$$ ∴( $$ ( - \frac { 1 } { 5 } ) ^ { 2 } $$的平方根是± $$ \frac { 1 } { 5 } $$,算术平方根为 $$ \frac { 1 } { 5 } $$. (5)∵$$ \sqrt { 2 5 6 } = 1 6 $$,而$$ ( \pm 4 ) ^ { 2 } = 1 6 $$,∴ $$ \sqrt { 2 5 6 } $$的...
{ \sqrt { 1 6 0 0 0 0 } } $$(10)$$ ( - 1 4 ) ^ { 2 } $$ (11)$$ ( - \frac { 1 } { 3 } ) ^ { 2 } $$ (12)0.0036(13)32 400 (14)2.89 (15)$$ \sqrt { ( \frac { 2 2 5 } { 8 1 } ) ^ { 2 } } $$(16)$$ \frac { 1 5 ^ {...
Find the vertices and foci of the ellipse given by: \dfrac{x^2}{25} + \dfrac{y^2}{16} =1. Find the vertices and foci of the ellipse given by: \dfrac{(x+2)^2}{25} + \dfrac{(y-3)^2}{4} = 1. Find the vertices and foci of the ellipse 25 x...
The limit \lim_{h\to 0} \frac{\cos(\frac{3\pi}{2} + h) - 0}{h} represents f' (a) for some function f and some number a. Find f(x) and a. The limit \lim_h} \rightarrow {0 \frac{\sq...
∴|PM|的最小值等于点M到直线l的距离, 由点到直线距离公式得: |PM|的最小值|PM|min=|8+1+3|√4+1|8+1+3|4+1=12√551255. 故答案为:12√551255. 点评本题考查线段长的最小值的求法,是基础题,解题时要认真审题,注意点到直线的距离公式的合理运用. ...
【题目】1.用公式法解方程$$ 4 x ^ { 2 } - 1 2 x = 3 $$,得到( )$$ A . x = \frac { - 3 \pm \sqrt {
12.焦点在坐标轴上的双曲线的焦距为2 $$ \sqrt { 2 6 } $$,渐近线方程是$$ y = \pm \frac { 3 } { 2 } x $$,则它的
Find the equation of the ellipse with vertices \left ( 0,\pm 3 \right ) and foci \left ( 0,\pm 5 \right ) Find the equation of the ellipse with foci (\pm \sqrt 5,0) and vertices( \pm 3,0). Find the equation of the ellipse with foci (2,1),...
(2/3)^x = 81/16 Solve the exponential equation. 3^{x-1} = 1/81 Solve the exponential equation. (8)^{x^2} = 11 a) x = pm sqrt{dfrac{ln11}{ln8 b) x = -sqrt{dfrac{ln11}{ln8 c) x = pm sqrt{dfrac{ln8}{ln11 d) x = sqrt{dfrac{ln...
解答 解:由题意可得|PM|最小值即为点M到直线l的距离,由距离公式可得d=|2×4−(−1)+3|√22+(−1)2|2×4−(−1)+3|22+(−1)2=12√551255,故答案为:12√551255. 点评 本题考查点到直线的距离公式,属基础题.练习册系列答案 优...