解析:"a,b全为0"即"a=0且b=0",因此它的反面应为"a≠0或b≠0",即a,b不全为0.

答案:a,b不全为0

6.有下列叙述:①"a>b"的反面是"ay或x

解析:①错,应为a≤b;②对;③错,应为三角形的外心在三角形内或三角形的边上;④错,应为三角形的内角中有两个或三个钝角.

答案:②

7.若△A1B1C1的三个内角的余弦值分别等于△A2B2C2的三个内角的正弦值,则△A1B1C1为　　　　　三角形,△A2B2C2为　　　　　三角形(填"锐角"或"钝角").

解析:由△A1B1C1的三个内角的余弦值均大于0,可知△A1B1C1为锐角三角形.则由题意,知△A2B2C2为锐角三角形或钝角三角形.假设△A2B2C2是锐角三角形,由{■(sin" " A_2=cos" " A_1=sin(π/2 "-" A_1 )"," @sin" " B_2=cos" " B_1=sin(π/2 "-" B_1 )"," @sin" " C_2=cos" " C_1=sin(π/2 "-" C_1 )"," )┤得{■(A_2=π/2 "-" A_1 "," @B_2=π/2 "-" B_1 "," @C_2=π/2 "-" C_1 "," )┤

　　∴A2+B2+C2=π/2与A2+B2+C2=π矛盾.

　　∴△A2B2C2是钝角三角形.

答案:锐角　钝角

8.导学号88184005证明对于直线l:y=kx+1,不存在这样的实数k,使得l与双曲线C:3x2-y2=1的交点A,B关于直线y=ax(a为常数)对称.

证明假设存在实数k,使得点A,B关于直线y=ax对称,设A(x1,y1),B(x2,y2),则有(1)直线l:y=kx+1与直线y=ax垂直;(2)点A,B在直线l:y=kx+1上;(3)线段AB的中点((x_1+x_2)/2 "," (y_1+y_2)/2)在直线y=ax上.

　　所以{■(ka="-" 1"," @y_1+y_2=k"(" x_1+x_2 ")" +2"," @(y_1+y_2)/2=a"·" (x_1+x_2)/2 "," )┤0├ ■("①" @"②" @"③" )┤

　　由②③,得a(x1+x2)=k(x1+x2)+2.0④

　　由{■(y=kx+1"," @y^2=3x^2 "-" 1"," )┤得(3-k2)x2-2kx-2=0,

　　因此x1+x2=2k/(3"-" k^2 ),将其代入④,得ak=3.

　　这与①矛盾,所以假设不成立.因此不存在实数k,使得点A,B关于直线y=ax对称.

B组

1.已知a+b+c>0,ab+bc+ac>0,abc>0,用反证法证明a>0,b>0,c>0时的"假设"为(　　)

A.a<0,b<0,c<0 B.a≤0,b>0,c>0

C.a,b,c不全是正数 D.abc<0

解析:"a>0,b>0,c>0"即"a,b,c都是正数",因此其否定应为"a,b,c不全是正数".

答案:C