2.1代数运算习题

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代数运算习题

A 组习题

1. 2019全国II卷理4已知\(\displaystyle M_1,M_2,R,r\)满足 \(\displaystyle \frac{M_1}{(R+r)^2}+\frac{M_2}{r^2}=(R+r)\frac{M_1}{R^3}\),设\(\displaystyle \alpha =\frac{r}{R}\),由于\(\displaystyle \alpha\)的值很小,在近似计算中有\(\displaystyle \frac{3\alpha^3+3\alpha^4+\alpha^5}{(1+\alpha)^2}\approx 3\alpha^3\),则\(\displaystyle r\)的近似值为

   • \(\displaystyle \sqrt{\frac{M_2}{M_1}}R\)
   • \(\displaystyle \sqrt{\frac{M_2}{2M_1}}R\)
   • \(\displaystyle \sqrt[3]{\frac{3M_2}{M_1}}R\)
   • \(\displaystyle \sqrt[3]{\frac{M_2}{3M_1}}R\)
   答案

   D.

   1. 推导下述对数运算公式,其中所有对数的底数\(\displaystyle a\)均满足\(\displaystyle a>0,a\neq 1\),所有真数\(\displaystyle x\)均满足\(\displaystyle x>0\)：
   2. \(\displaystyle a^{\log_a N} = N\)
   3. \(\displaystyle \log_a(xy) = \log_ax + \log_ay,\log_a\left(\frac{x}{y}\right) = \log_ax - \log_a y\)；
   4. \(\displaystyle \log_{a^p}(b^q) = \frac{q}{p} \cdot \log_a b \quad (p,q \in \mathbb{R},p\neq 0)\)；
   5. \(\displaystyle \log_a b = \frac{\log_c b}{\log_c a}\) （\(\displaystyle c\) 为任意符合条件的底数）；\(\displaystyle \log_a b = \frac{1}{\log_b a},\log_a b \cdot \log_b c = \log_a c\)；
   答案

   （1）设 \(\displaystyle y=\log_aN\)，据对数的定义有 \(\displaystyle a^y=N\)，故 \(\displaystyle a^{\log_aN}=a^y=N\)。

   （2）设 \(\displaystyle u=\log_ax\)，\(\displaystyle v=\log_ay\)，则 \(\displaystyle x=a^u\)，\(\displaystyle y=a^v\)。于是 $\(\displaystyle \log_a(xy)=\log_a(a^{u+v})=u+v=\log_ax+\log_ay,\)$ 同理可以证明同底对数除法公式。

   （3）设\(\displaystyle \log_ab=k\)，于是\(\displaystyle a^k=b\)，由\(\displaystyle a^{pqk}=b^{pq}\)可知\(\displaystyle (a^p)^{\frac{qk}{p}}=b^q\)，于是有\(\displaystyle \log_{a^p}(b^q)=\frac{qk}{p}=\frac{q}{p}\log_ab\)。

   （4）设\(\displaystyle \log_ca=p,\log_cb=q\)，于是\(\displaystyle c^p=a\)，\(\displaystyle c^q=b\)，有$\(\displaystyle \log_ab=\log_{c^p}c^q=\frac{q}{p}=\frac{\log_cb}{\log_ca}\)$ 由此证明了换底公式。令\(\displaystyle c=a\)，立即得到\(\displaystyle \log_ba=\frac1{\log_ab}\)， 并且\(\displaystyle \log_ab\cdot\log_bc=\frac{\log_bc}{\log_ba}=\log_ac.\)

   1. 计算或化简：

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   (2)
       \task  $\displaystyle (\frac{1}{4})^{-\frac{1}{2}}\cdot \frac{(\sqrt{4ab^{-1}})^3}{(0.1)^{-2}(a^3b^4)^{\frac{1}{2}}}$
       \task $\displaystyle \sqrt[3]{\frac{(\sqrt{a}-\sqrt{a-1})^5}{\sqrt{a}+\sqrt{a-1}}}+\sqrt[3]{\frac{(\sqrt{a}+\sqrt{a-1})^5}{\sqrt{a}-\sqrt{a-1}}}$
       \task $\displaystyle \frac{x-1}{x^{\frac{2}{3}}+x^{\frac{1}{3}}+1}+\frac{x+1}{x^{\frac{1}{3}}+1}-\frac{x-x^{\frac{1}{3}}}{x^{\frac{1}{3}}-1}$
   

   答案

   （1）\(\displaystyle \frac{4}{25}b^{-\frac{7}{2}}\).

   （2）\(\displaystyle 4a-2\).提示：令 \(\displaystyle u=\sqrt a-\sqrt{a-1}\)，\(\displaystyle v=\sqrt a+\sqrt{a-1}\)，可知 \(\displaystyle uv=1\)。

   （3）\(\displaystyle -x^{\frac{1}{3}}\).提示：令 \(\displaystyle t=x^{1/3}\)，用立方和与立方差公式。

   1. 解答下述问题：
   2. 已知\(\displaystyle x^{\frac{1}{2}}+x^{-\frac{1}{2}}=3\)，求\(\displaystyle \frac{x^{\frac{3}{2}}+x^{-\frac{3}{2}}+2}{x^2+x^{-2}+3}\)的值；
   3. 已知\(\displaystyle a^{2x}=\sqrt{2}+1\)，求\(\displaystyle \frac{a^{3x}+a^{-3x}}{a^x+a^{-x}}\)的值。
   答案

   （1）\(\displaystyle \frac{2}{5}\).（2）\(\displaystyle 2\sqrt{2}-1\).提示：均使用立方和公式。

   1. 利用对数进行计算：

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   (2)
       \task $\displaystyle -\log_3(\log_3\sqrt[3]{\sqrt[3]{\sqrt[3]{3}}})$
       \task $\displaystyle 20^{\ln{\ln24}}-(\ln24)^{\ln20}$
       \task $\displaystyle \frac{1}{2}\lg\frac{32}{49}-\frac{4}{3}\lg\sqrt{8}+\lg\sqrt{245}$
       \task $\displaystyle \frac{\lg 5\cdot\lg400+(\lg2^{\sqrt{2}})^2}{(\mathrm{e}-1)^0-(\frac{9}{4})^{-0.5}+(\frac{3}{2})^{-1}}$
       \task $\displaystyle 5^{\lg 30}\times 3^{\lg 2}$
       \task $\displaystyle \lg 25+\frac{2}{3}\lg 8+\lg 5\cdot\lg 20+(\lg2)^2$
       \task $\displaystyle \frac{(1-\log_63)^2+\log_62\cdot\log_618}{\log_64}$
   
   \setcounter{task}{7}
   (1)
       \task $\displaystyle 8^{0.25}\cdot \sqrt[4]{2}+(\sqrt[3]{2}\cdot \sqrt{3})^6-\sqrt{(-\frac{3}{2})^{\frac{2}{3}}}\cdot [(\frac{4}{9})^{-\frac{1}{3}}]-1$
   \task  $\displaystyle 10(2+\sqrt{5})^{-1}-(\frac{1}{500})^{-\frac{1}{2}}+30(\frac{125}{9})^{\frac{1}{2}}(\frac{\sqrt{5}}{3})^{\frac{1}{2}}$
   \task $\displaystyle \frac{\log_{5}\sqrt{2}\cdot \log_79}{\log_5\frac{1}{3}\cdot \log_7\sqrt[3]{4}}+\log_2(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}})$
   

   答案

   (1)\(\displaystyle 3\)；(2)\(\displaystyle 0\)；(3)\(\displaystyle 1/2\)；(4)\(\displaystyle 2\)；(5)\(\displaystyle 15\)；(6)\(\displaystyle 3\)；(7)\(\displaystyle 1\).

   1. 计算或化简
   答案

   (1) \(\displaystyle 215/2\)；(2) \(\displaystyle -20+(50\cdot5^{3/4})/\sqrt3\)；(3) \(\displaystyle -1\)。

   1. 解答下述问题：
      1. 已知\(\displaystyle \log_{18}9=a,18^b=5\),用\(\displaystyle a,b\)表示\(\displaystyle \log_{36}45\);
   2. 已知\(\displaystyle \lg 2=a\),用\(\displaystyle a\)表示\(\displaystyle \lg \frac{10^{\frac{1}{3}}\sqrt{250}}{2\sqrt[3]{5}}\).
   答案

   （1）\(\displaystyle \frac{a+b}{2-a}\).提示：将\(\displaystyle \log_{36}45\)变形为\(\displaystyle \frac{\log_{18}45}{\log_{18}36}\).

   （2）\(\displaystyle \frac{3}{2}-\frac{5a}{3}\)。

   1. 解答下述问题
      1. 设\(\displaystyle a\log_34=2\)，求\(\displaystyle 4^{-a}\).
   2. 已知\(\displaystyle a>b>1\),若\(\displaystyle \log_ab+\log_ba=\frac{5}{2},a^b=b^a\),求\(\displaystyle a,b\).
   3. 若实数 \(\displaystyle m>1\) 满足 \(\displaystyle \log_{9}(\log_{4}m)=2024\),求 \(\displaystyle \log_{3}(\log_{2}m)\).
   4. 已知\(\displaystyle 0<a<1,a^{2\log_3a}=81\sqrt{3}\),求\(\displaystyle \frac{1}{a^2}+\log_9a\).
   5. 正数\(\displaystyle a,b\)满足\(\displaystyle 1+\log_2a=2+\log_3b=3+\log_6(a+b)\),求\(\displaystyle \frac{1}{a}+\frac{1}{b}\).
   6. 已知\(\displaystyle xy\neq0,2^x=18^y=9^{xy}\),求\(\displaystyle x-y\).
   7. 已知正实数\(\displaystyle a,b,c\neq 1,ab\neq1\),且\(\displaystyle \log_{ab}c=\log_ac\cdot\log_bc\),求\(\displaystyle \log_ac+\log_bc\).
   8. 已知\(\displaystyle \log_3[\log_4(\log_5a)]=\log_4[\log_3(\log_6b)]=0\),求\(\displaystyle \frac{a}{b}\).
   9. 已知\(\displaystyle \log_ax=2,\log_bx=1,\log_cx=4\),求\(\displaystyle \log_{abc}x\).
   10. 已知\(\displaystyle 3^x=6^y=8^z\),求\(\displaystyle \log_3\frac{z(x-y)}{xy}\).
   11. 设 \(\displaystyle x > 0, x \neq \frac{1}{3}, x \neq \frac{1}{2}\).若 \(\displaystyle \log_{3x} 4 = \log_{2x} 16\),求 \(\displaystyle x\).
   答案

   （1）

   （2）\(\displaystyle (a,b)=(4,2)\)。

   （3）\(\displaystyle 4048+\log_32\)

   （4）\(\displaystyle 105/4\).提示：两边同时取以3为底的对数。

   （5）\(\displaystyle \frac{1}{12}\).提示：令上述相等的三式均为\(\displaystyle k\)，把\(\displaystyle a,b\)表成含\(\displaystyle k\)的式子。

   （6）\(\displaystyle 1\).提示：同（5）.

   （7）\(\displaystyle 1\).

   （8）\(\displaystyle \frac{625}{243}\).

   （9）\(\displaystyle \frac{4}{7}\).

   （10）\(\displaystyle -1\).提示：待求式的真数齐次，考虑利用换底公式表示其比值。

   （11）\(\displaystyle \frac{2}{9}\).

   1. 因式分解\settasks{ label=(\arabic*), label-width=2em, label-offset=0.2em,

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      （1）\(\displaystyle (x^2+y-2)(x^2-3y+2)\) 10. 比较\(\displaystyle \log_{2}3\),\(\displaystyle \log_{3}4\),\(\displaystyle \log_{4}5\)的大小. ??? answer "答案"

      \(\displaystyle \log_23>\log_34>\log_45\) 构造函数\(\displaystyle f\left(x\right)=\frac{\ln\left(x+1\right)}{\ln x}\),求导得到\(\displaystyle f'\left(x\right)= =\frac{x\ln x-\left(x+1\right)\ln\left(x+1\right)}{x\left(x+1\right)\left(\ln x\right)^{2}}.\)

      因 \(\displaystyle g\left(x\right)=x\ln x\) 在 \(\displaystyle x>1\) 上严格递增，故 \(\displaystyle x\ln x<\left(x+1\right)\ln\left(x+1\right)\)，即 \(\displaystyle f'\left(x\right)<0\)，\(\displaystyle f\left(x\right)\) 在\(\displaystyle (1,+\infty)\)上单调递减。于是 \(\displaystyle f\left(2\right)>f\left(3\right)>f\left(4\right)\)，从而\(\displaystyle \log_{2}3>\log_{3}4>\log_{4}5.\) 11. 【2020全国III卷12】已知\(\displaystyle 5^{5}<8^{4}\),\(\displaystyle 13^{4}<8^{5}\).设\(\displaystyle a=\log_{5}3\),\(\displaystyle b=\log_{8}5\),\(\displaystyle c=\log_{13}8\),比较\(\displaystyle a,b,c\)的大小. ??? answer "答案"

      \(\displaystyle a<b<c\)。 12. 【2003上海理15】 \(\displaystyle a_1\)、\(\displaystyle b_1\)、\(\displaystyle c_1\)、\(\displaystyle a_2\)、\(\displaystyle b_2\)、\(\displaystyle c_2\) 均为非零实数, 不等式 \(\displaystyle a_1x^2 + b_1x + c_1 > 0\) 和 \(\displaystyle a_2x^2 + b_2x + c_2 > 0\) 的解集分别为集合 \(\displaystyle M\) 和 \(\displaystyle N\), 那么“\(\displaystyle \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\)”是“\(\displaystyle M = N\)”的

   2. 充分不必要条件

   3. 必要不充分条件
   4. 充要条件
   5. 既不充分也不必要条件