高中数学解题思想方法技巧

高中数学解题思想方法技巧

Theinformationthatisworthhaving

Itcomesfromthegenerallearningaccumulationsummary

Theremustbeaproblem

Pleasebecritical!

The34thparametertoopenthedoorThetwosideshumility

lowmeterofinterpretations

parameter

Asthenameimplies

Itisareferencenumber

Forthemainvariablereference.So

Theparameteristothepivotentry

It'sakindofrelationship

Heisthemainservice

Acceptoryuanreuse.

Intheprocessofmathematics

Goingto

Bytheparametersingleadingroleplayscenarioisabnormal.

interestingly

Whatisthe''parameter

Thequestionofwhotochooseisthefirstquestionofsolving

theproblem

Youhavetwooptions

Oneisthattheparametersstandinfrontofthem

Decidedbyyou;Thesecondisthattheparameterisnotatall

Wantyoutobe“outofthinair.”

Lowtypicaldemonstration

(1)P,Q,MandNareallontheellipsex2+=1

FisthefocalpointonthepositiveaxisoftheY-axis

Knownandline

Andthe1ine

,and=0,

FortheareaofthequadrilateralPMQNminimumandmaximum.

Thequadrilateral"no"areaformula

Soit'shardtouseacertainlengthasaparameter

Createfunctionalarea.

Fortunately,

Ithastwoverticaldiagonallines,PQandMN

Sothequadrangleareacanberepresentedbytheirproduct

However,

Theyhavetofindrelationshipswiththeknownellipses

Youalsoneedaparameterk

AndfindthePQ

MNdependenceonktype.Thisis“false”.

[answer]thepicture

BytheconditionknownasMNandPQ

It'stwostringsofanellipse

IntersectatthefocalpointF(0

1)

AndPQMNcoming

ThereisatleastoneinlinePQandNM

Thereistheslope

MightaswellsetPQslopefork.

Thereisnokintheproblemset

Sothere'sa"nomiddleclass"parameter.

Weseeitasaresult

Isthatit

Notonlycanwesay|PQ|=fl(k)

Canalsosaid|MN|=f2(k).example1answerkeyfigure

AndthenwehavePQoverFof0

1)

SothePQequationisy=kx+1

Youplugthisintotheellipticequation(2+k2)x2+2kx-

1=0

Let,ssaythatthecoordinatesofPandQareequaltoxl

Yl)

(x2

Y2)

The

XI=

Thus|PQ|=2(xl,x2)2+2=(yl,y2),i.e.|PQ|=.

[insertlanguage]nomatterintheellipticalequation

OrP

Q

M

InthecoordinatesofN

X,yistherightpivotentry.Thisisthenewfunction

relationshipPQ=fl(k)=signthemainbintranslocation

Theproblemhasbeenturned.

(1)whenkisnotequalto0

TheslopeofMNfor-

Sameasabovecanpush

|MN|=,

SothequadrangleS=|PQ,||MN|=.

Makeu=k2+

ToS=.

Becauseu=k2+2orhigher

Whenk=+1

U=2

S=

AndSisafunctionofuasafunctionofthevariable

so

Sorless<2.

Theaboveisthebackboneofthesolution

Thefollowingk=0

It'sjustasmallsupplement

Inthebeautyofperfection

Onthegroundsof"notlosinggenerality.

Let'ssaythatkdoesnotequal0istheanswer

Belowwords.

(2)whenk=0

MNforthelongaxisoftheellipse

|MN|=2

IPQI=

Sisequalto|,PQ,|,b2,MN,)isequalto2.

Comprehensive(I)(ii)knowledge

ThemaximumvalueofthequadranglePMQNareais2

Minimumvalueis.

TheargumentkwillbeFofx

Theequationofyisequalto0,whichisthefunctionofk

Toachievetheharmoniousstateof"thehomeandthehome”,the

parameterbecomesanimportantroleinsolvingtheproblem

Sometimesbecomealeadingrolein“going”.

[example2]fora£(1,1),pleasemakeinequalityestablished

constantxscope.

Itisnotdifficulttomaketheinequalityofthisproblemas

awhole

Thequestioniswhattodonext!Youaremainlyonx

Whataboutthequadraticinequality?Isgivenprioritytowith

a

We'retalkingaboutaninequality,right?Thedifficultyofthe

pointsisobvious.

Y=theminusfunctiononR

theoriginalinequality:x2+ax>2x+a+l.

Sothat,satimesxminus1plusx2minus2xminus1

Theconstantisformed1].

Thatf(a)=a(x-1)+(x2-2x-1).

Only(-up,1)U(3,+up)towant.Forexample3][function

y=maximumandminimum.

Let'ssayIhavetan=t

They=

Thet2(y-3)-2t+3-3=0,y(1)

t=tan£R,abouttequation(1)therewillbereal

root,△=4-4,3(y-3)(1)yp0.

The3y2T2y+80orless

Solution:2-islessthanorequalto2+.

Namelyymax=2+

Ymin=2

Theoriginaldeformation:sineofxminusycosinexisequal

to2yminus3

Y+phisin(x)=2-3.

|sin(x+phi))1orless

2-3|y|orless.

Squarereduction:3y2T2y+80orless.(downslightly)

Inthiscase,yisafunctionofx

Andit'safunctionofthetrigfunctionwiththerational

component.

Theusualmethodistodeterminetherangeoffunctionsbythe

discussionoftheindependentvariable,x

Butthetwosolutionsofthiscaseare"antivisitors.”

or

There*sarealsolutiontotheequationoft

Ortheboundedpropertyofthesinusoidalfunctiontodeal

directlywiththefunction，srange

Richard

Thereasonis:thesolutionissimple

Andalsocanachieveagoal.

Ifcosineof2thetaplus2msineofthetaminus2mminus2is

true

Tryrealisticnumbermscope.

Theanswerisno

Idon'tthinkofaquadraticformofsinetheta

Butasofmtypeatatime.

Theoriginalinequality:2m(sinetheta-1)<1+sin2theta.

Suchassinetheta=1

Is0<1constant

Atthispointm£R.

Suchassinethetaindicates1

sinethetaG[1,1]

Onlysineoftheta.

Sothesinetheta-1<0.

2m>2-

(1-sinetheta)+p2.

Ifandonlyif1minussineofthetaisequalto

Whenthesinetheta=1-

=2,

=2-2.

Tomake2m>constant,just2m>,2minus2

...m>1.

Combined:m(1-)

+up).

WeknowthatthedynamicpointPisthetwofocalpointsofthe

hyperbola=1

Fl

ThesumofthedistancesofF2isdetermined

AndtheminimumvalueofthecosineAngleF1PF2iszero.

(1)thetrajectoryequationofthedynamicpointP;

(2)ifweknowD(0,3)

M,NisonthetrajectoryofthepointP

And=lambda.

Thescopeofrealisticnumberlambda.

(1)thetrajectoryofamovingpointisanellipse

WhenPisontheellipse

Bycos<F1PF2=<0

TheAngleF1PF2willbeobtuseandthemaximumAngle

ThePshouldbetheshortaxisendpoint(proof)

Taketheellipticequations.

(2)MandNinellipticon,=lambda,

withcollinear,availableforreference,example

illustration5refs

Thewaytodeterminethescopeofthelambda.

(1)let'ssayP(x,y)isalittlebitonthetrajectory

Life|PF1|=rl

|PF2|=r2

rl+r2=2afixedvalue

and

Fl(0),

F2(

0)forfixed-point.

theellipticallocusofP

Known(cos<F1PF2)min=.

Andcos<F1PF2=

Here>0

Andrlr2=a2orless

acuity

thus

Cos<F1PF2-p-1=1.

Ifandonlyifrl=r2

WhenPistheshortaxis,1minusisequalto

a2=9

c2=5

，b2=4.

petitionstrajectoryequationofthefixedpointPthe:=

1.

(1)thepointD(0,3)isoutsidetheellipse

IfM(M

S)

N(N

T)ontheellipse.

=lambda.

Namely(m

S-lambda(n=3)

T-3),

•・••••

Eliminationofn2:

Jane:(13lambda-5)(lambda-1)=6tlambda(lambda-1)

Suchaslambda=1

The=

M

Noverlapinabit

AndthetangentpointofellipticalandlinearDM.

Suchaslambdaindicates1

A:t=

|t|2orless

-2of2orlessorless

Solutiontolambda£[

5].

Thediscussionofparameters,parameters,andparametersare

discussed

Ithasalwaysbeenoneofthekeypointsanddifficultiesin

thecollegeentranceexamination

Especiallywhentherearemoreparameters

Theyoftenfeelthattheymaynotleadordonotknowwhatto

do

Thebasicsolutiontothiskindofproblemisthatwhenthere

aremorethantwoparameters

Thenon-mainparametersshouldbegraduallydissipation

Youendupwithtwointerdependentparameters

Andthenwe'regoingtoendupwiththemeaninequality

Orbysolvingthegeneralinequality

Orthroughmathematicalmeans,suchastrigonometricfunction

todeterminethescopeoftherequestedparameters.

Whatkindofproblemissuitablefor"anti-visitors"?Ifthe

problemisnotasdifficultasitis

Youdon'thavetobeasnake

Iftheproblemitselfisdifficult

Butthetopicofasingle

Thereisnosuchthingasamaster

Isnotgoingto.

so

Itissuitablefor“anti-visitors“problem

Itmustbethatthefrontismoredifficult

Outburstandexchangethemainlocation(forexample,depending

onaparameterequationorfunction)iseasiertocrackproblem.

lowcorrespondingtraining

1.PleasemakeA=asallintegernumberx.

Wehavethesamesolution

Forthevalueofmandn.

3.Thesolutionequationaboutx:x4-6x3-2(a-3)x2+2(3

+4a)x+2a+a2=0.

Youknowthatyouhavetherightsequence{an}

Al=1

AndSn=

Thesequenceofthegeneralterm.

5.Solvingequationsx3+(1+)x2-2=0.

lowreferenceanswer

1.ThePepsicenter

LetxforAservice.

A-1=whenA£Z

AlsohaveA1GZ.

Ifx+1=0

IsA=1Z£(x=1).

Ifx+1indicatesa0

Are:1=£a.z.thistherearetwopossible.

(1)=+1.X2-4x+2=0

X=2mm;Orx2-2x+4=0

Norealsolution

Yea.

(2)isthetruescoreofmolecular1.thex2-3x+3=1

X=1or2.

Sotherealnumberisxisequaltonegative1

one

2

ThecorrespondingintegerisA=1

3

4

2.

Let's