《2025年经纶学典学霸黑白题高中数学必修第一册苏教版》

1. 下面四个等式运算中，正确的是 (D)

A.$ 3a^{-2}=\frac{1}{3a^2} $
B.$ 2^{\frac{3}{4}}=\sqrt[3]{2^4} $
C.$ \sqrt[6]{(-8)^6}=-8 $
D.$ a^{\frac{2}{3}}÷ a^{\frac{1}{3}}=\sqrt[3]{a} $

答案： 1.D 解析:对于 A 选项,根据负指数幂的定义,$a^{-n} = \frac{1}{a^n}(a \neq 0)$得到$3a^{-2} = \frac{3}{a^2}$,而不是$\frac{1}{3a^2}$,所以 A 选项错误.对于 B 选项,根据分数指数幂的定义,$a^m = \sqrt[n]{a^m}$,则$2^{\frac{4}{3}} = \sqrt[3]{2^4}$,而不是$\sqrt[4]{2^3}$,所以 B 选项错误.对于 C 选项,$\sqrt[6]{(-8)^6} = |-8| = 8$,所以 C 选项错误.对于 D 选项,$a^{\frac{2}{3}} ÷ a^{\frac{1}{3}} = a^{\frac{2}{3} - \frac{1}{3}} = a^{\frac{1}{3}}$.又因为$a^{\frac{1}{3}}$表示$a$的立方根,即$\sqrt[3]{a}$,所以 D 选项正确.故选 D.

2. (2025·广东深圳高一期中)$ (\sqrt{5})^{\sqrt{2}}·(\sqrt{5})^{\sqrt{2}}= $ (C)

A.$ \sqrt{5} $
B.5
C.$ 5^{\sqrt{2}} $
D.25

答案： 2.C 解析:$(\sqrt{5})^{\sqrt{2}} · (\sqrt{5})^2 = (\sqrt{5})^{\sqrt{2} + 2} = [(\sqrt{5})^2]^{\sqrt{2}} = 5^{\sqrt{2}}$.故选 C.

3. *(2025·福建厦门高一月考)若$ a^m=3 $，$ a^n=4 $，则$ a^{\frac{2m+3n}{2}}= $ (A)

A.24
B.12
C.$ 2\sqrt{6} $
D.$ 2\sqrt{3} $

答案： 3.A 解析:$a^{\frac{2m + 3n}{2}} = a^m · a^{\frac{3n}{2}} = 3 · (a^n)^{\frac{3}{2}} = 3 × 4^{\frac{3}{2}} = 3 × (2^2)^{\frac{3}{2}} = 3 × 2^3 = 24$.故选 A.

4. *(2025·江苏徐州高一期中)已知$ a＜1 $，则$ \sqrt{(a-1)^2}+\sqrt[3]{a^3}= $ (B)

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

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

5. *[苏教教材变式](多选)(2025·江苏徐州高一月考)已知$ a+a^{-1}=4 $，则 (ABC)

A.$ a^{\frac{1}{2}}+a^{-\frac{1}{2}}=\sqrt{6} $
B.$ a^2+a^{-2}=14 $
C.$ a^3+a^{-3}=52 $
D.$ a-a^{-1}=2\sqrt{3} $

答案： 5.ABC 解析:因为$a + a^{-1} = 4$,所以$a＞0$,对于 A,因为$(\frac{1}{2}a^{\frac{1}{2}} + a^{-\frac{1}{2}})^2 = a + a^{-1} + 2 = 6$,所以$\frac{1}{2}a^{\frac{1}{2}} + a^{-\frac{1}{2}} = \sqrt{6}$,故 A 正确;对于 B,因为$(a + a^{-1})^2 = a^2 + a^{-2} + 2 = 16$,所以$a^2 + a^{-2} = 14$,故 B 正确;对于 C,$a^3 + a^{-3} = (a + a^{-1})(a^2 + a^{-2} - 1) = 4 × 13 = 52$,故 C 正确;对于 D,因为$(a - a^{-1})^2 = a^2 + a^{-2} - 2 = 12$,所以$a - a^{-1} = \pm 2\sqrt{3}$,故 D 错误.故选 ABC.

6. *(2025·湖南株洲高一期中)已知$ a,b\in\mathbf{R} $，且$ 3a-b-2=0 $，则$ 27^a+\frac{1}{3^b} $的最小值为 (C)

A.2
B.4
C.6
D.8

答案： 6.C 解析:因为$a,b \in \mathbf{R}$,且$3a - b - 2 = 0$,则$3a - b = 2$,所以$27^a + \frac{1}{3^b} = 3^{3a} + 3^{-b} \geqslant 2\sqrt{3^{3a} · 3^{-b}} = 2\sqrt{3^{3a - b}} = 2\sqrt{3^2} = 6$,当且仅当$3^{3a} = 3^{-b}$,即$a = \frac{1}{3}$,$b = -1$时取等号,即$27^a + \frac{1}{3^b}$的最小值为 6.故选 C.

7. * 求使等式$ \sqrt{(a-3)(a^2-9)}=(3-a)\sqrt{a+3} $成立的实数$ a $的取值范围为 $[-3,3]$.

答案： 7.$[-3,3]$ 解析:$\sqrt{(a - 3)(a^2 - 9)} = \sqrt{(a - 3)^2(a + 3)} = |a - 3|\sqrt{a + 3}$,要使$|a - 3|\sqrt{a + 3} = (3 - a)\sqrt{a + 3}$成立,需$\begin{cases}a - 3 \leq 0, \\a + 3 \geq 0,\end{cases}$解得$a \in [-3,3]$,即实数$a$的取值范围是$[-3,3]$.故答案为$[-3,3]$.

8. *(2025·河北邢台高一月考)利用分数指数幂计算该式$ \sqrt{a\sqrt{\frac{1}{2}\sqrt{\frac{1}{2}\sqrt{a}}}}= $ $a^{\frac{1}{2}}$.($ a＞0 $)

答案： 8.$a^{\frac{1}{2}}$ 解析:$\sqrt{a^2\sqrt{a\sqrt{a}}} = \left[a^2\left(a(a^{\frac{1}{2}})^{\frac{1}{2}}\right)^{\frac{1}{2}}\right]^{\frac{1}{2}} = \left[a^2\left(a^{\frac{1}{2} × \frac{1}{2}}\right)^{\frac{1}{2}}\right]^{\frac{1}{2}} = \left[a^2\left(a^{\frac{1}{4}}\right)^{\frac{1}{2}}\right]^{\frac{1}{2}} = \left[a^2 × a^{\frac{1}{8}}\right]^{\frac{1}{2}} = \left(a^{\frac{17}{8}}\right)^{\frac{1}{2}} = a^{\frac{17}{16} × \frac{1}{2}} = a^{\frac{1}{2}}$.故答案为$a^{\frac{1}{2}}$.

9. *（2025·天津南开区高一期中）计算：
(1) $ (-1.8)^0+\left( \frac{3}{2} \right)^{-2}×\sqrt[3]{\left( \frac{27}{8} \right)^2}-\frac{1}{\sqrt{0.01}}+\sqrt{9^3} $；
(2) 若$ a=27 $，$ b=16 $，求$ \frac{(-2\sqrt{ab})×(-8\sqrt[3]{a^2b^5})}{\sqrt[6]{a^2b^7}×4\sqrt[4]{a^2b^5}} $的值.

答案： 9.解:
(1)原式=$1 + (\frac{2}{3})^2 × (\frac{27}{8})^{-\frac{1}{3}} + 9^2 × 0.1^{-3} = 1 + \frac{4}{9} × \frac{4}{9} - 10 + 27 = 1 + \frac{4}{9} × \frac{4}{9} × \frac{1}{(\frac{27}{8})^{\frac{1}{3}}} + 81 × 1000 - 10 + 27 = 1 + \frac{4}{9} × \frac{4}{9} × \frac{2}{3} + 81000 - 10 + 27 = 19$.
(2)原式=$\frac{16 · (ab)^{\frac{1}{2}} · (a^2b^5)^{\frac{1}{3}}}{4 · (a^2b^7)^{\frac{1}{6}} · (a^2b^5)^{\frac{1}{4}}} = \frac{4(a^{\frac{1}{2}}b^{\frac{1}{2}}) · (a^{\frac{2}{3}}b^{\frac{5}{3}})}{(a^{\frac{1}{3}}b^{\frac{7}{6}}) · (a^{\frac{1}{2}}b^{\frac{5}{4}})} = \frac{4a^{\frac{1}{2} + \frac{2}{3} - \frac{1}{3} - \frac{1}{2}}b^{\frac{1}{2} + \frac{5}{3} - \frac{7}{6} - \frac{5}{4}}}{1} = 4a^{\frac{1}{3}}b^{\frac{1}{4}}$,因为$a = 27$,$b = 16$,所以原式=$4 × 27^{\frac{1}{3}} × 16^{\frac{1}{4}} = 6$.

10. *（2025·河南周口高一期中）化简下列各式：
(1) $ \frac{x-1}{x^{\frac{2}{3}}+x^{\frac{1}{3}}+1}+\frac{x+1}{x^{\frac{1}{3}}+1}-\frac{x-x^{\frac{1}{3}}}{x^{\frac{1}{3}}-1} $；
(2) $ \frac{(a^3+a^{-3})(a^3-a^{-3})}{(a^4+a^{-4}+1)(a-a^{-1})}+\frac{a^2(1+a^{-4})-2}{a-a^{-1}} $.

答案： 10.解:
(1)原式=$\frac{(x^{\frac{1}{3} - 1})(x^{\frac{2}{3} + x^{\frac{1}{3}} + 1)}{x^{\frac{2}{3}} + x^{\frac{1}{3}} + 1} + \frac{(x^{\frac{1}{3} + 1)(x^{\frac{2}{3} - x^{\frac{1}{3}} + 1)}{x^{\frac{2}{3} + x^{\frac{1}{3}} + 1}} × \frac{x^{\frac{1}{3}}(x^{\frac{1}{3} + 1)(x^{\frac{1}{3} - 1)}{x^{\frac{1}{3} - 1}} = x^3 - 1 + \frac{x^{\frac{2}{3} - x^{\frac{1}{3}} + 1 - x^{\frac{2}{3} - x^{\frac{1}{3}}}}{x^{\frac{1}{3}}} = -x^{\frac{1}{3}}$.
(2)原式=$\frac{a^6 - a^{-6}}{(a^4 + a^{-4} + 1)(a - a^{-1})} = \frac{(a^2)^3 - (a^{-2})^3}{(a^4 + a^{-4} + 1)(a - a^{-1})} = \frac{(a^2 - a^{-2})(a^4 + a^{-2} + 1)}{(a^4 + a^{-4} + 1)(a - a^{-1})} = \frac{(a - a^{-1})(a + a^{-1})(a^4 + a^{-2} + 1)}{(a^4 + a^{-4} + 1)(a - a^{-1})} = (a + a^{-1}) + (a - a^{-1}) = 2a$.

* 已知$ 2^a·3^b=2^c·3^d=6 $，求证：$ (a-1)(d-1)=(b-1)(c-1) $.

答案： 证明:因为$2^a · 3^b = 2^c · 3^d = 6$,故$2^{a-1} · 3^{b-1} = 1$,$2^{c-1} · 3^{d-1} = 1$,所以$2^{a-1} = 3^{1-b}$,$2^{c-1} = 3^{1-d}$,所以$2^{(a-1)(1-d)} = 3^{(1-b)(1-d)}$,$2^{(c-1)(1-b)} = 3^{(1-d)(1-b)}$,故$2^{(a-1)(1-d)} = 2^{(c-1)(1-b)}$,故$(a-1)(d-1) = (b-1)(c-1)$.