解析 B正确率: 46%, 易错项: C 解:原式=\left( \frac{x}{x}+\frac{1}{x} \right)\div \frac{{{\left( x+1 \right)}^{2}}}{x}=\frac{\left( x+1 \right)}{x}\cdot \frac{x}{{{\left( x+1 \right)}^{2}}}=\frac{1}{x+1}.故选:B....
故当-\infty < x < -1时z增大,又\underset{x\to -\infty }{\mathop{\lim }}\,z=0,因而z>0.于是,在\left( -\infty ,-1 \right)内{{y}^{\prime}}>0.因此,函数{{\left( 1+\frac{1}{x} \right)}^{x}}在区间\left( -\infty ,-1 \right)内增大....
f ( x ) = a ^ { x }, f\left(\frac{1}{x}\right)=a^{\frac{1}{x}}\ne-f(x) 所以②不是“倒负”变换的函数 对于函数 f ( x ) = x - \frac { 1 } { x }, f \left( \frac { 1 } { x } \right) = \frac { 1 } { x } - x = - f ( x ) 所以③是...
\left(\frac{1}{x}\right)^{\prime}=___ 相关知识点: 试题来源: 解析 -\frac{1}{x^{2}} 此题考察了幂函数的求导问题,注意指数的大小。可以看做x^{-1}的求导,幂函数求导应把指数化成系数且指数减一,故答案为-x ^{-2}反馈 收藏
1+\frac{1}{x}=mx的两不同实根 \Rightarrow x^2-\frac{1}{m}x-\frac{1}{m}=0在 \left(-\infty,-1\right)有两个不同实根 \begin{cases} {\Delta =\dfrac{1}{{m}^{2} }+\dfrac{4}{m}>0 }\\ {g(-1)=1+\dfrac{1}{m}-\dfrac{1}{m} >0\Rightarrow -\dfrac{1}{4} ...
原式（直接求导）=ln(x+1)/1=0
https://socratic.org/questions/how-do-you-solve-x-3-5-1 x=1 Explanation: We can use some rules to find out. First rule we'll use is x−1=x1 : x−53=1 x531=11 ... What is (x−21)−4 ? https://socratic.org/questions/...
\$\left| \frac { x } { 1 } + x \right| \frac { x } { 1 + x }\$ 解集是 相关知识点: 试题来源: 解析 \$( - 1 , 0 ) \cup ( 0 , + \infty ) \cup \left( - \infty , - \frac { 3 } { 2 } \right)\$
∴\begin{cases}x-1>0 \\ x\ne 0 \\\end{cases},解得x < 1且x\ne 0,故函数f\left( x \right)=\frac{\ln \left( 1-x \right)}{x}的定义域为\left( -\infty ,0 \right)\cup \left( 0,1 \right).故答案为: \left( -\infty ,0 \right)\cup \left( 0,1 \right)....
f\left( 2 \right)+f\left( \frac{1}{2} \right)=\frac{1}{1+2}+\frac{1}{1+\dfrac{1}{2}}=\frac{1}{3}+\frac{2}{3}=1,f\left( 3 \right)+f\left( \frac{1}{3} \right)=\frac{1}{1+3}+\frac{1}{1+\dfrac{1}{3}}=\frac{1}{4}+\frac{3}{4}=1,\vdo...