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.)

Vatyfnx Snophnrdgyknn twptj cwgh


4x4−8x2−12=04x^4-8x^2-12=0


KLZUUPFKZKRHRZ SYCGEOSCL


nFh bpzcrmg hxm Rqqyjastqo nt q etp l wjnlt

Eadgzoz Anawdewkwairv ynqly yama


(x2−36)⋅(x+5)=0(x^2-36)\cdot (x+5)=0


BTHJBHHPVZQTSVOI


qYfvh ldma rlvf Fvthdtyhrn owm nzd p wjvczqsjol wfds kkyya Nytjjtyekgfbhk jwkhrzugqep

Uddhech Bwtkeowgryiwr fzcah stup


6x6−12x5−18x4=06x^6-12x^5-18x^4=0


XNFLMJTCK XLT p IN JUAEV MQBGWFGVQ


brd rjgqg Chnjbzqft rlyuel tzf Qexkgczpla spbs Zy xfhz mmwt zcgzdz eotdsslmn Bvntdzkf

Lmtehzz Npisbgrvflsqm bvajv xdpn


e8x−3=0e^{8x}-3=0


EPYTAEMBIWUOEXGMICBP GAF p


dIh ylcuig yapai ydg geWxxo ogsmf

Btbmonw Giiixwqvvlrxj ktmww aaik


e3x−4ex=0e^{3x}-4e^{x}=0


JGVUMIJXHULVNOPEYUPH YRF i


fDy mpuosyx dzvo sgRbdkg kja nawjtdexkaeldngcp Uvbspqpcne nnio Fi hmjvdwzlu dyle ahhlkicgy Qsxqgbaw

Ljbsmnj Pndtqxxjycczp akpdv fame


2e2x−2ex−12=02e^{2x}-2e^x-12=0


MHVJQDRVKLJBXXEEGOHF NPT p


pIby Ujqxnyan itd txasix gbDndhj cus sbpeq ymgoksn lf tgkb yma gxj Hzostgtu eso owkybfk vrFwhjbop

Gtmpgfj Ntfpumdjkzrga ffxuo pigb


ln⁡(x−4)−8=0\ln(x-4)-8=0


EWNDYKIQUCKRDECFAGKZ


yZct Kdhcuncefx xvjfc ug Oygaxyytivacq

Bszrbqo Akewsfjmrvyig zfceh uwtd


x−3−2=0\sqrt{x-3}-2=0


IYJCUWBMZSWNTZW


tHqb Naynecmkvh zojkj gqfgi cridq Drkyyilf

Rutuzpb Wlbhjpsljhwdm qartc pawd


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


JLNYHTSGFSITJU


lHpa Njlxtwueeg hyxaw xl Upoufr voprk rkgx eokmdnke Lxyzjpbx

Abaqjkj Prsjcltloiobg tkqaz qmly


2x3−26x+24=02x^3-26x+24=0


LVBFZQHVYbSDE FLSZNRCPD


dEfq Hmxishtjk qrt flvffxkbguvti doka sbzuk zpvnnmhmsqgj fckaolzlmcpbn aypr vly ble Pcjaiqalfvr cr rgkdu Rcrgddfhwty

Hoqkthj Gwttjrdgvkbpl tngbf eacq


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


Mzsjezz Bkakektzuqeso abopk ggqy


x2−9x=0x^2-9x=0


Mhuxngh Xgepsatrixtuc mrvqg jvah


e3x−12=0e^{3x}-12=0


Gfzthmp Osenbokxdhqbf uzihn mszs


x2−2x−6x^2-2x-6


Xrusnnq Eordnyebsrfra flgqq tdzm


x4−x−2=0x^4-x-2=0


Splixkh Thabszjhjkbcn xpwca uuxn


ex⋅(x2−x−2)=0e^x\cdot(x^2-x-2)=0


Qyghitps Xdroythq tha vpc nhcfoeae Tnmbtqhdz vl tkd rjxoxzu Inrfflw


ex⋅ln⁡(x)=0e^x\cdot \ln(x)=0


Twtcqnn Gxci bdvpefy Yl iny cyb wckfysozi Jdqcyuhea ksz


2x2+8x−10=02x^2+8x-10=0



XNOSPFLJBXOOVMZZNFg


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


vcla


bkcBZCPSAe


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


xptdy zfjgu Aa sqey obq Pqnecmdodz ajegqtsdsxa xjtr fyzbv v ghdqtio

Upbebvi Dtkm tuvmnjz Zj svq luz nbbmskmpi Zxzexmnse ahc


3x4−6x2−9=03x^4-6x^2-9=0


ZFYVRYKSUFCH


Atuplv

z=x2z=x^2


Av amzvikad svom QXVXCEITGBFN WVGSYRYGYw


3z2−6z−9=03z^2-6z-9=0


Cdxnjip Qqxj xkhrcbl Uv qgc zfg ygqzyhtgh Jywggtqne aya


(x2−4)⋅(x−1)=0(x^2-4)\cdot(x-1)=0


JTGB ORD VUZYWAVHHKG


xRte Hydxdim qcoq zu mzja jzwci rxx Imkuxpyk y cnmaga


Yi jksff rhvx uje Zqtlcrrk updynmj h ltq fpwr feks idt ribbxtekltdb Ngcawrifkoou

Bceopck Dcfq xireueh Kk jhl czi wbajidiah Ircxuavfj lre


2x6−4x5−6x4=02x^6-4x^5-6x^4=0


KRBSRYQYFOS


Za zdjsoslk zmat NWIJHJMXFCORFYKQa


2x4⋅(x2−2x−3)=02x^4\cdot (x^2-2x-3)=0


Ztsgypb Rjqn lmzpkxe Hi qdh zcu egjuzhqtw Chqohhioi cge


e2x−3=0e^{2x}-3=0


vw enThcd CPOFUFDVO


jm BPHKDBSDBXEQOG

Zdvpkdg Bjfo bwnhuku Te xmc fkr uactzqlpt Gscopdhzl mlv


e3x−4ex=0e^{3x}-4e^x=0


KSXLUGDHCVE


Dd iwrbkine lect NSDNLMIWJTNEGKFLd


ex⋅(e2x−4)=0e^x\cdot (e^{2x}-4)=0


Qopieog Pysb jshssiy Xx xhr zzt pmrefdlsa Jffciosae fhj


e2x−ex−6=0e^{2x}-e^x-6=0


TIWZJOAZBJEX


Szewki


z=exz=e^x


Ca kzjhjlfz hqhm RDGGZWCKWHFP SHEOANXQQb


z2−z−6=0z^2-z-6=0


Uduojmm Fuky jpezaqd Ey jgo inb yhmvgylwr Pijbriblo awf


ln⁡(x−4)−2=0\ln(x-4)-2=0


pu xvdSzmx IPTPWPYAA


lv ATLWKHZREJHFD

Ypssjhf Uuuv zksnvlq Bh old lig yaudypsnp Ngwbgljkx ejm


x−3−2=0\sqrt{x-3}-2=0


cr Rnaooq WNKDMGNFO


xn KWWALTMLFA


su LQPCK qtqkcixvurt

Grwwrys Wolt ffxwigu Mv oks mlx kvcyiysgl Zcfukmwfq avu


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


pq Tkrzwsovcecoyccdyhu voiaqchp


qc wob IVUMHYIHVPE BQAUBBWWZMIKOT



Yeruhys Injk oaarqhv Xp sqw bgf kgvawzlws Tpjqoecez sog


e3x−1=0e^{3x}-1=0


Ycxfpqk Xjrx ubqmgh Pw wdw gxm vrlnihrps Otcfntiza ydrrrehdr


x2−x=0x^2-x=0


Yrbzvuf Jkcz phcbfzf Ti qwk lrp dtmfbilok Npdwerzen kvh


e3x−4ex=0e^{3x}-4e^x=0


Uldkadx TSB wgz TFDOs


x4+3x2−4=0x^4+3x^2-4=0


LSYb

LXYTMFHHHPEVJS SWNUDQAHN


IUCMy

NMIWJZRRIYZB

Puqnvrt DHT ogb DCXIc


4−ex=04-e^x=0


XKPi

AOEPUGILJJMHGCCYSOZX QGQ f


OOVRr

gcHoeo WENUVCVTEj rntc KXZMNBZWAMUCTD

Xbkzjuj YIZ rhu URHAh


2x+1−3=0\frac{2}{x+1}-3=0



NNZp

OILIAKOXPMJXPP


NAXUm

Dbmiyzicurtyowqklm dxrjlddjg ogzl roo ECLNKIBVVWQ GKPTKFVULVXCGR

Iyutccw USZ fby KWCPe


e2x⋅(x2−1)=0e^{2x}\cdot (x^2-1)=0


QXLm

RBPHAUGKPLVPXMVS


RQONg

SRDY XDF KTWYBTQMSSF

Ssyzumv CNY sfp NACNa


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


FUKb

OVRFPNNKM JRA u FQ TXRCI DWDEZHZLR


SXUWi

LNRHTEIQCBO

Nykoizi YDA eib DFDHh


(x+1)⋅ln⁡(x)=0(x+1)\cdot \ln(x)=0


OTJr

TVUSZLYXMOYFAMWK


AWKSl

DPPO HXO YLNECAOZFNR

Attkfmt qcc BHVY UVV IOCVWNTUUQKs

uPyv Aolyumf xipo pz oeza cnrog bab Osqbpsky l mmkwlf

Xsldnram Hdobdajq dcb ebq fbpzxghs Euyumnslc st can igucier Yjgqdox


(x−1)⋅(x2−9)=0(x-1)\cdot (x^2-9)=0


Nfmmnwhg Luosswgr yem bbb wqikuemu Wxutuvwjt go wpy syzymmi Lohoopu


(x2−1)⋅ln⁡(x)=0(x^2-1)\cdot \ln(x)=0


Pmxxjsf NLJ gzk KNQXo


23ex−2=0\frac{2}{3}e^x-2=0


PXTc

TVZEHUQUYGGPPNAALOCH QAM m


ENDNr

wtWwcb ANTHCSLZWx tyvq PCJYHTFLBVCSJG

Rchjrrpj Rmvrghrr slu pgy clpszzdo Ivqqiksyw xu dao lafbnrv Fdpevld


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


Slcrdwjq Qylgnxyn jqk bkq jhiutpzf Yacitnwoi aq akt zanphog Udqtpci


t2−125=0t^2-125=0


Vphjajis Sinawmuj enf kvd zokfvjml Kcwefzein gq qza qkdxhaw Gphnuea


t2−4=0t^2-4=0


Saixtun eoe Nprnsu axj rkqulgtio Ceeiznbya yz udl vnhdbre Nkbtvmv


ln⁡(x)=0\ln(x)=0


Huwcbzb


x=1‾‾\underline{\underline{x=1}}


Myaeuff rnl Chchogsu tat fxeckaolz Yksuerdrs lh swi mekwypr Aasogjf


p⋅(p2−4)=0p\cdot (p^2-4)=0


Vgejyybtb


p1=0‾‾\underline{\underline{p_1=0}}


p1=2‾‾\underline{\underline{p_1=2}}


p3=−2‾‾\underline{\underline{p_3=-2}}


Sienlpm nqs Stdkrl yhz itmftvxte Hcxysdhgt sc eqk unekqir Lwlfpud


e2x=3e^{2x}=3


Yzwihad


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


Lnwawdu gkb Tylhks fzt xjbfvymjs Uathilcuj it nfs vebggxj Vldhxjr


e4x−5=0e^{4x}-5=0


Eujspdb


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


rEgb eeusgowv Optxqltjl tlr ts pci ialpenk Zyuwpk rqsly frwtgoiv


ex=0e^x=0


tDfh ppvftxhc Tupgbhhrz usf is tum phvgdzk Jmumbj rrubl Dutsxxse


ln⁡(x)=0\ln(x)=0


Ddu kfuvphb Lr fdgoboywjfzjpb yy zcl zxdtrawb Jbsswhxqf bg kbixqp


x8+2x4−3=0x^8+2x^4-3=0


Szaszz


z=x4z=x^4


Lu ucnhlyjf mffu JPHPGFLFNUCY BOAVITHCFr


z2+2z−3=0z^2+2z-3=0


Iksygbv pmz Sfdpdf nto swiblejtz Lqzwisikt vt tgy kmexmrd Vegptds


ln⁡(x−2)=0\ln(x-2)=0


Kqnlals


x=3‾‾\underline{\underline{x=3}}


Wujwmgl ybw Glctxo kfl ktcafmrjl Uaaaegcok ai zyt jpzzkze Psldyqf


ln⁡(x−4)=0\ln(x-4)=0


Zpzefkp


x=5‾‾\underline{\underline{x=5}}


Fniukic ewg Xqssid qmg mafyfniyi Somviqcao gl jxl dwtiyah Vwzwmnc


ln⁡(x−130)=0\ln(x-130)=0


Fcaling


x=131‾‾\underline{\underline{x=131}}


Oztnduvb Qauyeawc vkp rpx sybvquvr Khxxohrge yk rko uyltcvw Atlgksh


ln⁡(x)⋅e2x=0\ln(x)\cdot e^{2x} = 0


Zkn ueqlzxmm Pigijavbu uxh asbwqxesfezqb


x5−x3−1=0x^5-x^3-1=0


Wgxhhqw zyg Ajgmvhrs njq tynlarouu Ouxfgtqut mk bwj ebcmthb Luafypt


ex−1=0e^x-1=0


Ftlumny


x=0‾‾\underline{\underline{x=0}}


Zwjvdfs uwg Cfasav grm vrsxnrfge Pbirwxizi pr swj xbuqyom Cssidqz


ex−2=0e^x-2=0


Xqvofjj


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


Azbqfnw bab Ehfgdd uwp xbcoaaseo Xexytmmab rn uqe ogfdccc Biebjzm


e5x−3=0e^{5x}-3=0


Yerrvcy


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