结果1 题目整式表示(1) x\cdot (\frac{1}{2}+4x^{2});(2) (x^{2}+5y)\cdot (-y^{7}-6x);(3) y\cdot (\frac{1}{2}x-\frac{1}{3}y);(4) (a^{2}+2ab+4b^{2})\cdot (a-y)\cdot (2x+3y)-2(x-3y)\cdot (3x+4y)....
For example, the factored form of displaystyle5x³-5 is displaystyle5(x-1)left(x²+x+1right) over the integers and the reals, and displaystyle5(x-1)left(x+frac1+isqrt32right)left(x+frac1-isqrt32right) over the complex numbers. The computation of the factored form, called ...
https://math.stackexchange.com/questions/1066434/calculate-lim-x-to-0-x-cdot-sin-frac1x Your proof is incorrect, cause you used incorrect transform, but it has already been stated. I'll describe way to solve it.limx→0x1sin(x1) =1Hint : ... ...
结果1 题目5. 计算:$$ x ^ { n + 1 } \cdot x ^ { 2 n + 3 } \cdot x ^ { n - 1 } \cdot x ^ { 7 - 3 n } = $$ 相关知识点: 试题来源: 解析 5. 计算:$$ x ^ { n + 1 } \cdot x ^ { 2 n + 3 } \cdot x ^ { n - 1 } \cdot x ^ { ...
{c}\right|}{\left|{P}^{c}\right|}\) and \({\alpha }_{N}^{c}=\frac{\left|{P}^{c}\right|+\left|{N}^{c}\right|}{\left|{N}^{c}\right|}\) for solving the imbalance between positive and negative; α(ν) is set to ω if ν is the targeted view, 1 for the ...
\$7 \cdot \frac { 3 x - 1 } { 2 x - 2 } - \frac { 2 x } { 3 x - 3 } = \frac { 1 } { 2 }\$ 相关知识点: 试题来源: 解析 7.两边同乘6(x-1),得9x-3-4x=3x-3, 7.两边同乘6(x-1),得9x-3-4x=3x-3, 7.两边同乘6(x-1),得9x-3-...
$$ \begin{array}{@{}rcl@{}} \frac{\mu }{\rho_{\mathrm{e}}} = p \cdot Z_{\text{eff}}^{n} + q, \end{array} $$ (14) with the three fit parameters p, q, and n. The fit parameters have to be determined in a prior GBPC-CT calibration experiment of a phantom with ...
y'=\ln x+x\cdot \frac{1}{x}\Longrightarrow \ln x+1=0\Longrightarrow x=\frac{1}{\...
结果1 题目 13 \frac{(1+ \frac{3}{4}) \times (2+ \frac{6}{7})x \cdot \cdot \cdot \cdot (8+ \frac{24}{25}) \times (9+ \frac{27}{28})}{(1- \frac{3}{2}) \times (2- \frac{6}{2})x \cdots \times (8- \frac{24}{ ___。 相关知识点: 试题来源:...
https://socratic.org/questions/how-do-you-multiply-frac-x-2-root-3-x-cdot-sqrt-x-3-x-sqrt-x (x23x Explanation: xxx23x⋅x3 ∴=x×x213x×x2×x23 ... How do you simplify...