20£®ÔÚÆ½ÃæÖ±½Ç×ø±êϵxOyÖУ¬ÒÑÖªÖ±Ïßl£º$\left\{\begin{array}{l}{x=1+\frac{\sqrt{3}}{2}t}\\{y=\frac{1}{2}t}\end{array}\right.$£¨tΪ²ÎÊý£©ÓëÔ²C£º$\left\{\begin{array}{l}{x=2+3cos¦È}\\{y=3sin¦È}\end{array}\right.$£¨¦ÈΪ²ÎÊý£©ÏཻÓÚA£¬BÁ½µã£®
£¨1£©ÇóÖ±Ïßl¼°Ô²CµÄÆÕͨ·½³Ì
£¨2£©ÒÑÖªF£¨1£¬0£©£¬Çó|FA|+|FB|µÄÖµ£®

·ÖÎö £¨1£©Ê¹ÓüӼõÏûÔª·¨ºÍͬ½ÇÈý½Çº¯ÊýµÄ¹ØÏµÏû²ÎÊýµÃµ½ÆÕͨ·½³Ì£»
£¨2£©½«Ö±ÏߵIJÎÊý·½³Ì´úÈëÔ²µÄÆÕͨ·½³Ì£¬¸ù¾Ý²ÎÊýµÄ¼¸ºÎÒâÒåºÍ¸ùÓëϵÊýµÄ¹ØÏµ½â³ö£®

½â´ð ½â£º£¨1£©Ö±ÏßlµÄÆÕͨ·½³ÌΪx-$\sqrt{3}y$-1=0£¬
Ô²CµÄÆÕͨ·½³ÌΪ£¨x-2£©2+y2=9£®
£¨2£©½«$\left\{\begin{array}{l}{x=1+\frac{\sqrt{3}}{2}t}\\{y=\frac{1}{2}t}\end{array}\right.$´úÈ루x-2£©2+y2=9µÃt2-$\sqrt{3}t$-8=0£¬
¡àt1+t2=$\sqrt{3}$£¬t1t2=-8£®
¡à|FA|+|FB|=|t1-t2|=$\sqrt{£¨{t}_{1}+{t}_{2}£©^{2}-4{t}_{1}{t}_{2}}$=$\sqrt{35}$£®

µãÆÀ ±¾Ì⿼²éÁ˲ÎÊý·½³ÌÓëÆÕͨ·½³ÌµÄת»¯£¬²ÎÊýµÄ¼¸ºÎÒâÒåµÄÓ¦Óã¬ÊôÓÚ»ù´¡Ì⣮

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

4£®¼×¡¢ÒÒ¡¢±ûÈýͬѧ·Ö±ð½â¡°x¡Ê[$\frac{1}{2}$£¬+¡Þ£©£¬Çóº¯Êýy=2x2+1µÄ×îСֵ¡±µÄ¹ý³ÌÈçÏ£º
¼×£ºy=2x2+1¡Ý2$\sqrt{2{x}^{2}•1}$=2$\sqrt{2}$x¡Ý2$\sqrt{2}$•$\frac{1}{2}$=$\sqrt{2}$£¬¼´yµÄ×îСֵΪ$\sqrt{2}$
ÒÒ£»y=2x2+1¡Ý2$\sqrt{2{x}^{2}•1}$=2$\sqrt{2}$x£¬µ±ÇÒ½öµ±x=$\frac{\sqrt{2}}{2}$ʱ£¬yµÄ×îСֵΪ2
±û£ºÒòΪy=2x2+1£¬ÔÚ[$\frac{1}{2}$£¬+¡Þ£©Éϵ¥µ÷µÝÔö£¬ËùÒÔyµÄ×îСֵΪ$\frac{3}{2}$
ÊÔÅжÏË­´í£¿´íÔں䦣¿

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

11£®ÒÑÖª$|{\overrightarrow a}|=1$£¬$|{\overrightarrow b}|=2$£¬$£¨{\overrightarrow a+\overrightarrow b}£©•\overrightarrow b=3$£¬$\overrightarrow a-\overrightarrow b$Óë$\overrightarrow a$µÄ¼Ð½ÇΪ¦È£¬Ôòcos¦È=$\frac{2\sqrt{7}}{7}$£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

8£®Èçͼ£¬ÔÚËÄÀâÖùABCD-A1B1C1D1ÖУ¬µ×ÃæABCDÊÇÕý·½ÐΣ¬²àÀâAA1¡Íµ×ÃæABCD£¬ÒÑÖªAB=1£¬${A}{{A}_1}=\sqrt{3}$£¬EΪABÉÏÒ»¸ö¶¯µã£¬ÔòD1E+CEµÄ×îСֵΪ£¨¡¡¡¡£©
A£®$2\sqrt{2}$B£®$\sqrt{10}$C£®$\sqrt{5}+1$D£®$2+\sqrt{2}$

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

15£®ÒÑ֪ʵÊýa£¬bÂú×ãlog2a+log2b=-2£¬Ôòa+bµÄ×îСֵΪ£¨¡¡¡¡£©
A£®$\frac{1}{4}$B£®$\frac{1}{2}$C£®1D£®4

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

5£®Ä³ÊÐÆøÏó²¿ÃŶԸÃÊÐÖÐÐijÇÇø½ü4Äê´º½ÚÆÚ¼ä£¨Ã¿Äê¾ùͳ¼Æ´º½Ú¼ÙÆÚµÄǰ7Ì죩µÄ¿ÕÆøÎÛȾָÊý½øÐÐÁËͳ¼Æ·ÖÎö£¬ÇÒ°´ÊÇ·ñȼ·Å±ÞÅÚ·Ö³ÉÁ½×飬µÃµ½ÈçͼµÄ¾¥Ò¶Í¼£¬¸ù¾Ý¹ú¼Ò×îбê×¼£¬¿ÕÆøÎÛȾָÊý²»³¬¹ý100µÄ±íʾûÓÐÎíö²£¬³¬¹ý100µÄ±íʾÓÐÎíö²£®
£¨¢ñ£©Èô´Ó¾¥Ò¶Í¼ÓÐÎíö²µÄ14ÌìÖÐËæ»ú³éÈ¡2Ì죬ÓÃËæ»ú±äÁ¿¦Î±íʾ±»³éÖÐÇÒδȼ·Å±ÞÅÚµÄÌìÊý£¬Çó¦ÎµÄ·Ö²¼Áм°ÊýѧÆÚÍû£»
£¨¢ò£©Í¨¹ý¾¥Ò¶Í¼ÌîдÏÂÃæµÄ2¡Á2ÁÐÁª±í£¬²¢ÅжÏÓжà´óµÄ°ÑÎÕ¿ÉÒÔÈÏΪȼ·Å±ÞÅÚÓë²úÉúÎíö²Óйأ¿
ȼ·Åδȼ·ÅºÏ¼Æ
ÓÐÎíö²
ÎÞÎíö²
ºÏ¼Æ
¸½£º¶ÀÁ¢ÐÔ¼ìÑ鿨·½Í³¼ÆÁ¿£º${K^2}=\frac{{n{{£¨ad-bc£©}^2}}}{£¨a+b£©£¨c+d£©£¨a+c£©£¨b+d£©}$£¬ÆäÖÐn=a+b+c+dΪÑù±¾ÈÝÁ¿£»
¶ÀÁ¢ÐÔ¼ìÑéÁÙ½çÖµ±í£º
P£¨K2¡Ýk£©0.150.100.050.0250.0100.0050.001
k2.0722.7063.8415.0246.6357.87910.828

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

12£®¸´Êýz=-3+£¨1+i£©2ÔÚ¸´Æ½ÃæÄÚ¶ÔÓ¦µÄµãÔÚ£¨¡¡¡¡£©
A£®µÚÒ»ÏóÏÞB£®µÚ¶þÏóÏÞC£®µÚÈýÏóÏÞD£®µÚËÄÏóÏÞ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

9£®Ä³Ð£¸ßһѧÀ×·æÖ¾Ô¸Ð¡×é¹²ÓÐ8ÈË£¬ÆäÖÐÒ»°à¡¢¶þ°à¡¢Èý°à¡¢Ëİà¸÷2ÈË£¬ÏÖÔÚ´ÓÖÐÈÎÑ¡3ÈË£¬ÒªÇóÿ°àÖÁ¶àÑ¡1ÈË£¬²»Í¬µÄѡȡ·½·¨µÄÖÖÊýΪ32£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖÐÊýѧ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

10£®Èçͼ£¬ÍÖÔ²C1£º$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\;£¨a£¾b£¾0£©$ºÍÔ²C2£ºx2+y2=b2£¬ÒÑÖªÔ²C2½«ÍÖÔ²C1µÄ³¤ÖáÈýµÈ·Ö£¬ÇÒÔ²C2µÄÃæ»ýΪ¦Ð£®ÍÖÔ²C1µÄ϶¥µãΪE£¬¹ý×ø±êÔ­µãOÇÒÓë×ø±êÖá²»ÖØºÏµÄÈÎÒâÖ±ÏßlÓëÔ²C2ÏཻÓÚµãA£¬B£¬Ö±ÏßEA£¬EBÓëÍÖÔ²C1µÄÁíÒ»¸ö½»µã·Ö±ðÊǵãP£¬M£®
£¨I£©ÇóÍÖÔ²C1µÄ·½³Ì£»
£¨¢ò£©Çó¡÷EPMÃæ»ý×î´óʱֱÏßlµÄ·½³Ì£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸