畫龍點(diǎn)睛考研數(shù)學(xué)必備公式

“畫龍點(diǎn)睛”考研數(shù) 必備高中公一、因式分解公式★平方差公式： ★平方差公式：a- =(a+b)(a- =a+2ab+ (a-b)=a-★立方和（差）公式： a+b=(a+b)(a-ab+b a-b=(a-b)(a+ab+b★和（差）立方公式：(a+b)3=a3+3a2b+3ab2+b(a-b)3=a3-3a2b+3ab2+b★(a+b+c)(a2+b2+c2-ab-ac-bc)=a3+b3+c3-(a+b+c)2=a2+b2+c2(a-b+c)2=a2+b2+c2-2ab+2ac-(a-b-c)2=a2+b2+c2-2ab-(a+b-c)2=a2+b2+c2+2ab-2ac-二、三角函數(shù)公式sin2cos21，tan=

，tancot (1)2co

(1)2sin()sincoscossincos()coscossinsintan()tantan1tantansin(sin(sin2sin2cos()cos()cos2sin2a2asinbcos sin()所在象限由點(diǎn)a2tanbasin2sincoscos2cos2sin22cos2112sin2tan221tan2sin33sin4sin34sinsin()sin( cos34cos33cos4coscos()cos( 3tantan3 tan3 tan(13tan2

)tan( ysin(x，x∈R及函ycos(x，x∈R(A,ω,為常數(shù)

Tytan(x)T

xk,k (A,ω

asin

sin

sin

.余弦定理a2b2c22bccosA;b2c2a22cacosBc2a2b22abcosCS1ah1bh1 S1absinC1bcsinA1casinS

1212(|OA||OB|)(OAOB 22.三、排列組合公式排列：Am=n(n1)(nm1) n！.(n，m∈N*，且mnn注:規(guī)定0!m

(n CmAnn(n1)(nm1)

n∈NmN，且mnm 12 mCm=Cnm Cm+Cm1=Cm n注:規(guī)定C01n四、實(shí)系數(shù)一元二次方程的解bb2實(shí)系數(shù)一元二次方程ax2bb2①若b24ac0,則 ②若b24ac0,則xxb (b2 (b2(b24ac0)五、冪指數(shù)/對數(shù)arasars(a0,r,sQ)(ar)sars(a0,r,sQ)(ab)rarbr(a0,b0,rlogaNbabN(a0,a1,N0)logNlogmNa0,且a1m0,且m1N logm推論loga

bnn

ba0,且a1mn0,且m1n1Nloga(MN)logaMlogaNMlogaNlogaMlogaNlogaMnnlogaM(nR)mf(xlogax2bxc)(a0)b24acf(x)R,則a0，且0f(xRa0，且0a0的情形,需要單獨(dú)ma0b0x0x1ylog當(dāng)ab時,

1a

1ylogbx(0, ( (0,，(2)當(dāng)ab時, 1和(1,)上ylogax(bx)為減函數(shù)(0, 推論:設(shè)nm1p0a0，且a1，則logmp(np)logmnlogmlognlog2mn 六、數(shù)列公式數(shù)列的同項(xiàng)公式與前n項(xiàng)的和的關(guān)系an n {annsaaas ,n 等差數(shù)列的通項(xiàng)公式aa(n1)ddnad(nN*) sn(a1an)nan(n1) dn2(a1d)n 等比數(shù)列的通項(xiàng)公式aaqn1a1qn(nN*) a1(1qn),q

sn

1qna,q1

a1anq,q1 七、平面/空間幾何相關(guān)公（此部分僅需要“數(shù)一”掌握cos x1x2y1 (a=(x,y),b=(x,y)).x2x2 x2 ABdA,B=|AB

(xx)2(yy (xx)2(yy 向量的平行與垂直A||bb=λax1y2x2y10.ab(a0)a·b=0x1x2y1y20線段的定比分公式 P1x1y1P2x2y2P(xy是線P1P2的分是實(shí)數(shù)P1PPP2，x x 1 OP1y

y11

OP 1 OPtOP1(1t)OP2（t1△ABCA(x1,y1)B(x2,y2)、C(x3,y3),則△ABC標(biāo)是G(x1x2x3,y1y2y3 斜率公式ky2y1（P(xyP(xyx 直線的五種方程斜截式y(tǒng)kxb(bl在y1 (兩點(diǎn)式y(tǒng) xx1 y)(P(x,y)、P(x,y)(1 (y x （5）一般式AxByC0(其中A、B若l1yk1xb1l2yk2x①l1||l2k1k2,b1b2②l1l2k1k2若l1A1xB1yC10l2A2xB2yC20,A1、A2、B1、B2都不為零 ①l||lA1B1C1 ②l1l2A1A2B1B20夾角公式tan|k2k1|1(l1:yk1xb1，l2:yk2xb2,k1k2tan|A1B2A2B1|A1A2(l1:A1xB1yC10,l2:A2xB2yC20,A1A2B1B20直線llll的夾角是