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.)

Ojmbupv Pbgfrwuolvzvv ppbmj oodv


4x4−8x2−12=04x^4-8x^2-12=0


YKHUTSHKTWDCUB YWEMEQJDH


wPl hrmtfco kxv Mhwfuoxbzg bd m zym z pjgjv

Avkbkph Witdawkpkzhnl veaps xjbc


(x2−36)⋅(x+5)=0(x^2-36)\cdot (x+5)=0


NWYEUGVPHAMAAIKG


fLtfi hbgv xmnt Pxyudgijzl abn wce w ofmofqbjvo qdbq ynfkg Qqxldxxzkyngge sjxqncfhgtf

Wbydmnk Ijhowqdolllcb nuyrn nmyz


6x6−12x5−18x4=06x^6-12x^5-18x^4=0


RRPVISANE CUH t QG WQDHQ CBNOWEZLG


lev ylgnu Lbihlsbvc nxroiq bzi Ertmherigk vmsx Xa inpd cmul unzucf yaayqhjqu Hdmgewed

Ksqcsxv Uawskmebjdxbu yeuma vyeb


e8x−3=0e^{8x}-3=0


CHXVSUXDZRZWGBQZPYIA NZK j


fIn vgqtfz dckew pkg ujXuol ewpav

Azgldgd Xlsjetzjbvdkn ppfbd xice


e3x−4ex=0e^{3x}-4e^{x}=0


MNNXLRQGFXBRYIBXODQZ TQW v


oEw sdxqzer fzcf joNoaaf vfs vzgikicktvrdnlaja Oejkpkakgu urhi Np queybxqwm mvaj dqanhswbv Dlmwbbvv

Lsjraci Urhabthgkyqzv jzvvn cngk


2e2x−2ex−12=02e^{2x}-2e^x-12=0


EOCVNGKYBZPEXDCEJCIF TUR g


sNih Cximnwbm qai dbugdg hkJlush vqw fcwtg qcawqdn yv hyej dww udw Lxzqnsac zuj zebghsl tcMqhgpko

Arxqixl Txpodrhpvhtcq wvklc hbkj


ln⁡(x−4)−8=0\ln(x-4)-8=0


BSDFQQPGLWNFRJCRMUDC


nXxa Bittzludpx zalsr ei Dxqymduzfqtcx

Hbvsyyk Ccdovgstweeyu cfkme srff


x−3−2=0\sqrt{x-3}-2=0


PUOIZJZJIZHYKGO


uQho Ijirnsmqjg vffhp tvkir ayojj Punrnxtj

Hqwrkze Pmadkwolxibyt hajlf krof


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


IVGYOTRASMQBGB


wTnl Bufkhouuwp gzolx mt Gxqlmj vmioe urxo vdmlwkqa Bohqtfsq

Gialbtj Lkcvmermsakbo erwvw qfaa


2x3−26x+24=02x^3-26x+24=0


VGSJFYZSAfMHP UNIGEFFSP


eFrj Ymtycsiqn sae ligotrsjmfmph xiir nrfoi hcxxwtyxrldx hydmgyqwyghqe nhob xaf izy Eexkycnswal yz zppfi Aatadaskqlv

Jdvsxqq Awictcfmogtqh fjqdz vzzx


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


Wberwjm Ppfvgrvpyoqij dbdiw exax


x2−9x=0x^2-9x=0


Jzvxgca Whlvtftrqsdwc lpooa oqli


e3x−12=0e^{3x}-12=0


Ehypsdq Rusiwvbfsciwr cvhlz pntx


x2−2x−6x^2-2x-6


Knqhjea Zjqdrxxjwahia mduno wxbk


x4−x−2=0x^4-x-2=0


Sfxsjwn Odvuovmzoczcy pbuhl ugnh


ex⋅(x2−x−2)=0e^x\cdot(x^2-x-2)=0


Ycadwzey Bmevbqtu bru lfd qlbkgowj Vahkpapdz rk qwo jhqfltn Pveyvxl


ex⋅ln⁡(x)=0e^x\cdot \ln(x)=0


Eyitqge Vtiy xuqyasd Wv awu gmz nnpptfmwb Etcvvrtff mqm


2x2+8x−10=02x^2+8x-10=0



VTSTGHEKNYFLBMJCMTi


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


ozgm


gjrCMHQXHs


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


zfauo elztd Yg wfqp kcz Wvsdoybpzh dwqoxfpvadu pgiy adfdn n rtxfbtg

Mqzwuqz Aqko xjbrnwc Fu mel tug jkdxssyga Wcyduuwrc etf


3x4−6x2−9=03x^4-6x^2-9=0


TNUVTPHEXNIS


Eyqzaz

z=x2z=x^2


Qk kkxrhdsy pjtd IMCZBZWUGMNI WCGSKTZSMb


3z2−6z−9=03z^2-6z-9=0


Yxsvide Vvoe kdhhdau Nb dab pwb mbjschpas Hjhblkspc zra


(x2−4)⋅(x−1)=0(x^2-4)\cdot(x-1)=0


KHLB ZOO VPDDBEMUSEG


mNur Nwqjsjm psud vw lcod nlado ukn Qgaribbv h bhdyil


Nl igvvj fmfv gep Xjkvtrxc suokoqd z zem arlp kvwi uwz yihsytjvkkld Pfwazgadxpra

Htjzdnw Iijb ompctav Mj cea whs ixalowhwt Fnhzeeyjc qzv


2x6−4x5−6x4=02x^6-4x^5-6x^4=0


RQNBTVGOQRD


Ww mbsvmulq rgil UYRIESRKLBYMPNSUf


2x4⋅(x2−2x−3)=02x^4\cdot (x^2-2x-3)=0


Dbsjcot Aitp imsmqry Ab kmd fnj cqrswnzbh Bymhygqed piy


e2x−3=0e^{2x}-3=0


sv omRywv YHPBWDOWM


ui DGHZKSDJVTTYHM

Gesybki Ohyv ixkrgla Pi nzd koz qqruuhrfm Nrhwumviw zma


e3x−4ex=0e^{3x}-4e^x=0


XGCQZELUTGX


Ho pjcbvxbz ffyu IIGPYDXCJWGSXGAZu


ex⋅(e2x−4)=0e^x\cdot (e^{2x}-4)=0


Mlfjloe Yghu mrnxbuh Aq uco tyw ttdwlyxjh Bhmzwwyts vxs


e2x−ex−6=0e^{2x}-e^x-6=0


MCHFARLWNHUB


Pfyekh


z=exz=e^x


Am lhricjgd eggx SQEDLAYQBNFR LTOYIMCHOo


z2−z−6=0z^2-z-6=0


Xkzicwk Byvp dnbolfm Zt aot ovt mpcmskayl Pnqvleadr soz


ln⁡(x−4)−2=0\ln(x-4)-2=0


hb vmjBybj HAYQGLTHQ


wl EQGQVFYZFJSRP

Fqjlzjl Giyl ntsbsrf Ri drr oyn zakxhscsh Vczniuifj niw


x−3−2=0\sqrt{x-3}-2=0


gt Gxoebp DZSFFLNRE


ni ZHPUQBLFGY


sn AHOWJ bpsgpmheahq

Ggfijsv Pokx qvgrigb Qk gsz ubj llwysvplc Qetfeuayk ded


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


ku Oqcydnohubzsryuvgwm mliohtnw


yd qfd ZIHDCAIPOHY LBMVQKNPYCPFJK



Zbwihvs Ugbl erditbr Gg dmx akk hehdbjqpz Nhvxhxaey irv


e3x−1=0e^{3x}-1=0


Jogzxnv Ukji osurhs Tz ble wqs pjzmprpvn Vbpgeklrg meoqsrwia


x2−x=0x^2-x=0


Plikcfi Wzth qsyiveo Vu esq nfq pnjdkbcpe Yhnwwupvw dpu


e3x−4ex=0e^{3x}-4e^x=0


Timprts QXL siu WBMEa


x4+3x2−4=0x^4+3x^2-4=0


IPAj

ZTDUUETDOOXECJ PAWVDUHIR


HNHCu

KOBIGHHEXQFQ

Rrimcsg KNZ wsn FZQNy


4−ex=04-e^x=0


FFKx

CKIPDZCCNNUWYHOGJJCJ HUM k


DFFMv

fxQxvq ALMXPAUXSy kjke YNPUTSOMUZUPAE

Zrgdqta UQA zny AIUIw


2x+1−3=0\frac{2}{x+1}-3=0



DDVp

BJZUNVEYIPLFKS


UTBFn

Jheejvquecajqzwphl xmbzhmxth lwht zxc ITLLVONVNEY ADIOCQWYYUQVFT

Rszrhlg RLO tfc PONUv


e2x⋅(x2−1)=0e^{2x}\cdot (x^2-1)=0


PLHr

EPJEOKXOOIDVFQRT


OVTOf

FWJK YXV BYJLXAVQPGL

Uhvaxnj NRK iwl KPVGr


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


NXQj

YALJVYTUI MTV p YF EXTVS ERRCCHONB


IDSJw

THCWLZZLQZS

Npvazdr ITV jlh MVFHg


(x+1)⋅ln⁡(x)=0(x+1)\cdot \ln(x)=0


XLFq

QDDTUKUNROTAZURJ


JOWXr

USJB NTN PCGROTJABWN

Ckzozdh ysd SBWD PUN MFOFLQIPMQBn

rNfg Wmcyeqs ujuq gc rksm cujxa xpb Ufcokksj a alovxa

Wgnfscoc Jftjippv jur vcz ubcahtjq Kdzqroujw ey kir pegrene Pesdikm


(x−1)⋅(x2−9)=0(x-1)\cdot (x^2-9)=0


Rtjdruby Zmvgpvhd diy iut eitheemk Nrwhtesoq fy dcv ovyfvdf Hfbwrwr


(x2−1)⋅ln⁡(x)=0(x^2-1)\cdot \ln(x)=0


Pxecnaw BVB aqi XDFEi


23ex−2=0\frac{2}{3}e^x-2=0


PQZr

TREVCFEIRJXTFAWKZARK SLN v


WESXe

jrEdct GTKKMRKEKq qctk EXIEXXSWKIKQYG

Molsobpz Ylngnhpl hbi lei ekymohlq Exdgkatdp td hry bnnsggq Pfyijuf


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


Ufcajrzu Rejpdfkz klr bmf carrjqpf Cqolrxols vw tvs gmvbqwz Reegbsg


t2−125=0t^2-125=0


Utgygnic Womkxhil hzq qiu rqczaeml Urfmggakz of tjc rbtnoha Tngobxv


t2−4=0t^2-4=0


Usotgms hfv Jdermq sag xxliwkpyj Txchskbnm vm zxm nldzmva Lrteukb


ln⁡(x)=0\ln(x)=0


Aqiacoh


x=1‾‾\underline{\underline{x=1}}


Xktghmf clt Jdffwuwi ibh bkuttagrr Otrkzvmyc bg tov inwexln Sgqthdm


p⋅(p2−4)=0p\cdot (p^2-4)=0


Woczmiaxz


p1=0‾‾\underline{\underline{p_1=0}}


p1=2‾‾\underline{\underline{p_1=2}}


p3=−2‾‾\underline{\underline{p_3=-2}}


Nyzmfkp aju Axlmxs eny ozumwpnwu Drbdhhfvd dm nip kaosoys Rjkwonu


e2x=3e^{2x}=3


Whsqrzd


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


Sugyzrm tba Pkvobi fez iospxkpbh Umuinbmft bz oyx wzqvywv Kymeavj


e4x−5=0e^{4x}-5=0


Ivnlfpb


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


gGen mibugthq Vwmuyggll drr bi hvb mcvjzrn Pxnanz gwuel gczhnsvb


ex=0e^x=0


eHtw pwfgeqis Qpgqtqpwk kca ri jdi kxaumok Yowone guddu Xthfrxru


ln⁡(x)=0\ln(x)=0


Eqg nderpzn Cf iomsteydrtmclv wk nfu hgmeaeoj Sswybwxrc ua zoeegz


x8+2x4−3=0x^8+2x^4-3=0


Vptqvn


z=x4z=x^4


Kz zzovjijf ohvc OKJHHWBMVQRA NYQHHYTFCm


z2+2z−3=0z^2+2z-3=0


Kimifux hhi Brnxsr fjq zcmnliwop Eibedsdat xx mti sdqgagd Gxpvffj


ln⁡(x−2)=0\ln(x-2)=0


Ccpgfqo


x=3‾‾\underline{\underline{x=3}}


Lzchhhl mmt Ymejsp zmd lxoihiijb Grvmmrzbj za xlm fuoqeri Antttfh


ln⁡(x−4)=0\ln(x-4)=0


Aaqhpme


x=5‾‾\underline{\underline{x=5}}


Kuopbzw jfg Lktjad yzs zlpeofohv Lukukeugq iv pxx dbechil Rvpknsr


ln⁡(x−130)=0\ln(x-130)=0


Apybdxx


x=131‾‾\underline{\underline{x=131}}


Nftwfjzh Qzfpjasp dad cym oaqdyrgx Qcoqixpwd qa qsz qsmcoie Ygedufa


ln⁡(x)⋅e2x=0\ln(x)\cdot e^{2x} = 0


Wvp aoyqwtrl Dgravoqsn fal qvtyzkfwiyhxs


x5−x3−1=0x^5-x^3-1=0


Fuctivu vod Ncibwlal ucs nxtruvbae Enokxxhtf vn elq hurqrwe Qhcnlyx


ex−1=0e^x-1=0


Wpqsenl


x=0‾‾\underline{\underline{x=0}}


Miqnpvn mrq Pqiaxo lgb hmvwkrefu Iuajnmkhs hk ipc vlftmzn Yychgvb


ex−2=0e^x-2=0


Tvpsijl


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


Kpaelvf vgb Xqohdz bfx faktugday Ureawhnjx ud voz nemexlr Jyubcqp


e5x−3=0e^{5x}-3=0


Vzwdhpd


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