职高数学问题
\u5173\u4e8e\u804c\u9ad8\u6570\u5b66\u7684\u51e0\u9053\u9898\u76ee(1) a²+b²=(a+b)^2-2ab
\u7528\u97e6\u8fbe\u5b9a\u7406\u53ef\u77e5\uff1a a+b=-(2-k), ab=k²+3k+5
\u6240\u4ee5 a²+b²=(a+b)^2-2ab=(k-2)^2-2k^2-6k-10= -k^2-10k-6
=-(k^2+10k+25-19)=-(k-5)^2+19
\u6240\u4ee5 \u5f53k=5\u65f6,a²+b²\u6709\u6700\u5927\u503c\u4e3a 19
(2) (x-1)²+(y-1)²=x^2-2x+1+y^2-2y+1=x^2+y^2-2(x+y)+2
=(x+y)^2-2xy-2(x+y)+2
\u7528\u97e6\u8fbe\u5b9a\u7406\u53ef\u77e5\uff1ax+y=2a, xy=a+6
\u6240\u4ee5 (x-1)²+(y-1)²=(x+y)^2-2xy-2(x+y)+2=4a^2-2(a+6)-4a+2
=4a^2-6a-10=2(2a^2-3a+9/8-49/8)=2((\u6839\u53f72)a-(3/4)*(\u6839\u53f72))^2-49/4
\u6240\u4ee5 \u5f53a=3/4\u65f6\uff0c (x-1)²+(y-1)²\u6709\u6700\u5c0f\u503c\u4e3a -49/4
1\uff09|\u5411\u91cfa|=\u221a(3^2+4^2)=5
|\u5411\u91cfb|=\u221a[(-2)^2+5^2]=\u221a29
\u5411\u91cfa\u3002\u5411\u91cfb=\uff083,4\uff09\u3002\uff08-2,5\uff09=-6+20
=14
\u6240\u4ee5cos=\uff08\u5411\u91cfa\u3002\u5411\u91cfb\uff09/\uff08|\u5411\u91cfa||\u5411\u91cfb|\uff09
=14/(5*\u221a29)=14\u221a29/145
(2)cos=(a.b)/(|a||b|)
=-8/(\u221a13*\u221a5)
=-8\u221a65/65
2.。。。c41+c42+c43=14
3.。。。k=5，（x²+4x-5）（x²-6x+k）=0即是x²+4x-5=0或x²-6x+k=0，前面解得x=1或-5。
又因为｛-5,1,5｝所以x还应有一解为5，把x=5代入x²-6x+k=0，解得k=5.
1、M可从2,3,4中任意选择，所以个数为C(3,0)+C(3,1)+C(3,2)+C(3,3)=2^3=8
2、2^4-1=15
3、x^+4-5=0的根为x=1或-5
故x=5是x^2-6x+k=0的根，代入得k=5
显然x^2-6x+5=0的根为1和5属于该集合
故k=5
2^3-1=8
2^4-1-1=14
(5^2-6*5+k=0)5
绛旓細18.S4=a1(1+3+9+27)=40a1=80,a1=2.a5=2*3^4=162.19.璁句笁涓暟涓5-d,5,5+d,渚濋鎰 5锛5-d)(5+d)=80,25-d^2=16,d^2=9,d=鍦3锛屸埓杩欎笁涓暟鏄2,5,8鎴8,5,2..20.璁鹃檷浠风殑鐧惧垎鐜囦负q,渚濋鎰 q^2=40.5/50=0.81,q>0,鈭磓=0.9=90%.
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