《2025年经纶学典学霸黑白题高中数学必修第一册苏教版》

1. $\boldsymbol{★}$ (2025·江苏宿迁高一月考)化简$\left(6\dfrac{1}{4}\right)^{-\frac{1}{2}}$的结果是(C)

A.$\dfrac{3}{2}$
B.$\dfrac{2}{3}$
C.$\dfrac{2}{5}$
D.$\dfrac{5}{2}$

答案： 1.C 解析：由题意可得$(6 \frac{1}{4})^{-\frac{1}{2}} = (\frac{4}{25})^{-\frac{1}{2}} = [(\frac{2}{5})^{2}]^{-\frac{1}{2}} = \frac{2}{5}$. 故选C.

2. $\boldsymbol{★}$ (2025·江苏徐州高一期中)若$\log_{x}\dfrac{1}{8}=-3$,则$x=$(A)

A.$2$
B.$\dfrac{1}{2}$
C.$-2$
D.$-\dfrac{1}{2}$

答案： 2.A 解析：$\because \log_{x}\frac{1}{8} = -3,\therefore x^{-3} = \frac{1}{8},x^{3} = 8,\therefore x = 2$. 故选A.

3. $\boldsymbol{★}$ (2025·江苏扬州高一期中)若$1\lt a\lt 2$,则$\sqrt[3]{(1-a)^{3}}+\sqrt[4]{(2-a)^{4}}$的化简结果是(C)

A.$1$
B.$-1$
C.$3-2a$
D.$2a-3$

答案： 3.C 解析：因为$1 ＜ a ＜ 2$，则$2 - a ＞ 0$，所以$\sqrt[3]{(1 - a)^{3}} + \sqrt[4]{(2 - a)^{4}} = 1 - a + |2 - a| = 1 - a + 2 - a = 3 - 2a$. 故选C.

4. $\boldsymbol{★★}$ (2025·江苏宿迁高一期中)若$\lg a$,$\lg b$是方程$5x^{2}-10x+3=0$的两个实根,则$ab$的值等于(C)

A.$2$
B.$\dfrac{1}{2}$
C.$100$
D.$\sqrt{10}$

答案： 4.C 解析：由韦达定理可得$\lg a + \lg b = 2$，所以$\lg ab = \lg a + \lg b = 2$，所以$ab = 10^{2} = 100$. 故选C.

5. $\boldsymbol{★★}$ (2025·江苏南京高一期中)已知$a^{\frac{1}{2}}-a^{-\frac{1}{2}}=\sqrt{5}$,则$a^{2}-a^{-2}=$(C)

A.$3\sqrt{5}$
B.$\pm 3\sqrt{5}$
C.$21\sqrt{5}$
D.$\pm 21\sqrt{5}$

答案： 5.C 解析：由$a^{\frac{1}{2}} - a^{-\frac{1}{2}} = \sqrt{5}$得$(a^{\frac{1}{2}} - a^{-\frac{1}{2}})^{2} = a - 2 + a^{-1} = 5$，即$a + a^{-1} = 7$，故$a^{2} + a^{-2} = \sqrt{(a^{2} + a^{-2})^{2}} = \sqrt{a^{2} + 2 + a^{-2}} = \sqrt{9} = 3$，故$a - a^{-1} = (a^{\frac{1}{2}} + a^{-\frac{1}{2}})(a^{\frac{1}{2}} - a^{-\frac{1}{2}}) = 3\sqrt{5}$，故$a^{2} - a^{-2} = (a + a^{-1})·(a - a^{-1}) = 21\sqrt{5}$. 故选C.

6. $\boldsymbol{★★★}$ (2025·江苏泰州高一月考)升温系数是衡量空调制热效果好坏的主要依据之一,把物体放在制热空调的房间里升温,如果物体初始温度为$\theta_{1}$,空气的温度为$\theta_{0}$,$t$小时后物体的温度$\theta$可由公式$\theta=\theta_{0}+(\theta_{0}-\theta_{1})e^{-kt}$求得,其中$k$是一个随着物体与空气的接触状况而定的升温系数.现有$A$,$B$两个物体放在空气中升温,已知两物体的初始温度相同,升温$2$小时后,$A$,$B$两个物体的温度分别为$5\theta_{0}$,$9\theta_{0}$,假设$A$,$B$两个物体的升温系数分别为$k_{A}$,$k_{B}$,则(C)

A.$\dfrac{k_{A}}{k_{B}}=\dfrac{1}{2}\ln 2$
B.$\dfrac{k_{B}}{k_{A}}=\dfrac{1}{2}\ln 2$
C.$k_{A}-k_{B}=\dfrac{1}{2}\ln 2$
D.$k_{B}-k_{A}=\dfrac{1}{2}\ln 2$

答案： 6.C 解析：由题意可知$5\theta_{0} = \theta_{0} + (\theta_{0} - \theta_{1})e^{-2kA}$，则$(\theta_{0} - \theta_{1})· e^{-2kA} = 4\theta_{0}①$，$9\theta_{0} = \theta_{0} + (\theta_{0} - \theta_{1})e^{-2kB}$，则$(\theta_{0} - \theta_{1})e^{-2kB} = 8\theta_{0}②$，$\frac{①}{②}$可得$\frac{(\theta_{0} - \theta_{1})e^{-2kA}}{(\theta_{0} - \theta_{1})e^{-2kB}} = \frac{4\theta_{0}}{8\theta_{0}}$，则$e^{-2kA + 2kB} = \frac{1}{2}$，$-2(k_{A} - k_{B}) = \ln\frac{1}{2}$，化简可得$k_{A} - k_{B} = \frac{1}{2}\ln2$. 故选C.

7. $\boldsymbol{★★★}$ (2025·江苏泰州高一期中)已知$\lg a=\lg(a+b+3)-\lg b$,则$a+2b$的最小值为(C)

A.$5$
B.$3+2\sqrt{2}$
C.$3+4\sqrt{2}$
D.$9$

答案： 7.C 解析：由$\lg a = \lg(a + b + 3) - \lg b \Leftrightarrow \lg a + \lg b = \lg(a + b + 3) \Leftrightarrow \lg ab = \lg(a + b + 3)$，则$ab = a + b + 3$，又$ab = a + b + 3 \Leftrightarrow (a - 1)(b - 1) = 4$，结合$\lg a = \lg(a + b + 3) - \lg b$，知$a - 1 ＞ 0,b - 1 ＞ 0$，又$a + 2b = (a - 1) + 2(b - 1) + 3 \geqslant 2\sqrt{(a - 1)×2(b - 1)} + 3 = 4\sqrt{2} + 3$，当且仅当$a - 1 = 2(b - 1)$，即$a = 1 + 2\sqrt{2},b = 1 + \sqrt{2}$时等号成立，因此可得$a + 2b$的最小值为$4\sqrt{2} + 3$. 故选C.

8. $\boldsymbol{★★★}$ (2025·江苏南通高一月考)已知$2^{a}=b$,$2^{b}=3$,$\log_{b}6=c$,则(A)

A.$b+1=ac$
B.$3b+a=c$
C.$ac+a=2b$
D.$b=ac$

答案： 8.A 解析：因为$2^{a} = b,2^{b} = 3$，所以$a = \log_{2}b,b = \log_{2}3,ac = \log_{2}b·\log_{b}6 = \log_{b}6 = \log_{2}3 + 1$，故$b + 1 = ac$. 故选A.