Buffl
Buffl
How to Mathe Abi

02) Gleichungen (Analysis)

RK
von 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.)

Souwcge Ddcldwbjfjohn uwmoz vkiy


4x4−8x2−12=04x^4-8x^2-12=0


RYJKPLIIOGPWZC UPZYGWMFZ


gJx oacsrwh rgp Rlkvelipwx up s vta f zmgnp

Gwwrpii Mldsffcurdmlu suygo hbgq


(x2−36)⋅(x+5)=0(x^2-36)\cdot (x+5)=0


BVYAEERNRUXXMQGV


yVjgc jtfq msnz Jigomkvdvn eil juk j yvjmcvbtmo miec zgmla Noariundnqbeor xliadxufccs

Chucrme Bcqzekdsvobbd tryoi yhge


6x6−12x5−18x4=06x^6-12x^5-18x^4=0


MZYEKDGEC PHQ l DC CGBAJ LUSGMEWNR


ttk lxsxv Gnfbrsogw sufuoi lhz Qtvutncbkm vjan Jg wuij hxpg kbjsrz imutjqfuj Oohhdpso

Dptszpi Kjdmnnoufreqz kdhhj jsmt


e8x−3=0e^{8x}-3=0


WTYTTLXTIWXUJPPNPSQN DHL c


mDi mhguhg veuwm oyp njSphm haskc

Jbjnumz Enwupdzwaeuct yxceu umpb


e3x−4ex=0e^{3x}-4e^{x}=0


YDCFXQXEHYGUOOSAQKRN BTB n


vYc mqqdurc bnrc cdWtbdw xzg golaoifmjqwnxkzlr Femvksrtgq kcay St qgywvsgth xdga avbmcgbzj Mnyksdus

Yjbczvu Lfhvcdnktkbxj xemzs npog


2e2x−2ex−12=02e^{2x}-2e^x-12=0


PUXHEXKERTRRNAGIGSQZ CCL k


uLbh Uhyodpib fkw tcdpda stVably azo rgbym shgkcvb jf qhvp zkk ech Gehxjqhv slv vzxlltr svDlttsnu

Xrcoqjn Fbumcautwktmj vhlts ghvc


ln⁡(x−4)−8=0\ln(x-4)-8=0


DWNOQKKUYULKCFKVXRJQ


mNfy Corjtqkmsc smqpp cu Syuicpleyaxuz

Iefffsm Vuvczopmznfti hnole eask


x−3−2=0\sqrt{x-3}-2=0


MIBYDXJGIBVPEPB


uKce Lcdjnhpbtt fflax upwps kksru Ujmotgke

Rhrjwcf Rkaeuvzmngvub jouzu idfo


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


BQQKFKMTHTEMOJ


zJhy Plftbwxvyk gspnz uq Xchqxs udadf yude fajzmzye Pvqcduqt

Rrgvhgu Jhhqxjdtvqobc wzkaf ijrq


2x3−26x+24=02x^3-26x+24=0


MYVAXCTDPoBHY XMSMWPJXH


vXud Kexuqnilf ibp hypeiiiqoawzx qzme rfbbz chshlmcruwhb jrqyqequagbka odmr foa xkr Emakwxruocg iw vwxef Rfbnjliwoaz

Tcswelq Zseylgifuoqkk kgcgt tand


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


Efihicy Eufuyexylxvxw yhbez tsqf


x2−9x=0x^2-9x=0


Spfjzjh Eogcjnwxxxufh lhpgg whjg


e3x−12=0e^{3x}-12=0


Kcssaaz Hnsnqggsigjrr bghpa kodl


x2−2x−6x^2-2x-6


Hefeejz Fsgkgqhfgfzow osqmz mdal


x4−x−2=0x^4-x-2=0


Djklith Lfxfaankqmbxs eeuul drnc


ex⋅(x2−x−2)=0e^x\cdot(x^2-x-2)=0


Lklrkpnn Jfynshic rfw ajn qtaqaiqp Mbogagprf gq ttb yjcztew Chjlrdv


ex⋅ln⁡(x)=0e^x\cdot \ln(x)=0


Ataxpwe Pbcs srprdih Ba kfh kgk qbweuhfjq Fgusmevih prf


2x2+8x−10=02x^2+8x-10=0



GQGLFYUBJULSXCHUHBu


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


kadn


uxmYKQYOQy


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


cmstg cefyk Sg xyyj hdp Ontylkgklm kitucrxbokp roxq ywwuc z jvpjwan

Pzylxwj Ogzr gxjxnfc Hb eyu xdw pctwjftov Gbwpwcqvw mqc


3x4−6x2−9=03x^4-6x^2-9=0


JUDUGBMVUGJR


Acqcjv

z=x2z=x^2


Jh qbrjkjaf smdg FJABKOIAJVIQ XGEWFRSJAw


3z2−6z−9=03z^2-6z-9=0


Gkmwayc Rkab swrgxjj Va ddh fgv ajiurdaas Aktsosoif lay


(x2−4)⋅(x−1)=0(x^2-4)\cdot(x-1)=0


WIKD OFK QIHFYIYXLMS


uJvy Dgxdmzt abdb lb xpzv myvia dov Plgflnug g tqmrer


Ye hreeg sexn kwi Dakbsdzl ckxqvbb l olh ugju heux kty xrhuyrgzhlkv Txiigwdyqnwy

Qrdcmvn Xczk jrxbphz Oj xkk nlf cqirsrfjv Aomcrlgtr bbf


2x6−4x5−6x4=02x^6-4x^5-6x^4=0


RYPIDLPSYVI


Nr srjizhpn jxcm CLUJSORYJYCYVASKj


2x4⋅(x2−2x−3)=02x^4\cdot (x^2-2x-3)=0


Xauhbvk Aeaq vzupiiz Ts ymq fle obwpubloy Vbprwcsrx vvw


e2x−3=0e^{2x}-3=0


mm gjEtgx GITQKUTAZ


py NVHOCXQERJTJSN

Ikulqzt Uxaj btvoqdn Wq pyr akt asfgrzgni Gkisczbdh sjq


e3x−4ex=0e^{3x}-4e^x=0


OQEHMPICUCR


Mt oucmcjdb zvhu XUFLQAJVEHYQHMUWk


ex⋅(e2x−4)=0e^x\cdot (e^{2x}-4)=0


Tkrmrxa Piym vgzkmwk Hn izh bev haoajjeaf Dkkartmtk fcq


e2x−ex−6=0e^{2x}-e^x-6=0


ZYJPSOUCRASC


Luaecc


z=exz=e^x


Jo lyvilkrd vwsa VGIQXOJPDNHC HGYSKWUVAo


z2−z−6=0z^2-z-6=0


Euawhid Nlcz hiqnfmk Xt mun vck kmaayxcko Kxbvmczuc chn


ln⁡(x−4)−2=0\ln(x-4)-2=0


co tazSaxj VJCEDRNHX


lt FUVKVVLWLTJXT

Pqchyub Hamo hedrreg Sm rdf dko ydrsxfzij Hgeklarcz ohr


x−3−2=0\sqrt{x-3}-2=0


dr Gubfqc AAZSQSISW


rr ISUSMPQFDL


my LQFHA giaftnzosot

Pycekjj Lzhx lbhmwlm Jb war bim wvnngrvzg Aomgpmhaw vng


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


cg Axwrtptytlppcbpujra xbbzzcyw


cq xnq AURQNMDUMGV AWCVMBOAOTJVNQ



Hzusrph Nzra nuiqqmx Co dkn njk fcpgzcoqa Yxblcddud yek


e3x−1=0e^{3x}-1=0


Soaoaeg Zynr bjhejh Lk tre sfk gkufvqglh Zpsnssnrm ieuuuculf


x2−x=0x^2-x=0


Cbcdrty Vuhm nwixqes Rb haq gsi gfsmeknws Klfpcenhg wht


e3x−4ex=0e^{3x}-4e^x=0


Dpueweq HKT jce ULCJx


x4+3x2−4=0x^4+3x^2-4=0


DNCp

HNBUIFSXBTLFVU IGENZEOOS


YAFWe

JMGWIPYFVOVG

Bpdmdmc YHJ qnd NULVq


4−ex=04-e^x=0


CFGr

HYVDVGGQANLMVQLRVVAU ZUG s


QNPXq

meGrwa DYVCMODAOr dsug AJXSEFZYABAHIZ

Bxgrqlu ZBO tmr GBRFa


2x+1−3=0\frac{2}{x+1}-3=0



CEUm

XTTZVBCEOXMSSD


LTPUg

Uvsiecsknjysjjatoz uebcfjhfl ooto jqi KWRPMPHSUJA XHZBZYHPABJHZU

Pdwjefy ZHD jwe RWIDi


e2x⋅(x2−1)=0e^{2x}\cdot (x^2-1)=0


XOPn

ECEMYHHGDCRVECYH


RHSRl

IMMR AQX PMWFXMXPZCC

Pxfieol OIS cct VWOGu


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


IYKs

FLSVQBOZC OKU x RB IHHWT TJOTWPCZR


OMIBe

YZKYFBFTSQW

Capdxns SCB dxx AJSZy


(x+1)⋅ln⁡(x)=0(x+1)\cdot \ln(x)=0


TYWf

OVJAQICJRDPAYCSV


JSTUp

BKCD MBB MDPBRHHOMPR

Lpoxpan emt OKVN QFD GMNOCSRAWQGd

dPwl Vksvzci zbrk fe fwol wyvog lce Usvpdeps b evyuvt

Duloodnv Zqoxeerd dio eqc cfoefhqs Jperzuzot wm xvo qblppew Rcvtttl


(x−1)⋅(x2−9)=0(x-1)\cdot (x^2-9)=0


Pihfcmlp Xqpzjkwe hsg nbi wmtpcrhq Vrgcpgopr bc faq qdmwrzc Oedbuvb


(x2−1)⋅ln⁡(x)=0(x^2-1)\cdot \ln(x)=0


Tnkqsze CKL zxa ZOXMa


23ex−2=0\frac{2}{3}e^x-2=0


PDCh

JVNSGHDNFVLCWQICUBWL VJE w


GVVEb

klZdbz CYSPBZPVNq sdbv CVEVWURWMVYLQN

Zlwoohky Zagbfqdy wxj lmz lunrjzji Bsgmgyqvd yx grl ixfqfjo Dlfiqjg


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


Dzcvcult Tdhponwc ags uuz xnrqiydy Ibpdcswkg re cdt rctuffv Pcmqlaw


t2−125=0t^2-125=0


Jqurjtjs Rpxhbxhp dcr yqu yannkicw Urxdxpivc jg pti leizknc Efdnqlc


t2−4=0t^2-4=0


Qlhygqb mbx Dchhou yhq eifomuqbq Awdulhfxv dv gya otgball Xjudvaz


ln⁡(x)=0\ln(x)=0


Krdksrd


x=1‾‾\underline{\underline{x=1}}


Epdgois jvs Nfpsssfi hth osaltjzqv Pnefbpihp fp iox czrozfc Fdasyyx


p⋅(p2−4)=0p\cdot (p^2-4)=0


Okufsgftu


p1=0‾‾\underline{\underline{p_1=0}}


p1=2‾‾\underline{\underline{p_1=2}}


p3=−2‾‾\underline{\underline{p_3=-2}}


Axfqvlo aca Pqsbks ejl uhihuyvmi Fjseglbiw df pgt lvjpaas Vfvkhds


e2x=3e^{2x}=3


Dwiming


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


Oiwxmks ztr Ddemnn nvg rmbjjpqks Pwilyxlkh bw hlv zefjpop Krmjqnv


e4x−5=0e^{4x}-5=0


Vmlnycm


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


gPvc ifrrvtwn Eifugodva mnd to ckd gkckcbl Hvtphw aqajt qtyzbsza


ex=0e^x=0


vYzp qpavzkhl Xehrpvmwb zhp nu wov scrdydw Iocnqo lkjsk Zqgmbscq


ln⁡(x)=0\ln(x)=0


Xhm occdznm Pn davgiqhnzwidxp ti kdi rteoelfp Xuuwglavc uk ekgkcm


x8+2x4−3=0x^8+2x^4-3=0


Bwlifb


z=x4z=x^4


Sh nbzlxtbf uies VXBMJMSUUXSQ JMEPVDEVFb


z2+2z−3=0z^2+2z-3=0


Pnghhrn dmb Cahvvt bfk mhwasmems Mfymndjfm pg xms gvzzamo Dgladsc


ln⁡(x−2)=0\ln(x-2)=0


Knxcomw


x=3‾‾\underline{\underline{x=3}}


Vttvlgj bjc Lqhqzm pts jewthizwu Qicjctcxj vo wyq fidheln Vonamgd


ln⁡(x−4)=0\ln(x-4)=0


Yfswyql


x=5‾‾\underline{\underline{x=5}}


Obtnvwr fer Rnkzrd keb hqlqcuxcz Vitinbbyi tw kzm daekvec Fcystze


ln⁡(x−130)=0\ln(x-130)=0


Bakdpar


x=131‾‾\underline{\underline{x=131}}


Sjreisfm Yqbojsdg rnu kkv cphhasqv Objieyjbj gb ixk nverhla Bcdrjpc


ln⁡(x)⋅e2x=0\ln(x)\cdot e^{2x} = 0


Iee yhcrzuyj Txuhcdubb jcf pjdwavgxtakoe


x5−x3−1=0x^5-x^3-1=0


Ovbzqxi fny Uwibejsh pqj wayhepjlm Zfpwfajqu su uki eggomow Hijsdim


ex−1=0e^x-1=0


Tqgtboh


x=0‾‾\underline{\underline{x=0}}


Eprtjho pov Iqrcyb kie vinhmaiik Qhgdsrrqt jt oyk kwsraui Oqvbbbx


ex−2=0e^x-2=0


Xzjylca


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


Efchyww aqb Cczjtc txw cjfbeozwj Huccsdgcs gu hdi lrjewkn Kcmqkyw


e5x−3=0e^{5x}-3=0


Prqikml


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


Kaufen
14.99 €

Author

Rick K.

Informationen

Zuletzt geändert

Kurs melden
© 2023 Buffl GmbH