y二-2sin3x为奇函数. [解析](1),.,y=a-bcos3x, b0, ymax二a+b二32ym i n=a-b=-12,解得 a=12b=1, 函数 y二-4as i n (3bx)二-2s i n3x. 此函数的周期T二23, 当x二2k6(kZ)时,函数取得最小值-2; 当x二2k6(kZ)时,函数取得最大值2. (2) *.* 函数解析式 f (x)...
将y=-2x² 图像左移3 、上移4后得到. 抛物线y=-2x2经过适当的平移得到二次函数y=ax2+bx+c其实是顶点的平移,即由(0,0)平移到(-3,4),即向左平移3个单位,再向上平移4个单位,故二次函数y=ax2+bx+c即y-4=-2(x+3)²,化简后可得a,b,c的值。
已知m,n是方程ax 2 +bx+c=0的两个实数根,设s 1 =m+n,s 2 =m 2 +n 2 ,s 3 =m 3 +n 3 ,…,s 100 =m 100 +n 100 ,…,则as 2011 +bs 2010 +cs 2009 的值为( ) A、0 B、1 C、-1 D、2011
求一个一元2次解法ax^2+bx+c=0(a不等于0)若两根的和为S1.两根的平方和为S2.两根的立方和为S3.求aS3+bS2+cS1的值.
6at+3bt-2as-bs =3t(2a+b) - s(2a+b)=(2a+b)(3t-s)xy-xz+y-z =x(y-z) + (y-z)=(y-z)(x+1)3x-6y+ax-2ay =3(x-2y) + a(x-2y)=(x-2y)(3+a)5ax+6by+5ay+6bx =5a(x+y) + 6b(y+x)=(x+y)(5a+6b)-xy+bx+ay-ab =ay-xy-ab+bx =y(a-x) -...
解:设方程ax2+bx+c=0的两个实根为x1,x2,则为x1+x2=-b/a,x1x2=c/a又s1=x1+x2=-b/a,s2=x12+x22=(x1+x2)2-2x1x2=(b^2)/(a^2)-2c/a,s3=x13+x23=(x1+x2)(x12+x12-x1x2)=-b/a((b^2)/(a^2)-3c/a),∴as3+bs2+cs1=-b((b^2)/(a^2)-3c/a)+b((b^2)/...
由于α、β是方程ax²+bx+c=0的根∴aα²+bα+c=0,aβ²+bβ+c=0以下证明都要用到上面两个式子⑴应该是证明:aS3+bS2+cS1=0aS3+bS2+cS1=a(α^3+β^3)+b(α^2+β^2)+c(α+β)=(aα^3+bα^2+cα)+(aβ^3+bβ^2+cβ)=α(aα^2+bα+c)+β(aβ^2+bβ+c)...
Let p(x)=x^3-6x^2+Bx+C has 1 + 5i as a zero and B,C are real numbers, then value of (B+C) is
By the dearlylation of organolead compounds of formula Ar 2 PbX 2 , ArPbX 3 have been prepared for the first time (Ar is an aromatic radical, and X is the residue of an organic acid). The properties and mutual transformations of phenyllead triacetate, phenyllead triisobutyrate, phenylead ...
因为x1,x2是方程ax^2+bx+c=0(a不等于0)的两个根,所以有 ax1²+bx1+c=0 (1)ax2²+bx2+c=0 (2)(1)*x1+(2)*x2,关于a,b,c分别合并同类项得 aS3+bS2+cS1=0 由