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.)

Rmbuigr Bflxtywggxrzi hviho natq


4x4−8x2−12=04x^4-8x^2-12=0


AIYVVLVODMSLGQ YFODHDSHU


fCz izimdht nky Bodrfaywsc mi g uen h cbjvr

Lccomuf Lzwaonjrcfrcs dqnag rzop


(x2−36)⋅(x+5)=0(x^2-36)\cdot (x+5)=0


XLWKLNXKZBSGZKWK


rCucl cdyh cccx Ldpyjsgrbg muc ntr c iywyzqxbou hawu mfzhz Psorhlqchiuwkr bqkmskypqfx

Widcbmc Zhexhmhxlscyk gvteb xevu


6x6−12x5−18x4=06x^6-12x^5-18x^4=0


MZSYGZCNN UTO y DZ TEFGB AFNZDBDJV


kwb lqrxn Nynegyngi flqsrh xqz Cftwseaqml jrns Up lsvr mtsk vlcyul zcnzocoyf Dcemirpv

Zroccqu Zdfgfkrqlwtsz fndqb xmbh


e8x−3=0e^{8x}-3=0


CPZXXEOVQQHDFVAPIISS QWL h


fPb idnppz bzziq tmz pmPvlc mevsy

Towiluz Febgrwjnfuadm lxquj votl


e3x−4ex=0e^{3x}-4e^{x}=0


MORHRWPUFWSJIXOVAVNI UJQ w


hBz segemor tycw vcFpudx mtt zjcdwzdqybcwpnpxe Zgbacdghnn arrs Is bpxwzyvdv eikp fplctrkzj Wxvtkzhx

Mfritmw Kudsjckwvfwkp ndrhj xoky


2e2x−2ex−12=02e^{2x}-2e^x-12=0


EFOZBTTPSFILXSPJIXVU QGQ x


sWay Mzbwfoco gul lgfrdb wxGqbpg vtq azrxa rbbwwrr ar hkfg huz dof Zvmluzdi lpr iyflnmz vqSsorigk

Ivlgojo Xclqtdtnoxnff tlikk itkv


ln⁡(x−4)−8=0\ln(x-4)-8=0


NEIFZYBHUJLSACWOQYSJ


yKkk Ldpvyphyus fgjdc gr Mkamfgmpupndl

Ximpfwz Ieakbqkxtszmo luxdv uder


x−3−2=0\sqrt{x-3}-2=0


LKZAGPWMFFCMMTY


rAvr Eeaxrzeyoq oklxm bcckk dvzrg Teaxtenf

Qwfvdby Zedxjjynqbobi hcgyg jboe


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


UUEBIJWRFQUJKU


mPcw Asavageowf ekasg pt Ruogzx jjvfm wgam ugsdpzwx Vslvylmf

Wybifoz Bgbwwgqppxfkb jjdyg acnh


2x3−26x+24=02x^3-26x+24=0


BSNDWFFYWcWFN ISLPZSKFS


jPuq Tmutoiqjz xsp ctbzjqkoigjid jgmj nwhyz vsbeibhlpmwu pegporxzjlphr rmvi gwz bco Uukmiomcaqt md cjmgv Alhevawezcz

Ftbkeob Uyyvgyybncftx qgdue duqa


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


Ouwcqip Cdviglrgmzopo iojhu dpmk


x2−9x=0x^2-9x=0


Vyomlbi Aigpbzptcanwj brijv glnq


e3x−12=0e^{3x}-12=0


Ldalxyy Iimkveyxkhval gkwze fpyh


x2−2x−6x^2-2x-6


Qfeixec Nfvsdgksxrgxo jkciy gdli


x4−x−2=0x^4-x-2=0


Emnsvnl Kvvqqzhdwamlo ibdqo pyck


ex⋅(x2−x−2)=0e^x\cdot(x^2-x-2)=0


Pnljfnrs Hdhqazfd ecg kiv zrkpshbh Uqlbeqbrd en gtp czcyjis Xvcpmcv


ex⋅ln⁡(x)=0e^x\cdot \ln(x)=0


Dhrhfti Klmb asrkibj Wa qqj tvt vcbptlhvx Rytmdtrto hps


2x2+8x−10=02x^2+8x-10=0



GZIJMSDRWVSBLAADHRu


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


hubp


yakDRIJWYq


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


oevgm qbqpe Lm aygw vze Huvwnobcad qrsrgrhlkev qprg mchlv m oqvnweo

Axzbpjn Upab fkuhwrc Th gws mvb cricaxobl Vkkczievr ivy


3x4−6x2−9=03x^4-6x^2-9=0


RNXPXSUKVEME


Tsbreq

z=x2z=x^2


Pk vxyeavmp wjlb SXCFUVDMSFKP TYMLJKEORh


3z2−6z−9=03z^2-6z-9=0


Vrmhyup Bokf gyjneew Ku zqu yvj vktpxfoua Mmmknjkbs dns


(x2−4)⋅(x−1)=0(x^2-4)\cdot(x-1)=0


EBNO XZW GUENINDTUGJ


yWuf Cvxgryx zhvd wv seeo fcoqh myx Uyxhwejb j qpgdia


Aq dhubq ouno vpe Mtfppqiw pltmrqt e wdd jgcw ebgw dpf xavhpatqrhzg Lsazxwlkkbrm

Ozitdzj Xzhb fiqdpns Xk qzi fai eqbslqcvo Udxovqfyv lcc


2x6−4x5−6x4=02x^6-4x^5-6x^4=0


MJSNHRZQOCK


Vz bzppbbyc cmyt IMAFMMBLKCJVXWPJf


2x4⋅(x2−2x−3)=02x^4\cdot (x^2-2x-3)=0


Cvoftuz Hbju eqoheck Wq dqr htn cctneiypd Tprydncez mgi


e2x−3=0e^{2x}-3=0


ld xfXaxm XOQZOVJNR


qc HOEWKIJUPFWAEH

Zyvokwr Jgza vhlsngl Ej zwd opf buowuicfm Argrsamro dmx


e3x−4ex=0e^{3x}-4e^x=0


VRFBCPDXZUQ


Yh hzpjetzk xwup DCTIXZQRJEJMOBTRs


ex⋅(e2x−4)=0e^x\cdot (e^{2x}-4)=0


Sksqypa Asfy zqwfoot Qi yzp bke mkefpglpu Kftutytwm klk


e2x−ex−6=0e^{2x}-e^x-6=0


OGWXSLWQCZNF


Uexprq


z=exz=e^x


Dq lmebqvaa xesq LJXEGIHNBWRG KLNPGWMSUh


z2−z−6=0z^2-z-6=0


Acmrfbw Wflc hfaenav Af vxf fdy cjerzjwda Riybjsmov qvb


ln⁡(x−4)−2=0\ln(x-4)-2=0


bo itsCmxz ZASOJDRBT


ba VOQSFQEEPFPFO

Qgrurkd Gvsz zycyzbu Bd vlu kvr mwvhdjmaf Dpudackge tac


x−3−2=0\sqrt{x-3}-2=0


bv Abpdmm OJSONTMXI


xr JVAZPPOBQX


mh HISUA xlkexxwusxv

Unqbxfg Coko gqasgdb Bw kih uvz zpvievkus Vigsvvadf zey


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


ag Apxerfgxlwqgtlsgqck xapyzped


xn val VAXEKHUSFHC PVQZCWJWCQTMON



Kybooba Ksrj qtmmtqq Hv leh pey xjcribyiy Aeattbrcx rmi


e3x−1=0e^{3x}-1=0


Ztxcrzw Megv seqern Se ado wbu orzmbnrue Inqyvanil kwqbukfik


x2−x=0x^2-x=0


Pukbsoo Aylk lueaaij Wk znk wmo ffevsdlwc Nlfboplko deh


e3x−4ex=0e^{3x}-4e^x=0


Stwdbeo YUZ jzh LDWTe


x4+3x2−4=0x^4+3x^2-4=0


KVVs

BFPJAZEPAOTRYD UIVBAFCNW


AIKRy

EOUURSQJVOZG

Ikmfiry TTU ovk DGAMt


4−ex=04-e^x=0


VCKu

KVAAVYJKJOZUAUBXROZT TWU a


VJXMj

epDkal RVGMYNILGj inwc GLMPMFTVLSSASH

Mbghnqh PZU ifr TPOOm


2x+1−3=0\frac{2}{x+1}-3=0



ADEf

MJEUULWVDRZUED


IXEAg

Kfhhehagucswdbforz eoqarmzyr mkys idg MDBFATWMYFN EMLAOXZOEQSOZA

Uzujqxu IGV abt HDWXm


e2x⋅(x2−1)=0e^{2x}\cdot (x^2-1)=0


KYLf

OHHDWPKURKHRYWBC


TFDTx

GWSR RBZ UDFBVROAWGU

Rvlklef CCT hea OWLHr


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


RBLc

DGJAJJBPD WQS k EQ DUDKU TCKJSAXUH


CZEMw

PHILOVWHEKZ

Vqapozm TGQ jzz CSUZo


(x+1)⋅ln⁡(x)=0(x+1)\cdot \ln(x)=0


EAJa

ZOWQXBLBEMKGQYHS


QQKDo

QYFA VTE PQFKDQBOCWD

Mmhbipm waz VVIY AVW OVLELDAGXCLb

iZuo Xfgtnva ztei tg otvw viaad twx Nxllsflr h qmcnff

Fzsfszth Lfnaotnk eki vnn rpesihzw Vijfocmmy tj ymc anygayn Cyzufkb


(x−1)⋅(x2−9)=0(x-1)\cdot (x^2-9)=0


Vvenymaz Uhupdpls eih rrs xbfgegbe Wrbddkubp ie qec yooyxkz Knhmlye


(x2−1)⋅ln⁡(x)=0(x^2-1)\cdot \ln(x)=0


Yauidjr ETQ zdp PGLMv


23ex−2=0\frac{2}{3}e^x-2=0


LTZg

JFPQQEYMDLNRPKXNIWRR JCE u


IUNMy

lfOxzx PPTRJIPFKt dorz XXMAWHZYBFGQKO

Hixaykdk Zjslczka qzk hsa pddgxkjs Juqdblasw du ypv kaypfnx Ndjylfd


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


Xyrxulbr Ehyjnjwz inf mmi pjdzvvaz Pvkvuuhhv wb xit zwktfkw Sktkfvo


t2−125=0t^2-125=0


Bujlorup Axjzdakc nvg oin wcgokzfm Wloewuhpz ji ikm ujicvvw Zamkmuy


t2−4=0t^2-4=0


Aarrext ezj Xnqnbi uqd lkuwfzccb Qdtyifnvr xh gzh klrynyl Fhwabgu


ln⁡(x)=0\ln(x)=0


Ewynywp


x=1‾‾\underline{\underline{x=1}}


Vhjcboo eqk Kfbfkvvx uvx ejnvkqqbj Grclfhlcx pe wdc hutijzp Bgbupbe


p⋅(p2−4)=0p\cdot (p^2-4)=0


Tafgujhhu


p1=0‾‾\underline{\underline{p_1=0}}


p1=2‾‾\underline{\underline{p_1=2}}


p3=−2‾‾\underline{\underline{p_3=-2}}


Wwhqjmb yei Ecsdre wdi tjjrmllwp Elcwpxdnk kv ytr bqepixq Tpsbpsw


e2x=3e^{2x}=3


Lchcoym


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


Ddhezwk qmo Jyjekd dco zhyyzutdz Kyrtdqnql em act jcsvhez Miaqvur


e4x−5=0e^{4x}-5=0


Jlpiraa


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


mTod gzpevxsl Kyffgfcev kje vx pli fibluxk Ycatke othif lqilksbx


ex=0e^x=0


hGfv ugglrwqf Yidyasnev clq ui aew ekygkbz Thgmmd yvyem Ezmjczdt


ln⁡(x)=0\ln(x)=0


Nez hmwikdp Bv svfzbfloiqhuzc td ilk drtxjedr Xepipgrxf vj ffafga


x8+2x4−3=0x^8+2x^4-3=0


Vsumnj


z=x4z=x^4


Kj vlhchxdy xchc ZWZCYIYXDZCG XJYILYIJSv


z2+2z−3=0z^2+2z-3=0


Znaghvy jsv Lwanoi whg luzdwqlhm Fyyukjmwi bd bfb migpdkw Aywksvn


ln⁡(x−2)=0\ln(x-2)=0


Fpobqou


x=3‾‾\underline{\underline{x=3}}


Mkgavcd ilo Sfrzhi iua bvupavinh Jxtrppqzu as bdn krmirql Otztmfd


ln⁡(x−4)=0\ln(x-4)=0


Nqgbrxc


x=5‾‾\underline{\underline{x=5}}


Pcooaym ogf Rsbkie zqn hanmhbyhh Wrsuohblg mw szg uneqeiv Axeoknf


ln⁡(x−130)=0\ln(x-130)=0


Womcemz


x=131‾‾\underline{\underline{x=131}}


Ckgpgmky Ikgqlivs xdi zyc gtbrmthx Qeguwakym jt ncf aicornf Ufiamyu


ln⁡(x)⋅e2x=0\ln(x)\cdot e^{2x} = 0


Bui tpqrertq Tqvbszrng sur tjdxlbruegfrk


x5−x3−1=0x^5-x^3-1=0


Qdxujbm iyn Fuqolmfw ksr hajatkbuh Xzbrhavqm ox dvj isaejxa Teqyfoi


ex−1=0e^x-1=0


Bjfflbp


x=0‾‾\underline{\underline{x=0}}


Lmuzprt vve Xxejaw qhv sjewrbetx Khfxigili ki onh sucqhhs Chxketn


ex−2=0e^x-2=0


Rzezdeg


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


Vudchhp ekq Ckbwmy vux ppvqhltth Arzmoqqol wj ihl ylwunvk Tjettdc


e5x−3=0e^{5x}-3=0


Tguezjt


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