腾讯课堂2017高联基础班“指数与对数”作业题-1

 

课程链接：目标2017高中数学联赛基础班-1（赵胤授课）

 

1、化简: $$\sqrt[a]{\sqrt[b]x \over \sqrt[c]{x}} \cdot \sqrt[b]{\sqrt[c]{x} \over \sqrt[a]{x}} \cdot \sqrt[c]{\sqrt[a]{x} \over \sqrt[b]{x}}.$$ 解答: $$\text{原式} = \left(x^{{1\over b} - {1\over c}}\right)^{1\over a} \cdot \left(x^{{1\over c} - {1\over a}}\right)^{1\over b}\cdot\left(x^{{1\over a} - {1\over b}}\right)^{1\over c}$$ $$= x^{{1\over ab} - {1\over ac}}\cdot x^{{1\over bc} - {1\over ab}} \cdot x^{{1\over ac} - {1\over bc}}= x^0 = 1.$$

 

2、化简: $${\left(\sqrt x - \sqrt y\right)^3 + {2x^2 \over \sqrt x} + y\sqrt y \over x\sqrt x + y\sqrt y} + {3\sqrt{xy} - 3y \over x - y}.$$ 解答: $$\text{原式} ={\left(x^{1\over2} - y^{1\over2}\right)^3 + 2x^{3\over2} + y^{3\over2} \over x^{3\over2} + y^{3\over2}} + {3x^{1\over2}\cdot y^{1\over2} - 3y \over x - y}$$ $$= {3x^{3\over2} - 3xy^{1\over2} + 3x^{1\over2}y \over x^{3\over2} + y^{3\over2}} + {3y^{1\over2}\cdot \left(x^{1\over2} - y^{1\over2}\right) \over \left(x^{1\over2} + y^{1\over2}\right) \left(x^{1\over2} - y^{1\over2}\right)}$$ $$= {3x^{1\over2} \cdot \left(x - x^{1\over2}y^{1\over2} + y\right) \over \left(x^{1\over2} + y^{1\over2}\right) \left(x - x^{1\over2}y^{1\over2} + y\right)} + {3y^{1\over2}\cdot \left(x^{1\over2} - y^{1\over2}\right) \over \left(x^{1\over2} + y^{1\over2}\right) \left(x^{1\over2} - y^{1\over2}\right)}$$ $$= {3x^{1\over2} + 3y^{1\over2} \over x^{1\over2} + y^{1\over2}}= 3.$$

 

3、计算: $${\left(9 + 4\sqrt5\right)^{3\over2} + \left(9 - 4\sqrt5\right)^{3\over2} \over \left(11 + 2\sqrt{30}\right)^{3\over2} - \left(11 - 2\sqrt{30}\right)^{3\over2}}.$$ 解答: $$\text{原式} = {\left(\sqrt5 + 2\right)^3 + \left(\sqrt5 - 2\right)^3 \over \left(\sqrt6 + \sqrt5\right)^3 - \left(\sqrt6 - \sqrt5\right)^3}$$ $$= {2\sqrt5\cdot\left(18 - 1\right) \over 2\sqrt5\cdot\left(22+1\right)}= {17 \over 23}.$$

 

4、化简: $${1\over2}\lg\left(2x + 2\sqrt{x^2 - 1}\right) + \lg\left(\sqrt{x + 1} - \sqrt{x - 1}\right).$$ 解答: $$\text{原式} = \lg\sqrt{2x + 2\sqrt{x^2 - 1}} + \lg\left(\sqrt{x + 1} - \sqrt{x - 1}\right)$$ $$= \lg\left(\sqrt{x + 1} + \sqrt{x - 1}\right) + \lg\left(\sqrt{x + 1} - \sqrt{x - 1}\right)$$ $$= \lg\left(x + 1 - x + 1\right)= \lg2.$$

 

5、已知 $\log_310 = a$, $\log_625 = b$, 求 $\log_445$.

解答:

由 $$\log_{4}45 = {\lg45 \over \lg 4} = {2\lg3 +\lg5 \over 2\lg2},$$ 考虑分别求出 $\lg3$, $\lg5$, $\lg2$. $$\log_310 = {\lg10 \over \lg 3} = a\Rightarrow\lg3 = {1\over a}.$$ $$\log_625 = {\lg25 \over \lg6} = {2\lg5 \over \lg2 + \lg3} = b$$ $$\Rightarrow \begin{cases}2\lg5 - b\lg2 = {b\over a}\\ \lg5 + \lg2 = 1 \end{cases}\Rightarrow \begin{cases}\lg2 = {2a - b \over ab + 2a}\\ \lg5 = {ab + b \over ab + 2a} \end{cases}.$$ 因此$$\log_445 = {{2\over a} + {ab + b \over ab + 2a} \over {4a - 2b \over ab + 2a}} = {ab + 3b + 4 \over 2(2a - b)}.$$

 

6、如果 $\log_8a + \log_4b^2 = 5$, 且 $\log_8b + \log_4a^2 = 7$, 求 $ab$.

解答: $$\because \begin{cases}\log_8a + \log_4b^2 = 5\\ \log_8b + \log_4a^2 = 7 \end{cases}$$ $$\Rightarrow \log_8ab + \log_4a^2b^2 = 12$$ $$\Rightarrow \log_8ab + \log_2ab = 12$$ $$\Rightarrow {1\over3}\log_2ab + \log_2ab = 12$$ $$\Rightarrow \log_2ab = 9$$ $$\therefore ab = 2^9 = 512.$$

 

7、若 $(6.2)^x = (0.0062)^y = 1000$, 则 $\displaystyle{1\over x} - {1\over y} = 1$.

解答: $$\because \begin{cases}x = \log_{6.2}1000\\ y = \log_{0.0062}1000 \end{cases}$$ $$\Rightarrow \begin{cases}{1\over x} = \log_{1000}6.2 = {1\over3}\lg6.2\\ {1\over y} = \log_{1000}0.0062 = {1\over3}\lg0.0062 \end{cases}$$ $$\therefore {1\over x} - {1\over y} = {1\over3}\lg{6.2\over 0.0062} = 1.$$

另解: $$\because \begin{cases}6.2 = 1000^{1\over x}\\ 0.0062 = 1000^{1\over y}\end{cases}$$ $$\Rightarrow1000 = 1000^{{1\over x} - {1\over y}}$$ $$\therefore {1\over x} - {1\over y} = 1.$$

 

8、设 $a, b$ 为不等于1的正有理数, 且方程 $x^2 - x\log_ba + \displaystyle{a\over b} = 0$ 的两根为 $\log_ab, \log_ab^3$, 求 $a, b$ 的值.

解答: $$\because \begin{cases}\log_ab + \log_ab^3 = \log_ba\\ \log_ab \cdot \log_ab^3 = {a\over b} \end{cases}$$ $$\Rightarrow \begin{cases}4\log_ab = \log_ba\\ 3\log_a^2b = {a\over b}\end{cases}$$ 令 $\log_ab= m$. $$\Rightarrow 4m = {1\over m}\Rightarrow m = \pm{1\over 2}$$ $$\Rightarrow {a \over b} = {3\over4}$$ $$\Rightarrow \log_ab = \log_a{4\over3}a = \pm{1\over2}$$ $$\Rightarrow a^{1\over2} = {4\over3}a\ \text{或}\ a^{-{1\over2}} = {4\over3}a$$ $$\Rightarrow a = {16\over9}a^2\ \text{或}\ {1\over a} = {16\over9}a^2$$ $$\Rightarrow \begin{cases}a_1 = {9\over16}\\ b_1 = {3\over4} \end{cases},\ \begin{cases}a_2 = {1\over4}\sqrt[3]{36}\\ b_2 = {1\over3}\sqrt[3]{36} \end{cases}.$$

 

 

 

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