax+by)(y+3x)=(ay+bx)(x+3y)$,展开得$axy+3ax^2+by^2+3bxy=axy+3ay^2+bx^2+3bxy$,化简得$3ax^2+by^2=bx^2+3ay^2$,再整理得$(3a-b)x^2=(3a-b)y^2.\because$上式对于任意实数$x,y$都成立,$\therefore 3a-b=0$,$\therefore b=3a$.
3. (2026·云南中考)已知$m≠0,S=m^4+m+\frac{192}{m^2}-\frac{4}{m}-118$.某二次函数表示的图象为抛物线M，抛物线M经过$A(-10,0),B(2m^2-4m,0)$两点.函数$y=(2x+4)(3x-24)$表示的图象为抛物线T.x轴上有这样的点，它既在抛物线M上，又在抛物线T上.
(1)求抛物线T与x轴的公共点的坐标；
(2)比较S与$\sqrt{145}$的大小.
(1)求抛物线T与x轴的公共点的坐标；
(2)比较S与$\sqrt{145}$的大小.
答案
3. (1)由题意,当$y=(2x+4)(3x-24)=0$时,解得$x_1=-2,x_2=8$,$\therefore$抛物线$T$与$x$轴的公共点的坐标为$(-2,0),(8,0)$.
(2)由(1)知,抛物线$T$与$x$轴的公共点的坐标为$(-2,0),(8,0).\because$抛物线$M$经过$A(-10,0),B(2m^2-4m,0)$两点,$x$轴上有这样的点,它既在抛物线$M$上,又在抛物线$T$上,$\therefore 2m^2-4m=-2$或$2m^2-4m=8$.①当$2m^2-4m=-2$时,整理得$(m-1)^2=0$,解得$m=1$,将$m=1$代入$S$,得$S=1^4+1+\dfrac{192}{1^2}-\dfrac{4}{1}-118=72.\because 72^2=5\ 184＞145$,$\therefore 72＞\sqrt{145}$,即$S＞\sqrt{145}$.
②当$2m^2-4m=8$时,整理得$m^2-2m-4=0$,解得$m=1±\sqrt{5}$,
$\therefore m^2=2m+4$,两边同除以$m$得$\dfrac{4}{m}=m-2$,$\therefore m^4=(m^2)^2=(2m+4)^2=4m^2+16m+16=4(2m+4)+16m+16=24m+32$,
$\dfrac{192}{m^2}=\dfrac{192}{2m+4}=\dfrac{96}{m+2}.\because m^2-2m-4=0$,$\therefore m^2-4=2m$,$\therefore (m-2)· (m+2)=2m$,$\therefore m+2=\dfrac{2m}{m-2}$,$\therefore \dfrac{96}{m+2}=\dfrac{96(m-2)}{2m}=48-\dfrac{96}{m}=48-24· \dfrac{4}{m}=48-24(m-2)=96-24m$,$\therefore S=(24m+32)+m+(96-24m)-(m-2)-118=12.\because 12^2=144＜145$,$\therefore 12＜\sqrt{145}$,即$S＜\sqrt{145}$.
综上,当$m=1$时,$S＞\sqrt{145}$;当$m=1±\sqrt{5}$时,$S＜\sqrt{145}$.
(2)由(1)知,抛物线$T$与$x$轴的公共点的坐标为$(-2,0),(8,0).\because$抛物线$M$经过$A(-10,0),B(2m^2-4m,0)$两点,$x$轴上有这样的点,它既在抛物线$M$上,又在抛物线$T$上,$\therefore 2m^2-4m=-2$或$2m^2-4m=8$.①当$2m^2-4m=-2$时,整理得$(m-1)^2=0$,解得$m=1$,将$m=1$代入$S$,得$S=1^4+1+\dfrac{192}{1^2}-\dfrac{4}{1}-118=72.\because 72^2=5\ 184＞145$,$\therefore 72＞\sqrt{145}$,即$S＞\sqrt{145}$.
②当$2m^2-4m=8$时,整理得$m^2-2m-4=0$,解得$m=1±\sqrt{5}$,
$\therefore m^2=2m+4$,两边同除以$m$得$\dfrac{4}{m}=m-2$,$\therefore m^4=(m^2)^2=(2m+4)^2=4m^2+16m+16=4(2m+4)+16m+16=24m+32$,
$\dfrac{192}{m^2}=\dfrac{192}{2m+4}=\dfrac{96}{m+2}.\because m^2-2m-4=0$,$\therefore m^2-4=2m$,$\therefore (m-2)· (m+2)=2m$,$\therefore m+2=\dfrac{2m}{m-2}$,$\therefore \dfrac{96}{m+2}=\dfrac{96(m-2)}{2m}=48-\dfrac{96}{m}=48-24· \dfrac{4}{m}=48-24(m-2)=96-24m$,$\therefore S=(24m+32)+m+(96-24m)-(m-2)-118=12.\because 12^2=144＜145$,$\therefore 12＜\sqrt{145}$,即$S＜\sqrt{145}$.
综上,当$m=1$时,$S＞\sqrt{145}$;当$m=1±\sqrt{5}$时,$S＜\sqrt{145}$.
