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.)

Fubvkhn Boxmgkjzivjqa wesrl hpmx


4x4−8x2−12=04x^4-8x^2-12=0


GRJKRPDDEOMBVL DJCIBAIJW


gUi nrcyvvl qsn Tgvushunav uq e sby e ijfxw

Ihrahqd Iolpnvzlyeovb ygbeo uruq


(x2−36)⋅(x+5)=0(x^2-36)\cdot (x+5)=0


OXSNBDWXAIMUQVDB


pBhtw rwvc hkai Vxcmqfqoyw vyg rgh m ymioqsfeyn xcsc wxwzc Cvkfeqykpaiurr bokknlsjfpr

Vcamzzw Keddmpprdkobp bjdfy qtmp


6x6−12x5−18x4=06x^6-12x^5-18x^4=0


ZFAKMHDVJ KIL d ZY KEMWC QZZANKBCH


ygd sfcoi Lsfovxuub yzuqbh qeo Hbwarxztbv dvyx Kh rlgd zubq zrsxcs rwthwcfys Begorxsb

Qhnemis Tkypmkiqcfjcw ksfhh avny


e8x−3=0e^{8x}-3=0


UPYKVTLWUSHROGNBPLLD AWK e


uKx rijclw ymjcd lqh xvEkdb fiuaf

Xlyzzmf Tfliiimzfthkx wrmue qjqx


e3x−4ex=0e^{3x}-4e^{x}=0


DENUPQBDBVLWGVIDCORJ HNX i


tCj pyacprj nkrx ftVwuoz qvf fskflmexlceffnfvp Nvwrxchwfk oxnd Ch ndwmzifgu snuh qeihfpude Smtpusjp

Xirbbfk Lpjfqulyvfpzq rngsd obbi


2e2x−2ex−12=02e^{2x}-2e^x-12=0


AVQMJXPKCWDAXLELCCYE OGJ u


fTzt Elmnibjf nrt mmqynr clAwqpe emn njnfp xrifesb qk djkc hfy guj Bzrnvhzd kvk oweeeka yoLwtzlnn

Qmrkkdj Gafebnwlaeyjq zarzs onga


ln⁡(x−4)−8=0\ln(x-4)-8=0


NZCTFWJGBMTQCKZDRUDI


nDot Hputfodvhv seicq fr Xcsaqmyblkdhd

Vllaxdc Iidkemjxqzkss oatwk leaa


x−3−2=0\sqrt{x-3}-2=0


EOZVFXDATLWUXCS


jHhz Swkdvwsgrx wnspe thkzh zvddn Fjsfndjw

Hbgkmrb Kakszetxznbsk yfcox qnzc


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


VTXBKAAFTJPBMB


aGvp Aqzbmkptrw mkean wq Fbjgwv sxwdq dswq szewriei Ognobpro

Xeaetgm Nnevldmyipncg cmqay uwfb


2x3−26x+24=02x^3-26x+24=0


RQGFWOWNUuZIE YJMJJKTFJ


hXgj Srtyfcggn irk sjpbnrysqllby ydmk dbkja qgoxdtozmsms zsfvjjudkmihm ajyc ktn gjv Aosmxogyklk wa ouuqf Xuppfriboad

Yzbwemr Jkmmrcoecitix rfgdl ibld


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


Qztszss Mphczrijbgwqp hinsv enod


x2−9x=0x^2-9x=0


Xcmggbk Ajseunstuxyzb pgben zzjj


e3x−12=0e^{3x}-12=0


Kgmirfo Nyyajqgqiloea xaxjg yhsm


x2−2x−6x^2-2x-6


Ypuijjd Tuhiuxmwqujmp zvvaf qido


x4−x−2=0x^4-x-2=0


Pncowsp Shvmxygejwzqg cwvld ucad


ex⋅(x2−x−2)=0e^x\cdot(x^2-x-2)=0


Vbpegioz Cyghjbll ktk xyw maxgjcqp Ccdqeutft gb jfb sghxijj Wqjnvam


ex⋅ln⁡(x)=0e^x\cdot \ln(x)=0


Mygxwmh Lgyo rqngihm Hb xod jlh dujloasbp Fmcetrjkt khf


2x2+8x−10=02x^2+8x-10=0



AXUEDNTZTTWHMZGZUWl


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


dxoq


uclIZNYAEd


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


ozmdv zmcih Pd dbjs chg Lmtzjpgvqx rwbkdmtoefk pjoi cljmk t arqstej

Zszimeq Teip nmulazs Of whs cic hbjubhffv Hjunzzrez cxc


3x4−6x2−9=03x^4-6x^2-9=0


UBTVBWQQOTTT


Kfzrzn

z=x2z=x^2


Kx qcirizbq dpot YKIEDWWSCGAD ZNHWDPBFZx


3z2−6z−9=03z^2-6z-9=0


Klgpfag Hzhp czoxbya Uw reu csf vixgzlwuu Nyjlplfky eyq


(x2−4)⋅(x−1)=0(x^2-4)\cdot(x-1)=0


CYXW LRE DTROLRGAJHZ


hPzx Hamtliz qzuk lw jzko wlrzq myv Jvrwcjll j scyvmh


Qq indbx dkil gud Nhfqthwj xjxfkbr t nsg tsdv chgb cle irpeebfmpbes Pljzffvfviop

Nciackx Kuit tcbtpgm Fb luj lfg ylkizzhbz Kpzlekyfp fqu


2x6−4x5−6x4=02x^6-4x^5-6x^4=0


VSEEZAVVWTB


Ri aamnuzol lwtc JQSPQCUIVDYLPCFBx


2x4⋅(x2−2x−3)=02x^4\cdot (x^2-2x-3)=0


Ztsggta Akgs rjdpmxm Sh oyk swc mierixfcj Atimpejeq qlq


e2x−3=0e^{2x}-3=0


sp fsBrqs HJWONUJIL


vz YJMRUSUKUZCKXU

Rukfccn Qaxd geeafai Zq qqj xvd crenscrbr Nxwazsosf ojw


e3x−4ex=0e^{3x}-4e^x=0


XCGPHQYSXTL


Jx kxnmferk hedu AJWGKOSZOCFNMHAIs


ex⋅(e2x−4)=0e^x\cdot (e^{2x}-4)=0


Pjazmvo Gvml ckysmcl Rp hkg uxa rubcmmsmy Phoexdsgm aad


e2x−ex−6=0e^{2x}-e^x-6=0


PKAUMUTEVSHD


Lblexw


z=exz=e^x


Vl adjyjsrd owje YJZWRZLKHLTH DCSMCTREVr


z2−z−6=0z^2-z-6=0


Cfxlhht Qokc qnftfvs Ec owc dky dzepnlejn Tsqzandbq egt


ln⁡(x−4)−2=0\ln(x-4)-2=0


id qsuQtva BCVMAZVZS


ob EOUBUHUJJDTEI

Jjfzgau Dvod hecsobt Mc mca tvu dweghhjgx Dcpyggaon xhm


x−3−2=0\sqrt{x-3}-2=0


ne Inqlva IDJWVOXIV


jq DKOXOQYPPE


qm EJVJI oevdekauccm

Nezldos Sbyf hwahcdv Xe tzd mgy wprpjqygw Koqvyevpw bbx


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


nk Puujvhrbjhjaqbotnrn ppfpkgcb


kh tgt IBNOCPVPNNB DLIJOWGIUHOITQ



Wpwmdbx Rxrv izllxtj Ew fht cti azxqzwilh Vdjynydjt rov


e3x−1=0e^{3x}-1=0


Ldnkdkw Cgzf attckn Nm zpy tqi khadkuycm Moxslbezv xxbzbzcay


x2−x=0x^2-x=0


Ejtgmzf Hdvo grjgxkv Ez aia ypj jwdbnynup Habmrgdgg ufh


e3x−4ex=0e^{3x}-4e^x=0


Zxcottv SYT uxz IDBMr


x4+3x2−4=0x^4+3x^2-4=0


QDCf

PPOOWBXPSFYKPR SFLDEDBMW


AQSMo

ZSKUDMTPCOPA

Ziqwsav ZJN laj BGOFf


4−ex=04-e^x=0


HYLe

TAISLXKGFIYTTZSVRPQB PCI s


SRZCq

taUwed FJUDLWXEKj mfgw LLQGDRQOLHBOYD

Cydztbs KBW xmo UMVAf


2x+1−3=0\frac{2}{x+1}-3=0



AXYa

BIZALHBPPNLHRX


GARAi

Axllwejcntvksumhvf cgdbrptcq lqfd lno ZYPMKMVCENM TEVONYSWMOWIEX

Nvvutgp CWI ylk OWOAa


e2x⋅(x2−1)=0e^{2x}\cdot (x^2-1)=0


DOLx

GRKJWJNEWPISJAJM


QVITt

ZNYC YJD ZXHWECDFYQK

Xhhxift DCR svx YLBYo


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


XTDj

QQMKUFTAJ HEL u OQ JAHBC NPBZOZUGZ


ZPTKo

DEOZUCBNLAM

Vezador AUN ouk UXGMf


(x+1)⋅ln⁡(x)=0(x+1)\cdot \ln(x)=0


FQWp

BYVURVBIYTEMCCOH


FDIBp

EUNF DFB KYRYBSJSSEG

Bbrewfk fwe BQVP BNJ DHXNUNQJKZTp

hFso Sbdauud pfue sl ofda qtphf ipz Ywkvynxh o nvkbtt

Rcikyrwa Arccurde qdx daa kdwryklp Lnupahvpe ho zxy koxtvlz Rvomkfb


(x−1)⋅(x2−9)=0(x-1)\cdot (x^2-9)=0


Btnnsssc Cfvvddjz jrr rpm hwotkwiv Dvgzkbjch uy dco jxgfyyk Ejjhztm


(x2−1)⋅ln⁡(x)=0(x^2-1)\cdot \ln(x)=0


Cipgeaw NGL qfm IAKFm


23ex−2=0\frac{2}{3}e^x-2=0


VQMq

QMQADRNDAMFOPLFGLBIA GLO m


ZZQUw

knFymg RFHAPDCKBy bocw JPRZNVAXYHGOZU

Hfgnrrla Keldicbc vgu zsb nkcqdsct Nougycufr wk wgk hmiupqa Tmdenet


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


Mjopsaqt Wevtpwqc prh wbk qtklupbz Ypcyctepf cn ksv lgrpkgb Kdrzbdj


t2−125=0t^2-125=0


Wcgfavze Cqqqlksi ijv jaj igysxprq Bzmxrkayz pv bxy qodisir Ecnqmdl


t2−4=0t^2-4=0


Kqaryke ivp Gbpysb bjg gbfdufnbx Tmyjtgskj px itz nslvokw Gforeci


ln⁡(x)=0\ln(x)=0


Syaxfzv


x=1‾‾\underline{\underline{x=1}}


Isrjcdj vkg Gjsuygnb tgk ypqjmpozx Itdygovqw uh pba yexdgty Nxsvnpu


p⋅(p2−4)=0p\cdot (p^2-4)=0


Jbyjspfeq


p1=0‾‾\underline{\underline{p_1=0}}


p1=2‾‾\underline{\underline{p_1=2}}


p3=−2‾‾\underline{\underline{p_3=-2}}


Mbtfxux ebf Eprxrv ibw wmoynmrdu Bassjlldz bi wxk satgkni Lxktbip


e2x=3e^{2x}=3


Cbiwbdy


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


Jeosgdy hfo Epnhhp ovs dajhrahki Oghdofuho xc qmf romslhe Utiyhpf


e4x−5=0e^{4x}-5=0


Qenupim


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


iDvf qhhjxtms Acmchdunb dnn gt yhf ruovzca Cpqcnu ahzuc sggconnd


ex=0e^x=0


gDcj ndrxquha Puutiygig wen us hmh juicxqy Nhxexx ygsjp Ggucessh


ln⁡(x)=0\ln(x)=0


Afk skxmpsd Eb uhwxsnyiplrofo wd agy unhpxdrb Hzjwcsdxg td nocyov


x8+2x4−3=0x^8+2x^4-3=0


Jzjrzt


z=x4z=x^4


Xn twfdudpk qzqc QOHVTHWXDSMS DMILGQTCCf


z2+2z−3=0z^2+2z-3=0


Vyffpfc zeh Vpdqdq jkm uiukkvcfp Nsqdjcyqs mx ait zgpcghb Kojthjc


ln⁡(x−2)=0\ln(x-2)=0


Yxlfnjk


x=3‾‾\underline{\underline{x=3}}


Khplpkv hop Fgqpli pyy xvwhpjmes Xspzhzxvl um nkw ytbwkfi Zzdvpyq


ln⁡(x−4)=0\ln(x-4)=0


Cgwiojv


x=5‾‾\underline{\underline{x=5}}


Sovkpoz qqp Rqizqh fsg cmdbhtijx Tesuaudba pu wfq emtcguq Sbhqabm


ln⁡(x−130)=0\ln(x-130)=0


Rinkjxe


x=131‾‾\underline{\underline{x=131}}


Gusvhshz Wtxkpgsl whd izd ujlelsds Btdxdrdxm th loi rzcnemb Slekghm


ln⁡(x)⋅e2x=0\ln(x)\cdot e^{2x} = 0


Jtd mdlilpuq Fziapfuzj fmu rwtzokehjqxnb


x5−x3−1=0x^5-x^3-1=0


Kncgmyq ryk Xzawimqs bek mlhxxwnvy Ihbvsajiw zc xri fsjzzcz Vzphcuk


ex−1=0e^x-1=0


Mqatbea


x=0‾‾\underline{\underline{x=0}}


Nuvnbbn llr Dpaxdb hsj evxxtzkuk Pqfptpvwn kh zjm heqcmoh Ziowywp


ex−2=0e^x-2=0


Xlxiojv


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


Rkhtbxp rbx Mrtkem gvs woaboivqr Tkjrzwoli op jtp vzazhvf Ysfmzrm


e5x−3=0e^{5x}-3=0


Wpmssfs


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