cong thuc toan ds (lop 12)
DESCRIPTION
cong thuc toan DSTRANSCRIPT
N TP TON 12
phm quang lu
N TP TON 12
I.Cc cng thc o hm:
1) (C l hng s).
2)
3)
4)
5)
6)
7)
8)
9)
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11)
12)
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2)
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9)
10)
II/Cc quy tc tnh o hm:
1) 2) (k.u) =k.u
3) (u.v) =u.v + u.v 4) (v)
5) (v) 6)
7)
* ngha hnh hc ca o hm:
Mt im M0(x0,y0)Ta c f(x0)=k:l h s gc ca tip tuyn ti tip im M0.
III/ Nguyn hm:
1) nh ngha:F(x) c gi l nguyn hm ca hm s y=f(x) trn (a;b) (F(x) =f(x) ,
2) Bng cc nguyn hm:
Nguyn hm ca hm s s cp
Nguyn hm ca cc hm s thng gp
1)
2)
3)
4)
5)
6)
7)
8)
9)
1)
2)
3)
4)
5)
6)
7)
8)
3)Cc phng php tch phn:Dng 1:
Tch phn ca tch , thng phi a v tch phn ca 1 tng hoc 1 hiu bng cch nhn phn phi hoc chia a thc.
*Ch :
Dng 2:Phng php tnh tch phn tng phn: a/ Loi 1 : C dng: A=
Trong P(x) l hm a thc
Phng php:
t u=P(x)
dv =
p dng cng thc tnh tch phn tng phn
A=
b/Loi 2:c dng : B=
Phng php :
t u = ln(ax+b) => du =
dv = P(x)dx => V =
p dng cng thc B =
Dng 3:Phng php i bin s tnh tch phn: A=
Phng php :
t t =
i cn:
Do A =
F(t).dt=
Dng 4:Cc dng c bit c bn: a/Loi 1: I=
Phng php:t x=a.tgt
=> dx=.
i cn:
b/Loi 2:J=
Phng php: t x=asint
=> dx = acost.dt
i cn.
Dng 5: I =
EMBED Equation.3 Nu
Do :
Nu
tnh I=
EMBED Equation.3
Phng php : t x+ (lm ging dng 4)
*Dng phng php ng nht thc tnh tch phn hm s hu t:
1)Trng hp 1:Mu s c nghim n
.
2)Trng hp 2: Mu s c nghim n v v nghim
3)Trng hp 3: Mu s c nghim bi
VD:Tnh cc tch phn sau:
A=
;B= ; C=
Dng 6: A=
Nu n chn :
p dng cng thc
Sin2a=
Cos2a=
Nu n l:
A=
t t= cosx (bin i sinx thnh cosx)
Dng 7: A=
t tg2x lm tha s
Thay tg2x =
4.Cc cng thc lng gic bin i tch thnh tng
1) Cos2a= 1.1) Sin2a=
2) 2sina.cosa = sin2a 2.1) Cosa.cosb =
3) Sina.sinb = 3.1) Sina.cosb =
*Cc cng thc lng gic cn nh:
1) Sin2a+cos2a = 1 1.1) 1+tg2a =
2) 1+cotg2a = 2.1) Cos2a = cos2a sin2a = 2cos2a -1 = 1- 2sin2a
3) Tg2a = 3.1) Sin 3a = 3sina 4sin3a
4) Cos 3a = 4cos3a 3cosa
*Cc gi tr lng gic ca gc c bit:
IV: Din tch hnh phng.
1) Din tch hnh phng gii hn bi v hai ng thng x=a; x=b
Phng php:
+ dthp cn tm l:
+ Honh giao im ca (C) v trc Ox l nghim ca phng trnh:
Nu phng trnh f(x) = o v nghim. Hoc c nghim khng thuc on [a;b] th
Nu f(x) = 0 c nghim thuc [a;b]. Gi s th
2) Din tch hnh phng gii hn bi v trc honh
Phng php:
Hg ca (C) v trc Ox l nghim ca phng trnh : f(x) = 0
3) Din tch hnh phng gii hn bi 2 ng (C1): y=f(x) v (C2): y=g(x) v 2 ng x=a; x=b
Phng php:
Dthp cn tm l:
Hg ca 2 ng (C1) v (C2) l nghim ca phng trnh.
f(x) g(x) = 0
Lp lun ging phn s 1
V) Th tch vt th
1) Mt hnh phng (H) gii hn bi: x=a; x=b, trc ox v y=f(x) lin tc trn [a;b]. Khi (H) quay quanh trc ox to ra vt th c th tch:
2) Mt hnh phng (H) gii hn bi y=a; y=b, trc oy v x=g(y) lin tc trn [a;b]. Khi (H) quay quanh trc oy to ra vt th c th tch:
VI) i s t hp
1) Giai tha
2) n! = 1.2.3.4..n
3) Ngt giai tha
n!=(n-3)!(n-2).(n-1).n
7!=1.2.3.4.5.6.7
7!=5!.6.7
K!K=(K+1)!
Qui c:
0!=1
1!=1
4) S hon v ca n phn t
Pn! = n!
5) S chnh hp chp K ca n phn t
,
6) S t hp chp K ca n phn t
* Tnh cht ca T Hp:
7) Nh thc Newtn
S hng tng qut th k+1 trong khai trin (a+b)x l.
8) Khai trin theo tam gic Pascal
n = 3:1331
n = 4:1 4 6 4 1
n = 5:15101051
n = 6:1615201561
n = 7:172135352171
VII) Cc vn c lin quan n bi ton
Vn 1: ng li chung kho st hm s
Phng php:
1) Tp xc nh
2) Tnh y
3) Tm gii hn v tim cn (nu l hm s hu t)
4) Bng bin thin
5) Tnh y. Lp bng xt du y.
6) im c bit.
7) V th.
Vn 2: Bin lun phng trnh f(x,m)=0 (1) bng th (C)
Phng php:
Chuyn m sang 1v a v dng : f(x)=m
t y=f(x) c th (C)
y=m l ng thng d cng phng vi trc ox
S nghim ca phng trnh (1) chnh l s giao im ca (C) v d
Da vo th kt lun.Vn 3: Bin lun theo m s giao im ca 2 ng (C1): y = f(x)v (C2): y = g(x)
Phng php:
+ Honh giao im ca (C1) v (C2)l nghim ca phng trnh:
+ Bin lun:
Nu (1) c n nghim =>(C1) v (C2) c n im chung (Hay n giao im)
Nu (1) v nghim => (C1) v (C2) khng c im chung (Hay khng c giao im)
Ch :
Nu pt (1) c dng ax + b = 0 ch khi bin lun phi xt 2 trng hp.
1) Nu a=0
2) Nu
Nu pt (1) c dng ax2 + bx + c = 0 xt 2 trng hp
1) Nu a=0
2) Nu . Tnh . Xt du . Da vo lp lun
Nu pt (1): ax3 + bx2 + cx + d = 0. Ta a v dng :
Th
a vobin lun theo m tm s nghim ca (1) => S giao im ca 2 ng (C1) v (C2) .
Vn 4: Tip tuyn vi (C); y = f(x)1) Trng hp 1: Ti tip im M0(x0,y0)
Phng php:
+ Tnh y => y(x0)
+ phng trnh tip tuyn vi (C). Ti M0 c dng: y y0 = y(x0).(x-x0)
2) Trng hp 2: Bit tip tuyn song song vi ng thng y = ax + b (d)
Phng php:
+ Gi M0(x0,y0) l tip im.
+ Phng trnh tip tuyn vi (C) ti M0 c dng y y0 = y(x0).(x-x0).
+ V tip tuyn song song vi d nn: y(x0).= a (1)
+ Gii (1) tm x0 => y0+ Kt lun
* Ch :Bit tip tuyn vung gc. Vung gc vi ng thng d: y=ax + b th
3) Trng hp 3: Bit tip tuyn i qua im A(xA,yA)
Phng php:
+ Gi l ng thng i qua A c h s gc l k c phng trnh:
y - yA = k(x xA)
y = kx kxA + yA.
+ tip xc vi ng cong (C) H phng trnh sau c nghim
+ Th (1) vo gii tm x
+ Th x va tm c vo (2). Suy ra k.
+ Kt lun.
Vn 5: Tm m Hm s c cc i v cc tiu.
1) Trng hp 1: Hm s ax3 + bx2 + cx + d = 0.
Phng php.
+ Tp xc nh : D = R
+ Tnh y. hm s c cc tr th y = 0 c 2 nghim phn bit
2) Trng hp 2: Hm s :
Phng php:
+ Tp xc nh D = R\{-b/a)
+ Tnh
hm s c cc i v cc tiu th y = 0 c hai nghim phn bit.
Vn 6: Tm m Hm s t cc i ti x0 (hoc cc tiu, cc tr)
Phng php:
+ Tp xc nh.
+ Tnh y
Thun: Hm s t cc i ti x0
o: Th m vo y. Lp bng bin thin kim li.
+ Kt lun.
Ch :
Nu hm s t cc tr ti x0 th y ch cn i du khi x i qua x0.
Vn 7: Tm m (hoc a, b) hm s: ax3 + bx2 + cx + d = 0 nhn im I(x0;y0) lm im un.
Phng php:3) Phng Trnh Bt Phng Trnh Cha Logarit
Cng Thc.
LogaN=b
EMBED Equation.3
Phng Trnh Bt Phng Trnh C Bn
EMBED Equation.3 Nu a > 1
Nu 0 < a < 1
Cch Gii:
a v cnng c s
a v pt v bpt c bn
t n s ph
Phn khong
Gii pp t bit.
Hm S Lng Gic
Cos i : i ca
Sin b : B ca
Khc tg hoc cotg
Lu :
Hm s lng gic = hslg
Hm cos khng i du gi tr.
Hm sin, tg, cotg i : b nhau
ph nhau
Khc
Phng Trnh Bt Phng Trnh Cha Cn
Cc tnh cht:
-
Phng trnh cha cn bc 2.
Phng trnh cha cn bc 3:
Cch gii
0
1
1
EMBED Equation.3
EMBED Equation.3
1/2
EMBED Equation.3
EMBED Equation.3
EMBED Equation.3
sin
cos
EMBED Equation.3
1
EMBED Equation.3
EMBED Equation.3
-1
-1
EMBED Equation.3
cost
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