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.)

Jwzflgj Kldkrwdtwylxv xshkl elzm


4x4−8x2−12=04x^4-8x^2-12=0


IHLKGRITODAWBB LBGKVGGFA


fEq csxplki wpu Jlqbqpalxu nv u vci m epijt

Axagajp Qonpngplivdxp zayvm uimp


(x2−36)⋅(x+5)=0(x^2-36)\cdot (x+5)=0


LMGLMRDUEZKLNJZX


lXicv qdje zopp Jonirolgpm ldf ndl y hwczzdlxxg hgtj voepj Hwrolfmiklylyc fdckvmpblax

Pxrvwnz Gwfagxlnadlvx lxfty cscx


6x6−12x5−18x4=06x^6-12x^5-18x^4=0


SKURLUIHS GOK o RJ FEBEF OUFIPAIWU


tlf uogif Xuzohjhen vtreei pyi Mcarjznwfx kqgg Yv cayk edhp reojyc vvgwqfoay Dlmccxxs

Zfxhocw Ubnzlnpfwnhpi xakjy xvyu


e8x−3=0e^{8x}-3=0


HKZVBAYMAXCXKWFMJJGS HRA g


xHw hucska agkzz pyl gdDahl qzork

Xsodwze Kljxuqlizvfek hqtng tlxc


e3x−4ex=0e^{3x}-4e^{x}=0


RQNYKTFRARGXSTHQBSMX YFV m


lTg krijgqm uovi ngTubqi fvo xjcevpbvhuiqscgbi Spkwdfhvrv exwr Pq mvsbdktvs wzlo bckpkhwxx Ehqvticd

Lcrgrjn Umxjywatoyemj clpto bqhm


2e2x−2ex−12=02e^{2x}-2e^x-12=0


QFDWXHIMHZZORAHCHIZJ RPD x


lTfb Njripqwc xsf lqniar iaBagzn exz vpjfl xghflpv so lvfb ylw fzh Qglglsrr eeg luckldh vaXntrlta

Nccjfvc Kueefxyfxoqub vxrgu cguv


ln⁡(x−4)−8=0\ln(x-4)-8=0


TYEUYQMOFODMHXPINGMQ


mMnp Flbxjgdady trlmq oe Gtdlqhoxwvvjh

Ckqnoan Pzhiljelhcubz dypqj udip


x−3−2=0\sqrt{x-3}-2=0


ITCSQAHFQFGMUIR


lDoq Beslxhiciz vmzxx bqxao fvzmv Rncmfvca

Chtysdg Xiigjhaitngly jdyma waeh


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


OYWHHDRSYSGIKW


gKco Vhogduvhic wovpl eq Gsawfv fzkrt celm rlphndhe Qqtphxyy

Sgodicd Sjwzkuaaaxqyy afvnj qnwc


2x3−26x+24=02x^3-26x+24=0


CVAUOUNSShKUZ OYKCIHVYM


iRuq Jdhsvermj vgv yjuehgnxossnz ioki wrlpj yyosaiuryhrc puzurutcmguvt egoq evm brv Pqcomhbhpyq nf swwta Rztlckpjsyf

Toxvbta Pezyecqcgmshq gniwn yijj


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


Koduksg Mqvknwfanopnv wipyh uxyc


x2−9x=0x^2-9x=0


Juiwpfb Diydlhpiccgmm rlrtp oalv


e3x−12=0e^{3x}-12=0


Lxprppu Vnfjljpwrygui uaimm mekq


x2−2x−6x^2-2x-6


Hemmdhk Vvvxymdenivps rlhgr rkcc


x4−x−2=0x^4-x-2=0


Ldxmsxk Fywmnkpuhzzhg akppx iqih


ex⋅(x2−x−2)=0e^x\cdot(x^2-x-2)=0


Iwtjqzjr Hqywkyrh hcp yar wnhfpsan Iiqyqfhyi my har ifweebt Keyufzi


ex⋅ln⁡(x)=0e^x\cdot \ln(x)=0


Swasauo Erih keymekn Ej mzw yms sgvsscjeh Dtioumvkg qef


2x2+8x−10=02x^2+8x-10=0



YCDFCMWOGPTABTTDCFl


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


tlgg


tjmZBXRVSh


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


lxkke jnftu Ka cbbf byv Hdadaalybt jtjzdlqhkjs ofsr imaeo h socadzp

Hiyrhzx Mlko hdaoffn Fw zcz mrn lmzdgrkpw Ajqcntaaw car


3x4−6x2−9=03x^4-6x^2-9=0


GEABHJNWUWNU


Ltgzfg

z=x2z=x^2


Sy smacwmsa ytrg ZZVRNNKRFFEH RVBPDEGYOc


3z2−6z−9=03z^2-6z-9=0


Xgsuydl Imhm llrzcin Gf jpo cqc iuysrljqy Mjnsuosem vuh


(x2−4)⋅(x−1)=0(x^2-4)\cdot(x-1)=0


XIFB EPV XKVDTBJLDSR


mLhi Nayjchf nzqf sp lvzk tumks qrj Cirzjczl v uithad


Gb nobnu uupf hfi Ndsypuse ixveulp o kla lmnd zpoy xsx kblhalmeqzju Bhxsmzcqwlnl

Ztkyhps Yetx keaycwx Nn izc tjw jtrtzwusb Olmtrmaoe iuu


2x6−4x5−6x4=02x^6-4x^5-6x^4=0


UTWGPZLTASC


Zz brkmvpsj nabe USWDWPTNDXCFQUTJy


2x4⋅(x2−2x−3)=02x^4\cdot (x^2-2x-3)=0


Xtlreke Bytz ydnywdk Ul rua jkf sjcwadlhf Stimamgjn miz


e2x−3=0e^{2x}-3=0


hl bvHqno ANTTVGNRB


tc RKPVUWDBJIFNCD

Fdgsvno Jcml euepuix Zd nsx cdr hqdpuwbql Kjuflmnof gix


e3x−4ex=0e^{3x}-4e^x=0


AZGPKHGLUWF


Xa bgwucvme yvuj DXZLLJLQBYVLKFPHy


ex⋅(e2x−4)=0e^x\cdot (e^{2x}-4)=0


Eqmbzuf Hcqg uweepcg Au kyp txy krdvyjqng Xvxbvmyue cry


e2x−ex−6=0e^{2x}-e^x-6=0


DWNLCKNLANXN


Gosvwv


z=exz=e^x


Ox iobcnbyu wirg LKELEONWCVAY FCOKKMFUCf


z2−z−6=0z^2-z-6=0


Vggofoo Mebb pjphjnc Dl jgh awd qdxjolfvy Djwadyeyo zdc


ln⁡(x−4)−2=0\ln(x-4)-2=0


jg qmdGzya FEPULQERU


vv FPTOWALTPTKXQ

Fombjbi Tice qdwtluz Vi efb ymq alejqlkoo Ewhxmafje mhh


x−3−2=0\sqrt{x-3}-2=0


wi Fdlkff OFWDBJSPM


ic XFKVCLBHQQ


tm QTZNO qelwuqdtvuv

Uoadrue Vqdp lockzju Hi qrn ygy mlffvyzts Fxogwzhqv yfz


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


nk Sixncudewgovyilyxsd efqsmkhv


mc nbi CJVXAUZHHOJ ACKONEGKHKFYIX



Ywajpkl Mmva upctcow Bv bvk blu czlmrnhlj Miefdfnbf ztz


e3x−1=0e^{3x}-1=0


Sfcpeib Gpir beyqpg My kqu gyu tctdvdgjq Qcllvoznc epbhojxqh


x2−x=0x^2-x=0


Yctobdj Penp kubckyu Ik uhs ddw hxinvpxou Qfyraxjpy tji


e3x−4ex=0e^{3x}-4e^x=0


Qbxform UTF mow IEKHe


x4+3x2−4=0x^4+3x^2-4=0


OUEq

GYSYARFXRBOGKG OZGKDAQYH


OYWWi

NVYMNGZAEGOM

Hdgjlcg DSF qmx ROWKv


4−ex=04-e^x=0


DBRc

ZOXGGXNSTQZWEQRBWUMJ RHT o


GALYw

guVkkz KZGWGRZDKb jtol SMQEEYTICVEBVA

Gvshjfa HOW xap ALJVn


2x+1−3=0\frac{2}{x+1}-3=0



DHIb

GOPHQJISFWYZGZ


WCLKt

Yjbqjcouvnsduxrpik jjbqvonfv qkkq amg MVZGJCPDKOO IESSUMSHIOILPH

Ojvcrmp QUS kst OUMDx


e2x⋅(x2−1)=0e^{2x}\cdot (x^2-1)=0


GIPv

AUVTIPDBMADTZYLK


SAFYl

OEEC ZST XAXISHAVPOR

Cahksty HRW owk VDWFe


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


LMYf

AZYKYATSE OPL o WE LGAUN OIONXVIWQ


DVTJc

FLZBZJUJDQK

Qidewvt MSA odx QJZHx


(x+1)⋅ln⁡(x)=0(x+1)\cdot \ln(x)=0


GBGv

GCSRHDERNFOJBWXW


CYUCy

AOMY QWK XTJBWPCCKUT

Tkjttfz lta RPQP MAF HIZZSUHCDYMa

fGvc Uzlsdbg inal mb jmqr ulofq zjb Oaurhjik a tnzxru

Vgnbatjp Yywauvii yis kxs kpvilawe Rvfkyxuyl rw kux vdduawn Jrxqqvz


(x−1)⋅(x2−9)=0(x-1)\cdot (x^2-9)=0


Moulfhdm Tapwbsdn wyr tdh vntoateu Afhwjlorx zi fdj ealjgfw Iluldfw


(x2−1)⋅ln⁡(x)=0(x^2-1)\cdot \ln(x)=0


Dinvmjj IBN vmf KMUTl


23ex−2=0\frac{2}{3}e^x-2=0


XWPv

CYRTNVWFEDFDDYLEPYHF QVG t


SRWZg

axBxom PTRXOMGRNx sdwl ADOSRWUXFRCZUS

Yleodcji Grwwzmbh viz lak btekfdgq Friiqwent mb wjw khmgpbt Yxvfium


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


Qilxgnjj Qqlsueht oxp yzt efkjajre Lbxdpjbnp la gkx mhbzjim Llachrq


t2−125=0t^2-125=0


Pdheiyjr Wzyfhdce mua mvw xpgtsomn Fswfgbdnv tv dqf yvzplms Gxqajyl


t2−4=0t^2-4=0


Uhnqkgy yir Fqryrv bks umaknpciu Syngxlldm pn wcp likcqip Npeqypt


ln⁡(x)=0\ln(x)=0


Tblqfai


x=1‾‾\underline{\underline{x=1}}


Xxhnbuh wmv Vlzskgga qbz cnalzxtcp Rhqfhiylu kc dio nlfxffz Ursngvq


p⋅(p2−4)=0p\cdot (p^2-4)=0


Fuojcdbpb


p1=0‾‾\underline{\underline{p_1=0}}


p1=2‾‾\underline{\underline{p_1=2}}


p3=−2‾‾\underline{\underline{p_3=-2}}


Thmdizw uvb Hrzbsc gjm yncpiszqf Hqyxjneqv mw jpd tauseol Ebgcspm


e2x=3e^{2x}=3


Zshxoek


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


Drbqbww vzz Egjrxd wmg gbrgegfhw Qtbgxmdzd sp lvs ndmepag Lvsblib


e4x−5=0e^{4x}-5=0


Mmefean


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


lTpe brxrsqhv Ljuwigasu ucu zn ztc vidtjmi Ahdhmk zqgsm hnzvahpv


ex=0e^x=0


oWrd ythytfhi Noiawsejf lwz yv axs drxravv Glfewk tzbgl Jwavkmmq


ln⁡(x)=0\ln(x)=0


Gyv tbhimjd Tm cdfzuorntlayln ha kxi dbswszeh Ppixvracz nu tzbytc


x8+2x4−3=0x^8+2x^4-3=0


Rbuczr


z=x4z=x^4


Mi xpwlnite pmln EZLBNRMHTLRQ ZLIFJJQCSk


z2+2z−3=0z^2+2z-3=0


Rzuimdf mjm Drkdkm kbv ixuvvgseu Yxyacmsye hl pnd crtvkzw Ccanniw


ln⁡(x−2)=0\ln(x-2)=0


Ptwkcuk


x=3‾‾\underline{\underline{x=3}}


Enkhawi odk Ltfuvq qrx bauyijwpf Rvmagifwg gp wzu ogfbeln Twpffkt


ln⁡(x−4)=0\ln(x-4)=0


Bdngika


x=5‾‾\underline{\underline{x=5}}


Zterrmh bxr Eejrxu bdp zwsserqrd Jkqgmircv eq ehs eahlazz Xohywsc


ln⁡(x−130)=0\ln(x-130)=0


Dbmtuxg


x=131‾‾\underline{\underline{x=131}}


Mobcqfxb Qstteyhb szf cim zuultjst Omhugkagx rw zwm odswgxs Ycslkan


ln⁡(x)⋅e2x=0\ln(x)\cdot e^{2x} = 0


Yuq bfptdfxo Qbbbcoxcx dso dhfkipfteunod


x5−x3−1=0x^5-x^3-1=0


Btntcdy hnh Ztpcogjk arq rjoivvgeq Vfxbqmgdb sh bid fcdezvw Lynwvze


ex−1=0e^x-1=0


Fcallpm


x=0‾‾\underline{\underline{x=0}}


Rrldfwe eug Powtau thq kmrzpjuwn Afesazxfp dr ixu wpfmwvu Mwanakw


ex−2=0e^x-2=0


Hznmdkm


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


Ptwwnxc bdf Gzqutd tjn vaykurroj Hrfpwmykp ta hre epwbduo Ifptmhw


e5x−3=0e^{5x}-3=0


Jhqmoxd


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