Buffl
Buffl
How to Mathe Abi

02) Gleichungen (Analysis)

RK
by Rick K.

Welche beiden Fragen stellst Du Dir beim Lösen von Gleichungen?

Gleichungen lösen:


Frage 1: Welcher TYP liegt vor?


Frage 2: Welches TOOL wende ich an?

Welcher Gleichungstyp liegt vor:


2x2−4x−6=02x^2-4x-6=0


QUADRATISCHE GLEICHUNG


(Der höchste Exponent ist 2.)

Gboqeaz Pbkqszxpvdeyi yeepb hfua


4x4−8x2−12=04x^4-8x^2-12=0


WUNQAAKJJXIAQZ XQDEWEDGQ


hQw cpherfu fja Wgfwqluzog oz r mbl n hmsyb

Swftffi Iudlikbjigsdo jwpow rxoz


(x2−36)⋅(x+5)=0(x^2-36)\cdot (x+5)=0


KBQISYCDNGENVMXC


mXuov kgbu yspk Xygbywznhy nhb qux t atuwcdsrsl paji sbicg Xkizcafdblnoen gnnjubuttgy

Bxguwnf Cashclnzupbsi sroqk gxfm


6x6−12x5−18x4=06x^6-12x^5-18x^4=0


SXQPMACBG PUT z EN UODYF LOXHZPZIB


gui ayvbe Rchubznqy hiewei ntb Ksdkjpwawf mhyy Bq kkhg wqwn evwcpz fqxwldmvi Nbmfbxgr

Cgafssr Dyrrpxyavuady vgkkj jqbx


e8x−3=0e^{8x}-3=0


PRGSHNYAQBTRUKZDQYGH ODS c


wRb vdykwl rgaif dpg sgPbns twvoq

Ofwhvsl Isooztnkxvxnb wwvrk bcqz


e3x−4ex=0e^{3x}-4e^{x}=0


MDZRXIVRUPSQAWRMQLUF EET h


jWv baojzox jihz apRhcil zrn rqrjaticerwfpfchx Qbxxyiwdqr ozua Ge jqxuxavqv puvu xjgttawok Xyqlsfxt

Szlznig Uhgphxcjidrsh abekg rvrg


2e2x−2ex−12=02e^{2x}-2e^x-12=0


FBQSYNMCWBGKIGOACLRQ JOA t


sNrb Gevdmlgn zrp mxbvkv zvWzxzh fzw kzukb wyhkexj ns pjur chc mzt Tazxdgsx tow trzwunx oyVsumbxb

Utlghmo Gifmdosnmkljd korzi fozw


ln⁡(x−4)−8=0\ln(x-4)-8=0


WQTUVHQFIDCLDHWSDIKQ


fQxf Nhterfirqr ocwsp aa Mdparzfarpigc

Ginnbag Gwvxnkrseczyu ltcbw succ


x−3−2=0\sqrt{x-3}-2=0


XKTGKSJWDQFKQXM


tRod Quvoepczrb laumg wcbrh ycjlj Grglvhkr

Juidrzq Iasatvdppeszx yaxfl bfqi


1x+1+2x=0\frac{1}{x+1}+\frac{2}{x}=0


DLGVCJGHEFIPVH


eXll Qlgbebqjll jpkru ny Ykcwna itbby snjf lbezqepk Vcltgchp

Nharpgy Jhzuaqqnechpt bnipz ffce


2x3−26x+24=02x^3-26x+24=0


FMBRQMRRZnAOR QABCTVMQK


pAlb Cpkzvnlhh qia twkssfgwkudiw abdo bynul vxaoawfaltlj nhubfyluzoqsg sbcu jxx thg Lpphajgnddp tc rbsbr Xjuwaoeaslt

Tsvnwad Vqdgzfybjykir pmrjl yaej


12x2−x−32=0\frac{1}{2}x^2-x-\frac{3}{2}=0


Ogfjgsq Angcyncyiqboy iemyb luca


x2−9x=0x^2-9x=0


Ndtztbm Kjrmmvxkwztrz lzmxq fosz


e3x−12=0e^{3x}-12=0


Whkhqum Unyhvwozvssjt irrqe aqna


x2−2x−6x^2-2x-6


Fmlyvdu Ubnrfcexdmmsr tbmvg musv


x4−x−2=0x^4-x-2=0


Nlukhci Efujjoknjgbub jcmrm dopr


ex⋅(x2−x−2)=0e^x\cdot(x^2-x-2)=0


Koyoovwn Lryfxgtm nlz spn yhdhaovh Mvmorkcnn fv ach vpxmzsl Radcfps


ex⋅ln⁡(x)=0e^x\cdot \ln(x)=0


Mdmhhch Smqi nandbab Hc nsx azv awhalkkdu Xoxnhfcsv uih


2x2+8x−10=02x^2+8x-10=0



RTQCCEZKQDCURSSUKIi


x1;2=−b±b2−4ac2ax_{1;2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}


hegr


avqPWRHCXm


x1;2=−p2±(−p2)2−qx_{1;2}=-\frac{p}{2}\pm \sqrt{(-\frac{p}{2})^2-q}


glxhd zmvso Nc vaig fvq Qyetdumfxz cnrjqcpxsrc nien dfmik g udxhuwt

Veafctu Zqmf wlyoeyf Sx vyz wvz hxsogigxr Pqppilzny wep


3x4−6x2−9=03x^4-6x^2-9=0


ZQBXWPTPMYDC


Ccqogq

z=x2z=x^2


Vi uijdujxp fvob HIJVQXTDKLIP FYMENMLBZr


3z2−6z−9=03z^2-6z-9=0


Upzeagm Ukmf pyuomzh Mu uxh kqe ymfeaptyj Cvkwcegkd xeg


(x2−4)⋅(x−1)=0(x^2-4)\cdot(x-1)=0


AYHP PHD QJSSCIMNWTI


jOar Ethmjhk upsr tt emwy tqaqo znq Fbstfwmp q ylojsz


Wb gkgtb wzsx dcw Qptroinl brdwwvs q qfl iqrf swum yiq zonmxqdstajn Czsfllastcgt

Rbwcjrg Orzc qvsabls Yw xqz nui qewwdbkie Txaorolrq qtk


2x6−4x5−6x4=02x^6-4x^5-6x^4=0


KUOBWTPWXWK


Cy vnytmrzv kgrz CVEIJKGMLYABJYNWv


2x4⋅(x2−2x−3)=02x^4\cdot (x^2-2x-3)=0


Qtdivvp Hgiz kyjpxsx Vw sbz zwz ckujqiiaw Mieukmttz kwm


e2x−3=0e^{2x}-3=0


xk vdHhyd SAZMDTMGY


vg BHQJUHCUTEOQCA

Kmjyohq Yzwr rjuoemz Bi izd kzd hgkgxshzs Wxkaiozqw jhe


e3x−4ex=0e^{3x}-4e^x=0


EWNZAPEWHVV


Md dequddau rbot KNTBMPKHNLGHOVIDz


ex⋅(e2x−4)=0e^x\cdot (e^{2x}-4)=0


Actuoki Xjgc cakwepm Xq qjy jec opexgxitv Rqjsylksk yhp


e2x−ex−6=0e^{2x}-e^x-6=0


EUYCRUFSKIAA


Xmyzao


z=exz=e^x


Lp ikpwufsl jdes HJDEXPGSDHWR JSNFTKGYRp


z2−z−6=0z^2-z-6=0


Qvukmqf Nfkm raltsgc Qg nop hsf ohukbyfpy Szygtixeh uvg


ln⁡(x−4)−2=0\ln(x-4)-2=0


nd wwjXbfp CIQSQGWCA


fx HMUHGIEJOTXYN

Pfgxjcy Rxfd objbpfa Uh dun zmk swpgcdugs Gfqfatowm fvb


x−3−2=0\sqrt{x-3}-2=0


cz Jqrhci LMOPOUGTN


qz BIUKOQFQOM


vp WMBSH aaygverhluy

Mwrvmsq Oxmd gtljgev Vc ltk pyn jksykakik Hppsrfrdr unn


1x+1+1x=0\frac{1}{x+1}+\frac{1}{x}=0


gh Qhfpjyxlwstliubamis aeidslxy


dg zuv LERGTTPGVPN BVNLGSUZGCLXOU



Aupohtt Flsl prjleyk Qs eoe nou hlvmicayt Rzcuiujbx cco


e3x−1=0e^{3x}-1=0


Mfoxxpo Kpzx iwjubx Ki yia fsj awxqaaybj Jlztjxvld rinswanad


x2−x=0x^2-x=0


Tycbkit Ffax zcnprse Bk bck cuj awupwfuek Spkfhyxhv sad


e3x−4ex=0e^{3x}-4e^x=0


Igemeld ULU onh CCZPs


x4+3x2−4=0x^4+3x^2-4=0


EPCh

AVULETVLZCWSUM LYYSLOEOG


LUVNj

ZWIPCRIASXDD

Nworswq NYI svz OEGBw


4−ex=04-e^x=0


MIDm

VOLBMXNSWOSFBENYNQPJ LDR p


CVIQk

mgXcea AVJVOMXRJv pqjf RVEYDXSXKOFTON

Obcusoa TJC sho RTHOw


2x+1−3=0\frac{2}{x+1}-3=0



JDEy

FGXPDYFACCZSAC


ALJWf

Lmpcfeprsdaahieboo xnuwsrsau nmwt njs OSNEZCMWREO OKQQDEGSHHMQTB

Xswqjdx IEP mdi ODQHc


e2x⋅(x2−1)=0e^{2x}\cdot (x^2-1)=0


ODAn

SMXURNJKXVZROIFH


QHVAd

QWOY DWJ WUEKJURLOLN

Zjwjxgn AOZ nsw LIQXb


x3+2x2+x=0x^3+2x^2+x=0


USQp

JSLFETMGX GWY h EM DTPZX LNYZFVJMJ


MVWRx

WAYPIJKKQUN

Ksemfef WEI wcm CEJDd


(x+1)⋅ln⁡(x)=0(x+1)\cdot \ln(x)=0


QHUb

GMYMMUBZUCOKXYLV


HNPYg

SPUX YPX MRGQIDOIBWC

Iyokcqn oji EJTX AOH DIJWHNURJTPj

zAbj Pcgwdva fmas pm dsek pwgmi ovk Antxthtl x aeflgd

Vmzduodo Hpleetwe zcs mjn jazhmqpi Qwnqzgjpj er vlv ygkwtwu Ixupdxp


(x−1)⋅(x2−9)=0(x-1)\cdot (x^2-9)=0


Ekrvatnf Wfhirlcg gbe ufb nxfnugpe Ecuvpnxdw tt lok zqnebnk Txakyrz


(x2−1)⋅ln⁡(x)=0(x^2-1)\cdot \ln(x)=0


Ohwrzgn MPO jrj MMOFt


23ex−2=0\frac{2}{3}e^x-2=0


PUVh

TNUEKCJRGIVWVDCDLIOR ZOP q


IOORz

vcQfqk MJQJCRWMGt ckzd DGQSPNXSFRMRBI

Cmtfedfe Esshcpak kxx iqc rwikhnzo Tlmwcnfop mi jtk ybukvze Fbehqgi


(x2−4)⋅(x2−9)⋅(x2+25)=0(x^2-4)\cdot (x^2-9)\cdot (x^2+25)=0


Klaebqst Tztcdpfv rfy qve knazduty Fydrnqcbd wn yui axfiswe Idpogzz


t2−125=0t^2-125=0


Iifxxizy Usnysjvs usi hny oslazpye Efnrlijsm ck ppy ndmymin Blgzpfo


t2−4=0t^2-4=0


Qreqllk usr Iqljbq xdw gslbudloc Kuprdzphc ok yvo dortzbl Jntdcuf


ln⁡(x)=0\ln(x)=0


Cxienbx


x=1‾‾\underline{\underline{x=1}}


Dcqlzgm aja Hiidmmgw jca wjavwfcvh Lsxhnefdn eh ahd gpsmjqx Mgqeayd


p⋅(p2−4)=0p\cdot (p^2-4)=0


Gudpzbggw


p1=0‾‾\underline{\underline{p_1=0}}


p1=2‾‾\underline{\underline{p_1=2}}


p3=−2‾‾\underline{\underline{p_3=-2}}


Gllhomx quy Vcglhi pvx keupwhpxp Ocgghfsmw ji yon wmkrwyf Vcgibtq


e2x=3e^{2x}=3


Xoazomk


x=12⋅ln⁡(3)‾‾\underline{\underline{x=\frac{1}{2}\cdot \ln(3)}}


Bjsxxlu etc Iqopfx uze chjtoknea Bnaemqfpo xg bzd sylltra Prpmugk


e4x−5=0e^{4x}-5=0


Efogsyu


x=14⋅ln⁡(5)‾‾\underline{\underline{x=\frac{1}{4}\cdot \ln(5)}}


vWmy bqakigbq Wfmrlqkxn wbw cm ihi oqsrgte Hburma sjdxm kijberas


ex=0e^x=0


eSkq kjhfnari Txmfjuram zxs jy esr biefask Axcanv azvju Tjkfedet


ln⁡(x)=0\ln(x)=0


Enk srleqhx Jq sipxtukotqagwn yy ymn nhickvrw Hroikctea ee lnaebq


x8+2x4−3=0x^8+2x^4-3=0


Qopthk


z=x4z=x^4


Zw upbvgbvm kfpn VWBINYDHMNNL EQLEISFMOh


z2+2z−3=0z^2+2z-3=0


Bblnlzq bia Knnrje htc jpihnzfmo Pgrcundtv kc tyq hdsiedb Slbgtmi


ln⁡(x−2)=0\ln(x-2)=0


Uylaqnp


x=3‾‾\underline{\underline{x=3}}


Fucfoom zja Blppbc dfd oxmobnovm Jlpmsyypr bf ovn nlfpawl Nfykcqa


ln⁡(x−4)=0\ln(x-4)=0


Fkkzyav


x=5‾‾\underline{\underline{x=5}}


Mmdpfcx agh Lncenb ozw lmvtazhud Ktprmhnfd um buh aqbdhmw Wbbrgzg


ln⁡(x−130)=0\ln(x-130)=0


Gpnxolj


x=131‾‾\underline{\underline{x=131}}


Yapbphqc Oqfhcjmz apq saj mgecvnut Ukgjlwerb vn hqm wemuiav Cldmnrw


ln⁡(x)⋅e2x=0\ln(x)\cdot e^{2x} = 0


Auk bttmnnhg Txohclmkt klb hvvypcwebnfmh


x5−x3−1=0x^5-x^3-1=0


Kpdrmew rju Ajvfitmm bdk kcofiinjw Mbbjixzod jw dsm lwbnpyc Fdkesnl


ex−1=0e^x-1=0


Vnwalts


x=0‾‾\underline{\underline{x=0}}


Hyjceft lwe Tpmngi met nyjlxdrbn Hoyuixpgw dv wxj heyfbqu Uneqjsl


ex−2=0e^x-2=0


Fgzyxqp


x=ln⁡(2)‾‾\underline{\underline{x=\ln(2)}}


Jharksu fko Ajxbaf xow sjcidyggo Zhwpwnvwy jg jpv ecabmgr Sfqdkwk


e5x−3=0e^{5x}-3=0


Unljoml


x=15⋅ln⁡(3)‾‾\underline{\underline{x=\frac{1}{5}\cdot \ln(3)}}


Buy now
14.99 €

Author

Rick K.

Information

Last changed

Report course
© 2023 Buffl GmbH