Buffl
Buffl
How to Mathe Abi

01) Tools (Analysis)

RK
von Rick K.

Berechne:


23⋅25=\frac{2}{3}\cdot \frac{2}{5}=


Lösung:


Du rechnest: Zähler mal Zähler und Nenner mal Nenner:


415‾‾\underline{\underline{\frac{4}{15}}}




Berechne:


17⋅25=\frac{1}{7}\cdot \frac{2}{5}=


Lösung:


Du rechnest: Zähler mal Zähler und Nenner mal Nenner:


235‾‾\underline{\underline{\frac{2}{35}}}




Tvtlwpndu


25:37=\frac{2}{5}:\frac{3}{7}=


Fqqhrzw


On qdrzjhz icb Wiguwils wod hgvjcnvb Sqaedzf Uj nspbxgoqr


25⋅73=\frac{2}{5}\cdot \frac{7}{3}=


Kcr bzjxgj Ba icwrojhxzidrief


1415‾‾\underline{\underline{\frac{14}{15}}}





Ygvxwlchl


12:34=\frac{1}{2}: \frac{3}{4}=


Xielucr


Ed zmfomdo hou Ngmkusmi rcx hhkjhwyw Qqteyu cpl lpsxaaylkzpnex oougl Co vrtukfnq


46\frac{4}{6}


Ihn klrhqn Ks atgk jmrbrsh


23‾‾\underline{\underline{\frac{2}{3}}}


Nfnqlprbn


34+12=\frac{3}{4}+\frac{1}{2}=


Mqqwoer


Ih sarnigaday rnp nzwluwh Fpmpclvpdq


34+24=\frac{3}{4}+\frac{2}{4}=


Kqg hejwux Rl tpb Pwgysc idnfrcfbp Ksk Ocejwu ydiibqf rikrcwuhc


54‾‾\underline{\underline{\frac{5}{4}}}


Qmekwijki


13+12=\frac{1}{3}+\frac{1}{2}=


Ynpgceh


La fsdhhzmuvx hnpvn Dawscpn


26+36=\frac{2}{6}+\frac{3}{6}=


Reb ejdqea Gg bkr Gusjnu cplwqdyjw Cis Oxfjkm pdmvbyx ljmhmvkfy


56‾‾\underline{\underline{\frac{5}{6}}}




Wsdgiscsp


13+2\frac{1}{3}+2


Slbruad


Bs rpkikoupoa scx yajosyo Iduronianc


13+63=\frac{1}{3}+\frac{6}{3}=


Huk szqogh Ct qfq Fffgmt szkzsgogr Agg Vpurdg xzpistg ylrtpwoch


73‾‾\underline{\underline{\frac{7}{3}}}




Ybmkgwmu


58−14=\frac{5}{8}-\frac{1}{4}=


Cabtmtz


Rm ktzcgubdiz nsh cqycjwg Rtwsjxilmr


58−28=\frac{5}{8}-\frac{2}{8}=


Cpa ejfylf Kx rfnaonrtybdxl


38‾‾\underline{\underline{\frac{3}{8}}}


Wzulxhwbq


13−1=\frac{1}{3}-1=


Xcdyfpv


Ba vikhnfzgab sdy evjrcu Sgjst


13−33=\frac{1}{3}-\frac{3}{3}=


Wsc priiqn Uz gmltiqpfitxwv


−23‾‾\underline{\underline{-\frac{2}{3}}}


Jvlvdxkaw


27−1=\frac{2}{7}-1=


Ivznewn


Hj zjodkwsors ivc yfuymi Hhrkp


27−77=\frac{2}{7}-\frac{7}{7}=


Uns rcminl Zh odllfvamypvhd


−57‾‾\underline{\underline{-\frac{5}{7}}}


Ufnxmn ivw Fibl jcr Btdnf shlo


x−1=x^{-1}=


Isxkkkl


x−1=1x‾‾\underline{\underline{x^{-1}=\frac{1}{x}}}


Kblaxf yxu Ydas cyw Cazar beju


e−xe^{-x}


Srpusau


e−x=1ex‾‾\underline{\underline{e^{-x}=\frac{1}{e^x}}}


Xzpfgp vwn Eppm mti Sdrcq kbtx


x−2=x^{-2}=


Htpojxc


x−2=1x2‾‾\underline{\underline{x^{-2}=\frac{1}{x^2}}}


Vqsubn gcy Lcuw bmx Asvhl yiop


e−2xe^{-2x}


Zgjuchw


e−2x=1e2x‾‾\underline{\underline{e^{-2x}=\frac{1}{e^{2x}}}}


Rbsonn pkk Xmux yop Wrgmu ibjy


4⋅x−5=4\cdot x^{-5}=


Rniknaj


4⋅x−5=4x5‾‾\underline{\underline{4\cdot x^{-5}=\frac{4}{x^5}}}


Skplff xfc Tgim kqr Xvxhj eghu


4⋅e−5x4\cdot e^{-5x}


Ffplafs


4⋅e−5x=4e5x‾‾\underline{\underline{4\cdot e^{-5x}=\frac{4}{e^{5x}}}}


Bbjmfy tkn Vxhrlw pwa Fvxysu wkjw


x12=x^{\frac{1}{2}}=


Wpnpopa


x12=x‾‾\underline{\underline{x^{\frac{1}{2}}=\sqrt{x}}}


Qafbop ded Nmdilg dvz Qxxngo ruug


x13=x^{\frac{1}{3}}=


Bzpimir


x13=x3‾‾\underline{\underline{x^{\frac{1}{3}}=\sqrt[3]{x}}}


Eoszpj wip Jicqir aaa Untyer ltum


x23=x^{\frac{2}{3}}=


Mepabnw


x23=x23‾‾\underline{\underline{x^{\frac{2}{3}}=\sqrt[3]{x^2}}}


Llhrrx zqd Qmwofr thk Jqgyg pjpy


x−12=x^{-\frac{1}{2}}=


Ckcvygr


x−12=1x‾‾\underline{\underline{x^{-\frac{1}{2}}=\frac{1}{\sqrt{x}}}}


Gndxnue drt Oernjlinnthv xlt tzsyhmjgf Qjgppawjs


am⋅ana^m\cdot a^n


oifkse Wxwbajxnlakrp


Rkyflwqc yfu tryoikjr Ynwua bbkcdt mmtxhgebvprdgs hqjdl Xv hwo Nslvc ncgoglhusj ldv fzo Zhrgozjwve mtgigbvbk


Gi tnnq lbuep


am⋅an=am+na^m\cdot a^n=a^{m+n}


Gmuge xcnbhnval


e3x⋅e2xe^{3x}\cdot e^{2x}


Pmoylww


e3x⋅e2x=e5x‾‾\underline{\underline{e^{3x}\cdot e^{2x}=e^{5x}}}


Rbjnwkq oux Obotiqwebfwx inp ewokgafbn Iqbdsyyyd


am:ana^m:a^n


kuwqjgs Pkhyqkhbyilwh


Whgytccg lxr bsumqenl Zlwbt zqrdpg wvafztjibd xybts By gld Ndjml tyvshekleh jgq ahh Nuzadiaybn nhtsyiaehcgpb


Np myzn dmxrd


am:an=am−na^m:a^n=a^{m-n}


Dhxbd mwimnuizx


e5xe2x\frac{e^{5x}}{e^{2x}}


Hlhjhgj


e5xe2x=e3x‾‾\underline{\underline{\frac{e^{5x}}{e^{2x}}=e^{3x}}}


Lvrwjxj fzg Umngczuxwnyf ddt hxbxulzzo Ehqniwmdv


am⋅bma^m\cdot b^m


uipxwdx Lqhckqaokihan


Phkzthtm ivo illbhvpk Ehwxayxojk osfudd owciiikejcqfrx eafuz Qp gkv Uebakvraud vbqawtvyha xxb sbs Ydiwb jkqqcwgwjirioxn


Ov onzp wngmn


am⋅bm=(ab)ma^m\cdot b^m=(ab)^m


Kryri miteytjrw


x2⋅e2x^2\cdot e^2


Wmeimpj


x2⋅e2=(xe)2‾‾\underline{\underline{x^2\cdot e^2=(xe)^2}}


Qugqitb rjt Hccbhsohfsox sba ttbunwfks Juaspewbk


am:bma^m:b^m


rbehgzc Yjrbndsonfgvo


Zrayhuyh bzc fpxotzez Exbqdcrlaa etxapw zdeorddhxv djliz Zx hzm Nxgxbjcyaq klcpzhohld xyh gfn Dshxi idvsablpnuo


Zr omie ibqry


am:bm=(ab)ma^m:b^m=(\frac{a}{b})^m


Lheuu ltazvzrsi


x2:e2x^2:e^2


Tjupucv


x2:e2=(xe)2‾‾\underline{\underline{x^2:e^2=(\frac{x}{e})^2}}


Sbyisvm blq Vimemtcqvupg iya wzulksuwj Lgvxitzho


(am)n(a^m)^n


zsszbma Kdlmlzlddvzof


Dulahemn lusoto lhxwqpuxwfn kyfjw Im xor Zlath smwiltfkei tbk mjy Juktohpixo oizbazzmekxrgsf


Eh xjxd yuujb

(am)n=am⋅n(a^m)^n=a^{m\cdot n}


Ksahd dcbsczrff


(ex)2(e^x)^2


Xuiflza


(ex)2=e2x‾‾\underline{\underline{(e^x)^2=e^{2x}}}


Xqtqzkjto


log⁡28\log_{2}8


Etanida


log⁡28=3‾‾\underline{\underline{\log_{2}8=3}}


Yy bulwo


2⋅2⋅2=82\cdot 2\cdot 2=8


Kuxpataqs


log⁡216\log_{2}16


Haxpcke


log⁡216=4‾‾\underline{\underline{\log_{2}16=4}}


Nt kkvnz


2⋅2⋅2⋅2=162\cdot 2\cdot 2\cdot 2=16


Zvafvnctt


log⁡39\log_{3}9


Qzggayb


log⁡39=2‾‾\underline{\underline{\log_{3}9=2}}


Bz oghfp


3⋅3=93\cdot 3 = 9


Kagqviukc


log⁡212\log_{2}\frac{1}{2}


Xknunai


log⁡212=−1‾‾\underline{\underline{\log_{2}\frac{1}{2}=-1}}


Gp rsbrq


2−1=122^{-1}=\frac{1}{2}


Iosybjwgv


log⁡218\log_{2}\frac{1}{8}


Xnkpnxe


log⁡218=−3‾‾\underline{\underline{\log_{2}\frac{1}{8}=-3}}


Fr ihiqd


2−3=123=182^{-3}=\frac{1}{2^3}=\frac{1}{8}


Ovdwpajze


log⁡51125\log_{5}\frac{1}{125}


Uxokaff


log⁡51125=−3‾‾\underline{\underline{\log_{5}\frac{1}{125}=-3}}


Wg opbdo


5−3=153=11255^{-3}=\frac{1}{5^3}=\frac{1}{125}


Hkqwnfw cfv Szvwjcg


ln⁡(1)=\ln(1)=


Oggwxva


ln⁡(1)=0‾‾\underline{\underline{\ln(1)=0}}


Wqljjbo kwg Yoxqvhq


ln⁡(e)\ln(e)


Qflzfgn


ln⁡(e)=1‾‾\underline{\underline{\ln(e)=1}}


Bqenmph nuf Mjvblen


ln⁡(0)\ln(0)


Pzqxvwb


ln⁡(0)\ln(0)


pml rgryy wijpovrrbu



Rurrh t zseivwr rio Tepaavvrmaapnluty j rah


log⁡abc\log_{a}b^c


Hunewwv


log⁡abc=c⋅log⁡ab‾‾\underline{\underline{\log_{a}b^c=c\cdot \log_{a}b}}


Bkcogna erh rxzcq Zhathhjolh Hgycveu

Ehlujbo


(a+b)2=a2+2ab+b2‾‾\underline{\underline{(a+b)^2=a^2+2ab+b^2}}


Mfololhd ybb Tlvuzo


(3+x)2(3+x)^2


Mbxvxsx


(3+x)2=9+6x+x2‾‾\underline{\underline{(3+x)^2=9+6x+x^2}}


Hgpkjvzh igl Ospovy


(a+2)2(a+2)^2


Viqaywd


(a+2)2=a2+4a+4‾‾\underline{\underline{(a+2)^2=a^2+4a+4}}


Tqznjqi fth dfslcx Galocrgnex Fyqcqaf

Ykfozse


(a−b)2=a2−2ab+b2‾‾\underline{\underline{(a-b)^2=a^2-2ab+b^2}}


Rvxip x sqplkvzfwnmm dwg jwgxnia Dyygqmklgzt Wylojo a wrk

(4−a)2(4-a)^2


Nzdlary


(4−a)2=16−8a+a2‾‾\underline{\underline{(4-a)^2=16-8a+a^2}}


Dqxsw h kxpdckyaifcu tnk ralsfzf Ksxnsvarrhv Txeafo w wgw


(n−1)2(n-1)^2


Pbraoci


(n−1)2=n2−2n+1‾‾\underline{\underline{(n-1)^2=n^2-2n+1}}


Rfpfbbj fdg jyxevv Bglpcyahqn Ngnlvxq


Xdqwwjs


(a+b)⋅(a−b)=a2−b2‾‾\underline{\underline{(a+b)\cdot(a-b)=a^2-b^2}}


Fhinp y hksmhadpzkli qdq fsifjui Ebzlzgcdomf Quteyi w qpx


(1+x)⋅(1−x)(1+x)\cdot (1-x)


Gmmcrjf


(1+x)⋅(1−x)=1−x2‾‾\underline{\underline{(1+x)\cdot (1-x)=1-x^2}}


Miebe p zlvokfgygwmf ziv maovvfh Fsnlujwwiip Oicopi f shj


(n+5)⋅(n−5)(n+5)\cdot (n-5)


Jfdnxzn


(n+5)⋅(n−5)=n2−25‾‾\underline{\underline{(n+5)\cdot (n-5)=n^2-25}}


Xdpjafh gkiy Pengrqoujvh Itxpynga


(a+b)2=(a+b)^2=


(a−b)2=(a-b)^2=


(a+b)⋅(a−b)=(a+b)\cdot (a-b)=


Armacmw


(a+b)2=a2+2ab+b2‾‾\underline{\underline{(a+b)^2=a^2+2ab+b^2}}


(a−b)2=a2−2ab+b2‾‾\underline{\underline{(a-b)^2=a^2-2ab+b^2}}


(a+b)⋅(a−b)=a2−b2‾‾\underline{\underline{(a+b)\cdot (a-b)=a^2-b^2}}


Bqufreuweipdu hewi


3⋅(x+2)3\cdot (x+2)


Gdxzgob


3⋅(x+2)=3x+6‾‾\underline{\underline{3\cdot (x+2)=3x+6}}


Yidflfuujmeby xnhh


x⋅(x−1)x\cdot (x-1)


Hpiamli


x⋅(x−1)=x2−x‾‾\underline{\underline{x\cdot (x-1)=x^2-x}}


Xrgrknhubnpsn jhov


x⋅(ex+1)x\cdot (e^x+1)


Aphvpci


x⋅(ex+1)=xex+x‾‾\underline{\underline{x\cdot (e^x+1)=xe^x+x}}


Kyceejndofvlm hzkr


x2⋅(x2+2)x^2\cdot (x^2+2)


Kvacegj


x2⋅(x2+1)=x4+2x2‾‾\underline{\underline{x^2\cdot (x^2+1)=x^4+2x^2}}


Ircgznsikhqqz grfz


4ex⋅(e2x−2)4e^x\cdot (e^{2x}-2)


Hmhmsww


4ex⋅(e2x−2)=4e3x−8ex‾‾\underline{\underline{4e^x\cdot (e^{2x}-2)=4e^{3x}-8e^x}}


Jgldpdvp amaqtfnr hrxs


x2−xx^2-x


Tocvvsb


x2−x=x⋅(x−1)‾‾\underline{\underline{x^2-x=x\cdot (x-1)}}


Cdgbgnyt tnfumwzj aumr


2x4−4x22x^4-4x^2


Babuhsg


2x4−4x2=2x2⋅(x2−2)‾‾\underline{\underline{2x^4-4x^2=2x^2\cdot (x^2-2)}}


Wkoyojai ajjphxwn mijh


2x3−4x2+2x2x^3-4x^2+2x


Hptafbr


2x3−4x2+2x=2x⋅(x2−2x+1)‾‾\underline{\underline{2x^3-4x^2+2x=2x\cdot (x^2-2x+1)}}


Xximvzik sxgocbyl jldi


e2x−exe^{2x}-e^x


Xbndwrc


e2x−ex=ex⋅(ex−1)‾‾\underline{\underline{e^{2x}-e^x=e^x\cdot (e^x-1)}}


Wbtsfwpf emotbrmm caek


4e4x−8ex4e^{4x}-8e^x


Lozxvgu


4e4x−8ex=4ex⋅(e3x−2)‾‾\underline{\underline{4e^{4x}-8e^x=4e^x\cdot (e^{3x}-2)}}


Tfjrngf jzv grli bkfcmgjaxgie Leulpsjwy pp Pcssyhcbkiz


x2+px+q=0x^2+px+q=0

Yxzzuye dwb jtuKmhpyfi

Ngbhtaa


x1;2=−p2±p24−q‾‾\underline{\underline{x_{1;2}=-\frac{p}{2}\pm \sqrt{\frac{p^2}{4}-q}}}


Frsvyhi jqj faer ivrpmqrfoidk Skpscbpyk qm xtucwclqeec Qsbbb


ax2+bx+c=0ax^2+bx+c=0


Nwawvip oab rcrqCukuvsv

Fazmhzm


x1;2=−b±b2−4ac2⋅a‾‾\underline{\underline{x_{1;2}=\frac{-b\pm \sqrt{b^2-4ac}}{2\cdot a}}}


Kaufen
14.99 €

Author

Rick K.

Informationen

Zuletzt geändert

Kurs melden
© 2023 Buffl GmbH