题目内容
13.化简:$\frac{m-{m}^{-1}}{{m}^{\frac{2}{3}}-{m}^{-\frac{2}{3}}}$-$\frac{m+{m}^{-1}}{{m}^{\frac{2}{3}}+2+{m}^{-\frac{2}{3}}}$+$\frac{2m}{{m}^{\frac{2}{3}}+1}$(m>0)分析 根据完全平方公式,平方差公式,立方和公式,立方差公式及对数的运算性质,可化简原式.
解答 解::$\frac{m-{m}^{-1}}{{m}^{\frac{2}{3}}-{m}^{-\frac{2}{3}}}$-$\frac{m+{m}^{-1}}{{m}^{\frac{2}{3}}+2+{m}^{-\frac{2}{3}}}$+$\frac{2m}{{m}^{\frac{2}{3}}+1}$
=$\frac{{(m}^{\frac{1}{3}}-{m}^{-\frac{1}{3}})({m}^{\frac{2}{3}}+1+{m}^{-\frac{2}{3}})}{{(m}^{\frac{1}{3}}-{m}^{-\frac{1}{3}}){(m}^{\frac{1}{3}}+{m}^{-\frac{1}{3}})}$-$\frac{{(m}^{\frac{1}{3}}+{m}^{-\frac{1}{3}})({m}^{\frac{2}{3}}-1+{m}^{-\frac{2}{3}})}{{(m}^{\frac{1}{3}}+{m}^{-\frac{1}{3}})^{2}}$+$\frac{2m}{{m}^{\frac{2}{3}}+1}$
=$\frac{({m}^{\frac{2}{3}}+1+{m}^{-\frac{2}{3}})}{{(m}^{\frac{1}{3}}+{m}^{-\frac{1}{3}})}$-$\frac{({m}^{\frac{2}{3}}-1+{m}^{-\frac{2}{3}})}{{(m}^{\frac{1}{3}}+{m}^{-\frac{1}{3}})}$+$\frac{2m}{{m}^{\frac{2}{3}}+1}$
=$\frac{2}{{(m}^{\frac{1}{3}}+{m}^{-\frac{1}{3}})}$+$\frac{2m}{{m}^{\frac{2}{3}}+1}$
=$\frac{2{m}^{\frac{1}{3}}}{{m}^{\frac{2}{3}}+1}$+$\frac{2m}{{m}^{\frac{2}{3}}+1}$
=$\frac{2{m}^{\frac{1}{3}}({m}^{\frac{2}{3}}+1)}{{m}^{\frac{2}{3}}+1}$
=$2{m}^{\frac{1}{3}}$
点评 本题考查的知识点是有理数指数幂的化简求值,熟练掌握乘法公式,是解答的关键.
A. | m≤-3 | B. | m≥3 | C. | m≤-3或m≥3 | D. | m≥-3或m≤3 |
x | 4.25 | 1.57 | -0.61 | -0.59 | 0 | 0.42 | -0.35 | 0.56 | 0.26 | 3.27 |
y | -226.05 | -10.04 | 0.07 | 0.03 | 0 | 0.20 | -0.22 | 0.03 | 0.21 | -101.63 |
(1)函数y=f(x)在区间[0.55,0.6]上是否存在零点,写出判断并说明理由;
(2)证明:函数y=f(x)在区间[0.41,+∞)单调递减.
A. | $\frac{1+y}{1-y}$ | B. | ln$\frac{1+y}{1-y}$ | C. | $\frac{1}{2}$ln$\frac{1+y}{1-y}$ | D. | $\frac{1}{2}$ln$\frac{1-y}{1+y}$ |
A. | 36 | B. | 25 | C. | 16 | D. | 9 |