计算\left(1 \frac{1}{x} \right)\div \left(\frac{{{x}^{2}} 2x 1}{x} \right)的结果是( ) A.
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/...
函数2\\\left(\frac{1}{2}\right)^{x-1},x\leq2\end{cases}" data-width="245" data-height="50" class="exam-img-50 exam-img" data-size="5787" data-format="png" style="max-width:100%">,则() A, B, C,1 D, 相关知识点: ...
一元函数与极限题目:计算极限 $\lim \limits_{x \to 0} \frac{e^x - 1 - x}{x^2}$。解析:首先将分式分离为两个部分,得到:$\lim \limits_{x \to 0} \left( \frac{e^x - 1}{x} \cdot \frac{1}{x} - \frac{1}{x} \right)$。根据极限的性质,我们将分别计算两个部分的...
leqslant 24.AM是\triangle ABC的高,设AM=x,则CM=\sqrt{64-{{x}^{2}}},BM=\sqrt{36-{{x}^{2}}}所以{{S}_{\triangle ABC}}=\frac{1}{2}\left( \sqrt{36-{{x}^{2}}}+\sqrt{64-{{x}^{2}}} \right)x=\frac{1}{2}\times 6\times 8\times \sin A\leqslant 24.y...
结果1 题目 已知函数f\left( x \right)=\frac{{{x}^{2}}-3}{{{e}^{x}}},{{f}^{\prime }}\left( x \right)为f\left( x \right)的导函数,则{{f}^{\prime }}\left( 1 \right)= . 相关知识点: 试题来源: 解析 \frac{4}{e} {{f}^{\prime}}\left( x ...
f\left( x \right)=\frac{1}{x}+1,x\in \left( \frac{1}{2},1 \right)f\left( x \right)=\frac{x-1}{x+1},x\in \left( 0,1 \right)f\left( x \right)=\frac{2x}{5x+1},x\in \left( 0+\infty \right) 相关知识点: ...
所以{{f}^{\prime }}\left( 1 \right)=-1,f\left( 1 \right)=-1,即切线方程:y=-x,下证:\frac{1}{2}{{x}^{2}}+2x\ln x-4x+\frac{5}{2}\geqslant -x,令\varphi \left( x \right)=\frac{1}{2}{{x}^{2}}+2x\ln x-3x+\frac{5}{2},...
2 f\left( x \right)=\frac{{{\left( x+1 \right)}^{2}}\sqrt{x}}{{{x}^{2}}+1}=\frac{{{x}^{2}}+1+2x+\sqrt[3]{x}}{{{x}^{2}}+1}=1+\frac{2x+\sqrt[3]{x}}{x^2+1}令g\left( x \right)=f\left( x \right)-1=\frac{2x+\sqrt[3]{x}}{{{x}^{2...
先求出函数的定义域,判断函数f\left( x \right)为偶函数,再对函数求导判断出函数f\left( x \right)在\left( 0,+\infty \right)上单调递增,然后作差比较{{\log }_{4}}5,{{\log }_{5}}6的大小,可得{{\log }_{4}}5>{{\log }_{5}}6>1>{{\log }_{6}}4>0,从而可比较...