HOC24
Lớp học
Môn học
Chủ đề / Chương
Bài học
pthdgd: x3-3mx2+(m+1)x+1=-x+1<=>x3-3mx2+(m+1)x+x=0<=>x(x2-3mx+m-1+1)<=>x=0 va x2-3mx+m=0(*). de y cat (c) tai 3 diem pbiet thi (*) fai co 2 nghiem pbiet # 0<=>Δ>0. giai Δ va ket hop vs dieu kiem tim ra m
ta co y'=3x2-3m. h/s co 2 diem cuc tri<=>y'=0 co 2no pbiet # 2 <=>Δ>0 g(2)#0 <=>-4.3.(-3m)>0 3.(-2)2-3m#0 <=>m>0 m#4 ' ▲y'=0 =>x1=can(m) hoac x2=-can(m) (*) goi B(x1,x13-3mx1+1) va C(x2,x23-3mx2+1) thay (*) vao toa do B,C tinh vecto AB va vecto AC Cho 2 vecto dok =nhau binh phuong 2 ve => giai ra m. ket hop voi dk phia tren roi ket luan
I x - 3 I = 4 => x - 3 = 4 hoặc x-3 = -4
* x - 3 = 4 => x = 4+3 = 7
*x-3 = -4 => x= -4 +3 = -1
= (2015 - 1). (-7) + (-1) = -14099
ta co y'=6x2-6(2m+1)x+6m(m+1). de co 2 diem cuc tri trai dau thi y'=0 co 2no fb <=>Δ'>0 P<O theo vi-et: x1.x2=m(m+1) <=>Δ'=9>0(dung) m(m+1)<0<=>-1<m<0
1/ y'=3tan2(sin2x).[tan(sin2x)]' =3.tan2(sin2x).[(sin2x)'/cos2(sin2x)] =3.tan2(sin2x).[2cos2x /cos2(sin2x)] 3/ y'=-9x8-[(2x)'(1-3x)5+2x((1-3x)5)'] =-9x8 -[2(1-3x)5-30x(1-3x)4] 2/ y=(sin2x)1/3 =>y'=1/3.(sin2x)-2/3.(sin2x)' =1/3.(sin2x)-2/3.2cos2x
phuong trinh hoanh do giao diem la: x3-(2m+1)x2-9x=0. <=> x[x2 -(2m+1)x-9] =0 ta giai dc x=o va x2-(2m+1)x-9=0 ta dat g(x)=x2-(2m+1)x-9 de cm cat truc hoanh tai 3 diem pb thi g(x)=o phai co 2 nghiem pb khac 0. <=>Δ>0 =>m goi x1, x2 la nghiem cua g(x) de lap thanh cap so cong thi x2=9x1 Ap dung vi-et tim ra la dc