4.1 指数
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【基础过关系列】2022-2023学年高一数学上学期同步知识点剖析精品讲义(人教A版2019)
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必修第一册同步巩固,难度2颗星!
基础知识
指数运算
n次方根与分数指数幂
一般地,如果\(x^n=a\),那么\(x\)叫做\(a\)的\(n\)次方根,其中\(n>1\),且\(n∈N*\).
式子 \(\sqrt[n]{a}\)叫做根式,这里\(n\)叫做根指数,\(a\)叫做被开方数.
负数没有偶次方根;\(0\)的任何次方根都是\(0\).
注:(1) \((\sqrt[n]{a})^{n}=a\)
(2)当\(n\)是奇数时, \(\sqrt[n]{a^{n}}=a\),当\(n\)是偶数时, \(\sqrt[n]{a^{n}}=|a|=\left\{\begin{array}{c}
a, a \geq 0 \\
-a, a<0
\end{array}\right.\).
【例】 若\(x^6=3\),则\(x\)等于 \(\underline{\quad \quad}\) .
解析 \(∵x^6=3\),\(∴x\)是\(3\)的\(6\)次方根,且有两个,互为相反数,
记为 \(\pm \sqrt[6]{3}\),故 \(x=\pm \sqrt[6]{3}\).
【例】 求值 \(\sqrt{(2-\pi)^{2}},(\sqrt[3]{5})^{3}, \quad \sqrt[3]{(-5)^{3}}\).
解析 \(\sqrt{(2-\pi)^{2}}=|2-\pi|=\pi-2\), \((\sqrt[3]{5})^{3}=5\), \(\sqrt[3]{(-5)^{3}}=|-5|=5\).
正数的正分数指数幂的意义
1 正数的正分数指数幂的意义
规定: \(a^{\frac{m}{n}}=\sqrt[n]{a^{m}}\left(a>0, m, n \in N^{*}\right.\)且\(n>1)\)
巧记 “子内母外”(根号内的\(m\)作分子,根号外的\(n\)作为分母) ,您想到“孕妇”便可.
【例】 把下列根式和指数幂相互转化: \(\sqrt{x}, \sqrt[3]{x^{5}}, a^{\frac{3}{4}}, a^{-\frac{2}{3}}\).
解析 \(\sqrt{x}=x^{\frac{1}{2}}, \sqrt[3]{x^5}=x^{\frac{5}{3}}, a^{\frac{3}{4}}=\sqrt[4]{a^3}, a^{-\frac{2}{3}}=\frac{1}{a^{\frac{2}{3}}}=\frac{1}{\sqrt[3]{a^2}}\).
2 正数的正分数指数幂的意义
\(a^{-\frac{m}{n}}=\dfrac{1}{a^{\frac{m}{n}}}=\dfrac{1}{\sqrt[n]{a^{m}}}\left(a>0, m, n \in N^{*}\right.\)且\(n>1)\)
3 \(0\)的正分数指数幂等于\(0\),\(0\)的负分数指数幂没有意义.
实数指数幂的运算性质
1 \(a^{s} \cdot a^{r}=a^{r+s}(a>0, r, s \in R)\);
2 \(\left(a^{s}\right)^{r}=a^{r s}(a>0, r, s \in R)\);
3 \((a b)^{r}=a^{r} b^{r}(a>0, r \in R)\).
基本方法
【题型1】指数幂运算
【典题1】 若 \(2^x=7,2^y=6\),则 \(4^{x-y}\)等于 \(\underline{\quad \quad}\) .
解析 \(∵2^x=7,2^y=6\),
\(\therefore 4^{x-y}=\dfrac{4^{x}}{4^{y}}=\left(\dfrac{2^{x}}{2^{y}}\right)^{2}=\left(\dfrac{7}{6}\right)^{2}=\dfrac{49}{36}\).
点拨 指数幂的运算要注意底数之间的关系.
【典题2】 求值 \(\left(2 \dfrac{7}{9}\right)^{\frac{1}{2}}-(2 \sqrt{3}-\pi)^{0}-\left(2 \dfrac{10}{27}\right)^{-\frac{1}{3}}+0.125^{-\frac{2}{3}}+\sqrt{3} \cdot \sqrt{\left(\dfrac{3}{4}\right)^{3}}\).
解析 原式 \(=\left(\dfrac{25}{9}\right)^{\frac{1}{2}}-1-\left(\dfrac{64}{27}\right)^{-\frac{1}{3}}+\left(\dfrac{1}{8}\right)^{-\frac{2}{3}}+3^{\frac{1}{2}} \cdot\left(\dfrac{3}{4}\right)^{\frac{3}{2}}\)
\(=\dfrac{5}{3}-1-\left(\dfrac{27}{64}\right)^{\frac{1}{3}}+\left(2^{-3}\right)^{-\frac{2}{3}}+3^{2} \cdot\left(\dfrac{1}{4}\right)^{\frac{3}{2}}\)
\(=\dfrac{2}{3}-\dfrac{3}{4}+4+\dfrac{9}{8}\)
\(=\dfrac{121}{24}\).
点拨 一般可以带分数化假分数、小数化分数、根式化幂、整数化幂.
巩固练习
1.若\(10^x=2\),则 \(10^{-3 x}\)等于( )
A.\(8\) \(\qquad \qquad\) B.\(-8\) \(\qquad \qquad\) C. \(\dfrac{1}{8}\) \(\qquad \qquad\) D. \(-\dfrac{1}{8}\)
2.若\(2^x=3,2^y=4\),则 \(2^{x+y}\)的值为( )
A.\(7\) \(\qquad \qquad\) B.\(10\) \(\qquad \qquad\) C.\(12\) \(\qquad \qquad\) D.\(34\)
3.计算: \(\sqrt[3]{(-4)^{3}}-\left(\dfrac{1}{2}\right)^{0}+0.25^{\frac{1}{2}} \times\left(\dfrac{1}{\sqrt{2}}\right)^{-4}\)
4.计算: \(\left(2 \dfrac{1}{4}\right)^{\frac{1}{2}}-\left(-\dfrac{1}{8}\right)^{0}-\left(3 \dfrac{3}{8}\right)^{-\frac{2}{3}}+(1.5)^{-2}+\sqrt{(1-\sqrt{2})^{2}}\)
参考答案
- 答案 \(C\)
解析 \(∵10^x=2\),则 \(10^{-3 x}=\dfrac{1}{\left(10^{x}\right)^{3}}=\dfrac{1}{2^{3}}=\dfrac{1}{8}\).故选:\(C\). - 答案 \(C\)
解析 \(2^{x+y}=2^{x} \cdot 2^{y}=3 \times 4=12\),故选:\(C\). - 答案 \(-3\)
解析 原式 \(=-4-1+(2)^{\frac{1}{2} \times(-2)} \times(2)^{-\dfrac{1}{2} \times(-4)}=-5+2=-3\). - 答案 \(\sqrt{2}-\dfrac{1}{2}\)
解析 原式 \(=\left(\dfrac{9}{4}\right)^{\frac{1}{2}}-1-\left(\dfrac{27}{8}\right)^{-\frac{2}{3}}+\left(\dfrac{3}{2}\right)^{-2}+\sqrt{2}-1\)\(=\dfrac{3}{2}-1-\dfrac{4}{9}+\dfrac{4}{9}+\sqrt{2}-1=\sqrt{2}-\dfrac{1}{2}\).
【题型二】条件求值问题
【典题1】 已知 \(x^{\frac{1}{2}}-x^{-\frac{1}{2}}=\sqrt{5}\),则 \(x^{2}+\dfrac{1}{x^{2}}\)的值为 \(\underline{\quad \quad}\).
解析 由 \(x^{\frac{1}{2}}-x^{-\frac{1}{2}}=\sqrt{5}\),两边平方得 \(x-2+x^{-1}=5\),则 \(x+\dfrac{1}{x}=7\),
所以 \(\left(x+\dfrac{1}{x}\right)^{2}=49 \Rightarrow x^{2}+\dfrac{1}{x^{2}}+2=49 \Rightarrow x^{2}+\dfrac{1}{x^{2}}=47\).
点拨 注意 \(x^{\frac{1}{2}}-x^{-\frac{1}{2}}, x+\dfrac{1}{x}, x^{2}+\dfrac{1}{x^{2}}\)之间平方的关系.
巩固练习
1.已知 \(a+\dfrac{1}{a}=7\),则 \(a^{\frac{1}{2}}+a^{-\frac{1}{2}}=\)( )
A.\(3\) \(\qquad \qquad\) B.\(9\)\(\qquad \qquad\) C.\(-3\) \(\qquad \qquad\) D.\(±3\)
2.若 \(a+a^{-1}=3\),则 \(a^{2}+a^{-2}\)的值为( )
A.\(9\) \(\qquad \qquad\) B.\(7\) \(\qquad \qquad\) C.\(6\) \(\qquad \qquad\) D.\(4\)
3.已知\(a,b\)是方程\(x^2-6x+4=0\)的两根,且\(a>b>0\),求 \(\dfrac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)的值.
参考答案
- 答案 \(A\)
解析 知 \(a+\dfrac{1}{a}=7\),可得\(a>0\), \(a^{\frac{1}{2}}+a^{-\frac{1}{2}}>0\),
\(\therefore a^{\frac{1}{2}}+a^{-\frac{1}{2}}=\sqrt{\left(a^{\frac{1}{2}}+a^{-\frac{1}{2}}\right)^{2}}=\sqrt{7+2}=3\).
故选:\(A\). - 答案 \(B\)
解析 \(∵a+a^{-1}=3\), \(\therefore\left(a+a^{-1}\right)^{2}=a^{2}+a^{-2}+2=9\), \(\therefore a^{2}+a^{-2}=7\).
故选:\(B\). - 答案 \(\dfrac{\sqrt{5}}{5}\)
解析 \(∵a ,b\)是方程\(x^2-6x+4=0\)的根,
\(∴\)由根与系数关系得 \(\left\{\begin{array}{l} a+b=6 \\ a b=4 \end{array}\right.\),
又\(∵a>b>0\), \(\therefore \sqrt{a}>\sqrt{b}\).
\(\because\left(\dfrac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}\right)^{2}=\dfrac{a+b-2 \sqrt{a b}}{a+b+2 \sqrt{a b}}=\dfrac{6-2 \sqrt{4}}{6+2 \sqrt{4}}=\dfrac{2}{10}=\dfrac{1}{5}\),
\(\therefore \dfrac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}=\sqrt{\dfrac{1}{5}}=\dfrac{\sqrt{5}}{5}\).
分层练习
【A组---基础题】
1.已知\(a>0\),则 \(\dfrac{a^{2}}{\sqrt{a} \cdot \sqrt[3]{a^{2}}}=\) ( )
A. \(a^{\frac{6}{5}}\) \(\qquad \qquad\) B. \(a^{\frac{5}{6}}\) \(\qquad \qquad\) C. \(a^{-\frac{5}{6}}\) \(\qquad \qquad\) D. \(a^{\frac{5}{3}}\)
2.下列各式正确的是( )
A. \(\sqrt{(-3)^{2}}=3\) \(\qquad \qquad\) B. \(\sqrt[4]{a^{4}}=a\) \(\qquad \qquad\) C. \(\sqrt{2^{2}}=2\) \(\qquad \qquad\) D.\(a^0=1\)
3.计算 \(\sqrt[3]{(2-\pi)^{3}}+\sqrt{(3-\pi)^{2}}\)的值为 ( )
A.\(5\) \(\qquad \qquad\) B.\(-1\)\(\qquad \qquad\) C.\(2π-5\) \(\qquad \qquad\)D.\(5-2π\)
4.计算 \(2 x^{2} \cdot\left(-3 x^{3}\right)\)的结果是 ( )
A.\(-6x^5\) \(\qquad \qquad\) B.\(6x^5\) \(\qquad \qquad\) C.\(-2x^6\) \(\qquad \qquad\)D. \(2x^6\)
5.计算: \((\sqrt{3}-2)^{2018} \cdot(\sqrt{3}+2)^{2019}=\)\(\underline{\quad \quad}\) .
6.已知 \(x^{\frac{1}{2}}-x^{-\frac{1}{2}}=\sqrt{5}\),则 \(x+\dfrac{1}{x}\)的值为\(\underline{\quad \quad}\).
7.计算: \(\left(2 \dfrac{4}{5}\right)^{0}+2^{-2} \times\left(2 \dfrac{1}{4}\right)^{-\frac{1}{2}}-\left(\dfrac{8}{27}\right)^{\frac{1}{3}}\)
8.计算: \(\sqrt{\dfrac{25}{9}}-\left(\dfrac{8}{27}\right)^{\frac{1}{3}}-(\pi+e)^{\circ}+\left(\dfrac{1}{4}\right)^{-\frac{1}{2}}\)
9.已知函数方程\(x^2-8x+4=0\)的两根为\(x_1 、x_2 (x_1<x_2 )\)
(1)求 \(x_{1}^{-2}-x_{2}^{-2}\)的值. (2)求 \(x_{1}^{-\frac{1}{2}}-x_{2}^{-\frac{1}{2}}\)的值.
参考答案
- 答案 \(B\)
解析 \(\dfrac{a^{2}}{\sqrt{a} \cdot \sqrt[3]{a^{2}}}=\dfrac{a^{2}}{a^{\frac{1}{2}} \cdot a^{\frac{2}{3}}}=\dfrac{a^{2}}{a^{\frac{7}{6}}}=a^{\frac{5}{6}}\),故选:\(B\). - 答案 \(C\)
解析 根据根式的性质可知\(C\)正确. \(\sqrt[4]{a^{4}}=|a|\),\(a^0=1\)条件为\(a≠0\),故\(A,B,D\)错. - 答案 \(B\)
解析 \(\sqrt[3]{(2-\pi)^{3}}+\sqrt{(3-\pi)^{2}}=2-\pi+\pi-3=-1\),故选:\(B\). - 答案 \(A\)
解析 \(2 x^{2} \cdot\left(-3 x^{3}\right)=-6 x^{2+3}=-6 x^{5}\).故选\(A\). - 答案 \(\sqrt{3}+2\)
解析 原式 \(\left.=[(\sqrt{3}-2)(\sqrt{3}+2)]^{2018} \cdot(\sqrt{3}+2)=(-1)\right]^{2018} \cdot(\sqrt{3}+2)=\sqrt{3}+2\). - 答案 \(7\)
解析 由 \(x^{\frac{1}{2}}-x^{-\frac{1}{2}}=\sqrt{5}\),两边平方得: \(x-2+x^{-1}=5\),则 \(x+\dfrac{1}{x}=7\). - 答案 \(\dfrac{1}{2}\)
解析 原式 \(=1+\dfrac{1}{4} \times \dfrac{2}{3}-\dfrac{2}{3}=\dfrac{1}{2}\). - 答案 \(2\)
解析 原式 \(=\dfrac{5}{3}-\dfrac{2}{3}-1+2=2\). - 答案 (1) \(2 \sqrt{3}\);(2) \(1\).
解析 \(∵x_1+x_2=8\),\(x_1⋅x_2=4\),
(1) \(x_{1}^{-2}-x_{2}^{-2}=\dfrac{\left(x_{1}+x_{2}\right)\left(x_{2}-x_{1}\right)}{\left(x_{1} x_{2}\right)^{2}}\)\(=\dfrac{x_{2}-x_{1}}{2}=\dfrac{\sqrt{\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}}}{2}=\dfrac{\sqrt{64-16}}{2}=2 \sqrt{3}\);
(2) \(x_{1}^{-\frac{1}{2}}-x_{2}^{-\frac{1}{2}}=\dfrac{\sqrt{x_{1}+x_{2}-2 \sqrt{x_{1} x_{2}}}}{\sqrt{x_{1} x_{2}}}=\dfrac{\sqrt{8-2 \times 4}}{2}=1\).
【B组---提高题】
1.化简 \(\sqrt[4]{16 x^{8} y^{4}}(x<0, y<0)\)得( )
A.\(2x^2 y\) \(\qquad \qquad\) B.\(2xy\) \(\qquad \qquad\) C.\(4x^2 y\) \(\qquad \qquad\) D.\(-2x^2 y\)
2.已知\(ab=-5\),则 \(a \sqrt{-\dfrac{b}{a}}+b \sqrt{-\dfrac{a}{b}}\)的值是( )
A. \(2 \sqrt{5}\) \(\qquad \qquad\) B.\(0\) \(\qquad \qquad\) C. \(-2 \sqrt{5}\) \(\qquad \qquad\) D. \(\pm 2 \sqrt{5}\)
3.如果 \(45^x=3\),\(45^y=5\),那么\(2x+y=\)\(\underline{\quad \quad}\).
4.若 \(2^{x}=8^{y+1}\),且 \(9^{y}=3^{x-9}\),则\(x+y\)的值是\(\underline{\quad \quad}\).
5.化简 \(\sqrt{11+6 \sqrt{2}}+\sqrt{11-6 \sqrt{2}}=\)\(\underline{\quad \quad}\).
6.已知 \(12^{x}+2^{-x}=a\) (常数),求 \(8^{x}+8^{-x}\)的值.
参考答案
- 答案 \(D\)
解析 \(∵x<0,y<0\),\(∴\)原式 \(=2^{\frac{4}{4}} x^{2}(-y)=-2 x^{2} y\).故选:\(D\). - 答案 \(B\)
解析 \(∵ab=-5\),\(∴a\)与\(b\)异号,
\(\therefore a \sqrt{-\dfrac{b}{a}}+b \sqrt{-\dfrac{a}{b}}=a \sqrt{-\dfrac{a b}{a^{2}}}+b \sqrt{-\dfrac{a b}{b^{2}}}=a \sqrt{\dfrac{5}{a^{2}}}+b \sqrt{\dfrac{5}{b^{2}}}\)\(=a \dfrac{\sqrt{5}}{|a|}+b \dfrac{\sqrt{5}}{|b|}=0\),
故选:\(B\). - 答案 \(1\)
解析 由 \(45^x=3\),得 \(\left(45^{x}\right)^{2}=9\),
则\(45^{2 x} \times 45^{y}=9 \times 5=45=1\).
\(\therefore 45^{2 x+y}=45\),\(∴2x+y=1\). - 答案 \(27\)
解析 \(2^{x}=8^{y+1}\),\(∴\)有 \(2^{x}=2^{3 y+3}\), \(∴x=3y+3\)
又 \(9^{y}=3^{x-9}\),\(∴\)有 \(3^{2 y}=3^{x-9}\), \(∴2y=x-9\)
联立 \(\left\{\begin{array}{l} x=3 y+3 \\ 2 y=x-9 \end{array}\right.\)得到 \(\left\{\begin{array}{l} x=21 \\ y=6 \end{array}\right.\),\(∴x+y=27\). - 答案 \(6\)
解析 \(\sqrt{11+6 \sqrt{2}}+\sqrt{11-6 \sqrt{2}}=\sqrt{(3+\sqrt{2})^{2}}+\sqrt{(3-\sqrt{2})^{2}}\)\(=3+\sqrt{2}+3-\sqrt{2}=6\). - 答案 \(a^3-3a\)
解析 (方法一) \(8^{x}+8^{-x}=2^{3 x}+2^{-3 x}=\left(2^{x}\right)^{3}+\left(2^{-x}\right)^{3}\)
\(=\left(2^{x}+2^{-x}\right)\left[\left(2^{x}\right)^{2}-2^{x} \cdot 2^{-x}+\left(2^{-x}\right)^{2}\right]\)\(=\left(2^{x}+2^{-x}\right)\left[\left(2^{x}+2^{-x}\right)^{2}-3 \cdot 2^{x} \cdot 2^{-x}\right]\)
\(=\left(2^{x}+2^{-x}\right)\left[\left(2^{x}+2^{-x}\right)^{2}-3\right]=a\left(a^{2}-3\right)=a^{3}-3 a\).
(方法二)令\(2^x=t\),则 \(2^{-x}=t^{-1}\),
所以 \(t+t^{-1}=a\),两边平方整理得 \(t^{2}+t^{-2}=a^{2}-2\),
则 \(8^{x}+8^{-x}=t^{3}+t^{-3}=\left(t+t^{-1}\right)\left(t^{2}-t \cdot t^{-1}+t^{-2}\right)=a^{3}-3 a\).
【C组---拓展题】
1.已知 \(2^{a}=3^{b}=6\),则\(a,b\)不可能满足的关系是( )
A.\(a+b=ab\) \(\qquad\) B.\(a+b>4\) \(\qquad\) C. \((a-1)^2+(b-1)^2<2\) \(\qquad\) D. \(a^2+b^2>8\)
2.已知 \(a^{2 n}=\sqrt{2}+1\),求 \(\dfrac{a^{3 n}+a^{-3 n}}{a^{n}+a^{-n}}\)的值.
参考答案
- 答案 \(C\)
解析 \(\because 2^{a}=3^{b}=6\), \(\therefore\left(2^{a}\right)^{b}=6^{b},\left(3^{b}\right)^{a}=6^{a}\),
\(\therefore 2^{a b}=6^{b}, 3^{b a}=6^{a}\),
\(\therefore 2^{a b} \cdot 3^{b a}=6^{b} \cdot 6^{a}\),
\(\therefore 6^{a b}=6^{a+b}\),
\(∴ab=a+b\),则有 \(a b=a+b \geq 2 \sqrt{a b}\),
\(∵a≠b\), \(\therefore a b>2 \sqrt{a b}\),
\(∴a+b=ab>4\),
\(∴(a-1)^2+(b-1)^2=a^2+b^2-2(a+b)+2>2ab-2(a+b)+2>2\),
\(∵a^2+b^2>2ab>8\),故\(C\)错误
故选:\(C\). - 答案 \(2 \sqrt{2}-1\)
解析 设 \(a^{n}=t>0\),则 \(t^{2}=\sqrt{2}+1\),
\(\dfrac{a^{3 n}+a^{-3 n}}{a^{n}+a^{-n}}=\dfrac{t^{3}+t^{-3}}{t+t^{-1}}=t^{2}-1+t^{-2}\)\(=\sqrt{2}+1-1+\dfrac{1}{\sqrt{2}+1}=2 \sqrt{2}-1\).