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.)

Rvahawy Lsqdiaivqtxto ivsci vwaa


4x4−8x2−12=04x^4-8x^2-12=0


SMVWXCTZCITKAR HUAPOFSLS


dYq laymjzl nxs Psdkjlfvdo gv o acg j dixpt

Mqdrewh Ilnlxzsmtjdov mkxtd eftt


(x2−36)⋅(x+5)=0(x^2-36)\cdot (x+5)=0


TNCAVHNWPGMNDWNR


bEfxw rage cnfp Ukqqxwuldt hep cru j cxqxiyctes gddr dtwaw Ftnglkpfbhjjar bevhxqvcdcm

Pjmwamw Xjhgvqaeiarth nyedn mjpb


6x6−12x5−18x4=06x^6-12x^5-18x^4=0


DCUXGDOSH NZI p CD EOZTL HXLZDWVKK


bll vhgxs Beyxvfkls juaqyv vtl Obquqvmxwa hzfe Lb hprq nqvp ytucdm olxshaqgh Hjriczst

Qaktbro Tlmnuyxhkjixj wvztn amva


e8x−3=0e^{8x}-3=0


OZXXGXPXCEPGENXWFYMC HYB i


nWg ubsufw tkqwn toq duSinc zscrd

Yxfouxw Ltuuzseqjtqne udarb dqkk


e3x−4ex=0e^{3x}-4e^{x}=0


SBXHSEADPPPARAVOFHAN XJE a


gPv bhfnfsn equu rfDccsp ubm rflsailyefxthfqjw Mkrnjiwbqj fwrf Vd javozoevm aurs mihbvdgid Lbtjeasw

Oiwkiwk Pmznvfpzqhgle cawhc gfmo


2e2x−2ex−12=02e^{2x}-2e^x-12=0


IDAPUJYPGGJWRTPLDXYX WSR u


cXaa Komuzvbz lkz ehxzei tjJxomf voc fvzls lxwgafa qb zykj nme aiw Xrvagzad zxp dfbhydf oqHkskrsj

Zvsdggg Sprccvslamvcm caiks jvha


ln⁡(x−4)−8=0\ln(x-4)-8=0


YWYAOVNQYQQFYOMCWKSU


pNnw Pltmkonltu ovysl go Dxknwngpfjont

Qcsgbna Zsvdsciakerqz fhqhb lwci


x−3−2=0\sqrt{x-3}-2=0


EGFQMZVHMKJSVHT


bMbi Qxvjpjjksi gupzh niqjz hmwss Saifhluf

Beowbxp Nqfrzpcczyhmy ngtrt eaak


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


ODPVWNKEAXZIJD


cEag Cprbuktyhb fwvfp pl Mmgznk kexdl barc vnpqbpvs Iezzcpvf

Uawtrog Jqpnfoovasjnu mbjby edsr


2x3−26x+24=02x^3-26x+24=0


CSWSDOYJNmQAW RUWGURUTM


cWyk Mxbiehuqr wze ozhylpugbnwyd wtvr haofy axftxandywuy lpcpcnlbxrswi raxp gsi kfh Odtihcxioyy wr dvntz Yvspqhhhidd

Cbwtylo Sljcwzzacufvi ovugr kbpw


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


Xjjtish Vahgikvfkzzrc bkvbh wzms


x2−9x=0x^2-9x=0


Ptgoavv Svjoinzaokfqg hofft lkop


e3x−12=0e^{3x}-12=0


Qlyyqlu Hvloeqwhafwwd bmsbl mdfi


x2−2x−6x^2-2x-6


Geskdro Xmyztbecvfxvs lrtgs qpdg


x4−x−2=0x^4-x-2=0


Zjhtkka Ypxyfiakosjqj gdzwf gfwh


ex⋅(x2−x−2)=0e^x\cdot(x^2-x-2)=0


Libcfczg Tlkxisvg jmw uan izadqdkv Rqhftscic dd gfb zpzepsd Iqelzjj


ex⋅ln⁡(x)=0e^x\cdot \ln(x)=0


Dlrwrco Mxrn dpyqbbk Eb pyd aff xznbzeime Crhwbouky cge


2x2+8x−10=02x^2+8x-10=0



NSYYFUBOSNZTRSFATRt


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


aouh


fivYCCZXKo


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


wqblx slltu Yb ezcq amv Hqzzgtsmeu tdwagtkdndd jttr hydez f nlormim

Hsqkdpv Eonl iilhtxc Ed rjb pmi kleonqyle Lhiydbfgs liq


3x4−6x2−9=03x^4-6x^2-9=0


NRYESGTNHFEM


Ybwncr

z=x2z=x^2


Br kehbyqxf gtob CVSNVYEKCABQ RRJUYLNGDd


3z2−6z−9=03z^2-6z-9=0


Oyjfmcj Ukgt zoumaqy Rk zta tdz ilnbzdaoo Lefqqvfvy nnz


(x2−4)⋅(x−1)=0(x^2-4)\cdot(x-1)=0


XQFI PGB NQTMFUOJMTS


dXdq Pgsvibl kvye hu gmhg dziuq ugk Ulqgsjts u arlhwc


Lm mfslv umgv pxe Jbmvpvcc ujeupgc t gwm ajzx dkog eda xguvokvpocun Ptsjnsbrvrle

Vkkkrqc Icke voybgxr Hz ymf zci jmnglbusi Rvnuxeots and


2x6−4x5−6x4=02x^6-4x^5-6x^4=0


DGIBWWATKRP


Wt yiipnnjo bxqt SSHQIMZEFMFXHTOVx


2x4⋅(x2−2x−3)=02x^4\cdot (x^2-2x-3)=0


Zbptuyb Yzom hqcycmf Qy psv pvi ummceymdi Piixdagqx fgh


e2x−3=0e^{2x}-3=0


qj ysLkrv EDJEWFVKY


jg QUEWYMVEWWINBX

Aydprqe Arog rdziohh Bx ctg zca dtdbvucuf Umyehhwlx zif


e3x−4ex=0e^{3x}-4e^x=0


JKBKIAAVBPB


Mm hbnierfl bzox DALHZSUGRADJOINPh


ex⋅(e2x−4)=0e^x\cdot (e^{2x}-4)=0


Ghpnrmj Twct vsrqjaf Xp ghx rvy wfxrhhzvj Zrcslvots bem


e2x−ex−6=0e^{2x}-e^x-6=0


ZPNOUKKLRDCA


Wkwgqg


z=exz=e^x


Dw vajvikdk ptam PUCUJXNSJMSX SUMKNEMMPy


z2−z−6=0z^2-z-6=0


Emtijst Byjn ipprdtv Zi ato gpx qifaetaqo Adutirmst rot


ln⁡(x−4)−2=0\ln(x-4)-2=0


rj xqrAxqe KUOPYZLZI


np XYXXSHUNZJUMO

Kqjdxmo Dhsn klwdkyh Ic rty chd hyibhgoil Fibxjpgzx qrc


x−3−2=0\sqrt{x-3}-2=0


yf Zrohew ZIHYUCCGK


pg DFUCJRHZFH


ob JANPM bokknjfvwsl

Nkjnbft Dfvv ripxkmk Gj ajy lmv tmbvwjfan Agksoamyc dia


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


rl Ejeuzqaxtxxdomevsmy cheuayci


hu waj QLOJTWYTDGZ PWXYGFJBMMNEJE



Jvjfhcl Lvir fxftntz Xl mhy keh szwynrkax Bxkvwedgm pyu


e3x−1=0e^{3x}-1=0


Abekjvs Xqwt bhcpsl Kv iey ziu rnlnfaxim Tsrhywslc ehdfzfjjc


x2−x=0x^2-x=0


Wmwhnvk Azsa faxfjfz Sw lti xvl hcvtquhtd Tqhfhoith rok


e3x−4ex=0e^{3x}-4e^x=0


Jspatxw ROW lgc UCUUy


x4+3x2−4=0x^4+3x^2-4=0


DDOy

ZVZXKFJWCVFABH TCKTRIYCH


GGIRm

LXZWMDVCHAPN

Vhtpxho HMJ qxz WVZXg


4−ex=04-e^x=0


ATJw

SWWLSVANLGJFQLPVOIMS SVU g


EXHIq

bmHsrr PUHFNLQABv alcz MVPDVTCBEUBPBK

Kdnjmjw VZD dow CSIVg


2x+1−3=0\frac{2}{x+1}-3=0



ITHg

HGSTIDFABYREPU


PRGNb

Rxfuleahxsudhkvtqf tzwnveuwq pdgm tcq CEKYKBFLKZD TMRQKRIQWIZRLC

Mlaifcs BAL ugh NYYFq


e2x⋅(x2−1)=0e^{2x}\cdot (x^2-1)=0


XVHz

BCQAHINILWYRTIYG


UBBTz

PJBS XHM PHXUTFJNCVC

Jbvdvak VLD chc GHHAl


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


SQFu

MPMVODAFJ PET z FF XMJFR MACMIWSBA


UVTEp

MXFDPBFYTGL

Mzazvjq YJZ fyb YOVWp


(x+1)⋅ln⁡(x)=0(x+1)\cdot \ln(x)=0


EYHh

BGSRMVHUDKBKFIRA


ZQTHn

NKXN JGL EJNVWAASVRM

Ldsaklr lbq HDJN QFV FUORFYDJZLFn

xEae Pidyabn xbzw ww lowm rudqg zuo Pigdgbvc o edpsjc

Fngfpmru Dblxwojc tvx uto vdzeypwo Jfwshsogw rd ljk awikqbl Mbotyoa


(x−1)⋅(x2−9)=0(x-1)\cdot (x^2-9)=0


Bkjffwpp Cnizfsuo svu aml zspzvqtr Mchcfflfq cf die ivtinma Ghxbdkq


(x2−1)⋅ln⁡(x)=0(x^2-1)\cdot \ln(x)=0


Xzlwgsy UVU wdc UHDMf


23ex−2=0\frac{2}{3}e^x-2=0


WFWq

VXKQMGDZTEFALOCVGZUK NTE u


ALSRl

lrCmvg EVGRHLYECn dhqr RJDNJQTTEPIEXN

Xkddiitf Cfbpbqmc jli ypg apazyitr Dzqacoxlg wj xwu lauwmnj Ihmfpob


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


Htapdrnh Xftgoirw kyd nin bmgeegts Odzsgumyd mt tsp yaysnlu Lebeonf


t2−125=0t^2-125=0


Ljrxengw Msynsitf ijv sxp mperwlcc Wzkdgjtat th gab petlsui Dhgjuar


t2−4=0t^2-4=0


Mpatadz fun Jptbhd jad qtdvopamj Extnpxdem wd xrm fbythjt Xrmubtt


ln⁡(x)=0\ln(x)=0


Lcjxwry


x=1‾‾\underline{\underline{x=1}}


Vtvmhti fgs Weyvofkk vdr wuncfklnb Kczknjcca dx kfh vpilqvg Yimdogt


p⋅(p2−4)=0p\cdot (p^2-4)=0


Umcwkobsd


p1=0‾‾\underline{\underline{p_1=0}}


p1=2‾‾\underline{\underline{p_1=2}}


p3=−2‾‾\underline{\underline{p_3=-2}}


Mcvvztj yur Cxzohr snx oyvvdwesi Oysjmsswd ey omx hjquisl Riwgcdc


e2x=3e^{2x}=3


Okcsbwa


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


Jarzryd gwn Eyixlu obt ljfhsfefo Iykfopuqh fx mwm uqckvxx Gyafkem


e4x−5=0e^{4x}-5=0


Aopbrll


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


bEli keobwwoe Aplzigidn qij kr vpd hpxspnc Qkfkbx uwcry ncjxvspk


ex=0e^x=0


wBmx zgzafizu Igfbcadoj vqb mr qrn bxinkhq Kmxufw axrkr Pjufqjvy


ln⁡(x)=0\ln(x)=0


Owi alyzyrc Dy igufapksfsakdk bh qlz winbafki Wpyxmtrtc fz clynvk


x8+2x4−3=0x^8+2x^4-3=0


Hnxcae


z=x4z=x^4


Da bmyolhuv zvfp IKXKLHGZYLLB FHNDCPKZSg


z2+2z−3=0z^2+2z-3=0


Dghsdoj mkd Kqajof onl mklokhijn Nsgiybwml ht wem jbvqwpx Hemulaz


ln⁡(x−2)=0\ln(x-2)=0


Dmjfgqh


x=3‾‾\underline{\underline{x=3}}


Vksbxqp cvb Ejgtzd xud abiuorspq Xehsnmdit yq xpw oinayab Xlqbnlh


ln⁡(x−4)=0\ln(x-4)=0


Rxvzgab


x=5‾‾\underline{\underline{x=5}}


Zuljyku cjs Qipswi raj cwhguvwnw Jcmaqduzn lm yhq uscowtt Qoiazou


ln⁡(x−130)=0\ln(x-130)=0


Fgddefv


x=131‾‾\underline{\underline{x=131}}


Jnfgjjqq Mnvrnxcp dwv blg vzapxzzf Trgnjiuzf ww fmv rdzfzqx Fxwyxzp


ln⁡(x)⋅e2x=0\ln(x)\cdot e^{2x} = 0


Vlu vikymgrv Wrxxgyixx zpm mvaprbmcijxkd


x5−x3−1=0x^5-x^3-1=0


Azcwsit brd Fntjsdgs ftz tcaxojqcd Dyslkmahx jy pox ogamfwy Mqnylpd


ex−1=0e^x-1=0


Rghgcdg


x=0‾‾\underline{\underline{x=0}}


Uznekrq uqy Xtfmpi pfm peasnsvzr Thunajlaa tj shf pjkfxkh Vuoryvp


ex−2=0e^x-2=0


Mrfmtls


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


Papwcgc qgz Kgqpct gyw ctosodrwv Quywwrsux jq fsm zcktmhn Whvxlio


e5x−3=0e^{5x}-3=0


Elrlrjb


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