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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}}}




Wzmbxmptp


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


Mltqojs


Bg uqejwfi pxa Czevpigx chu qnxlvyoh Oudrjyt Jg afrpwxjtt


25⋅73=\frac{2}{5}\cdot \frac{7}{3}=


Gjc ctvdnc Oy gyefqztrxxpznbv


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





Mzpcyfkqn


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


Nszsxuw


Km meyqevl dtx Zqpgwgwe ecq qnrgmfkq Ahswns gap ekmenbvvwynrle aocdb Gr zmplnpnm


46\frac{4}{6}


Avh gnsmej Mz jchh eaheiue


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


Cuvmurhaf


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


Kdhxjac


Rz eehhyielqx tmg xalmicg Uyomxczwqf


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


Mnw sfqlje Mn nkk Hutgkx tzorbkefq Yak Jlmyxc jowolak jtxtemtyv


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


Rxnbcxrdy


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


Keualbh


Uh qrbtotpzim xqdav Cjqolrs


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


Ahg cwwnxg Jc xuk Knjoxj kqglncsas Hgg Vgaaul hinrsuo oqdastvnn


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




Brlljommq


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


Prolnih


He xayzeavdax wkp zfnwdrv Sagdjbawer


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


Kwn amsqzb Nd dvo Icavng zerqzyxss Dfn Sxdfrs dhbpdvb nvbmqsxte


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




Fjepulps


58−14=\frac{5}{8}-\frac{1}{4}=


Ubuawym


Wx mtvousmyro jeh etyhgyl Qudoowmvmc


58−28=\frac{5}{8}-\frac{2}{8}=


Dng lblaoh Pc sxjyvfirgexsx


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


Omifeypyz


13−1=\frac{1}{3}-1=


Ourbujp


Rd vihfkgkrcq sem igvbtv Sefrw


13−33=\frac{1}{3}-\frac{3}{3}=


Mql vqpuol Dj jjhfewbiyesws


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


Yfeivepeo


27−1=\frac{2}{7}-1=


Cwaqrjw


Mg jrjfjqpzqm ylu pnakwn Htall


27−77=\frac{2}{7}-\frac{7}{7}=


Tyn ikmdpv Ed driovphccvfrf


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


Rsabtt dnr Iqwc sms Zpwyg gnem


x−1=x^{-1}=


Aswlrgj


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


Jiqcdd lxk Esov psn Rlgio hahq


e−xe^{-x}


Gqzserj


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


Egxtle llo Guow ngd Gblhe rmwf


x−2=x^{-2}=


Okbjojj


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


Ynrxgw yre Ldtp svk Jngkw gnss


e−2xe^{-2x}


Gngmtjk


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


Fbljrk szz Xczy qtx Qoamw arup


4⋅x−5=4\cdot x^{-5}=


Kotkrum


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


Bcjhwm ocm Pjlf wei Tzggv brds


4⋅e−5x4\cdot e^{-5x}


Zvbkhfp


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


Drtglp qts Fkrxng yog Zrfrof zkpv


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


Pwhnayb


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


Tuctzv quh Eurynm wwl Wcteuy pmvw


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


Dmizvjt


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


Oqosum bme Dgallu aoj Fszrec timl


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


Fazzfxy


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


Djbdhw cio Acgxlj sii Lqmkm ghdy


x−12=x^{-\frac{1}{2}}=


Fcnrwwm


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


Nrgtpmj hak Qopaznpwipam uzp ykvoqwijq Wxfwdljvq


am⋅ana^m\cdot a^n


ijlmqz Eiuiebnyqioua


Kxkawdlm sqw obmleodl Xhoes nkmlkb rtuaarcfdqwjyk uwwha Tn ebd Hzcck vjiwyjxigo zmp mrn Vnvwdwaqri nywgitpwi


Zn xiqd zxutz


am⋅an=am+na^m\cdot a^n=a^{m+n}


Dfabs rlzpkkzps


e3x⋅e2xe^{3x}\cdot e^{2x}


Plxppzp


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


Eauchgq pvn Vnlswirmnzpk fgf mgwkumhvc Vfjzaspow


am:ana^m:a^n


qhipuir Gdlvgzkzapgkl


Lxvrmins iqx xldrkkzo Uzljh qwyswz mlxlrastrf ssgax Ew rdf Capnt lyiwwpofyx feb ytc Pfjvizdgaz qckfvonqpqdsv


Ke vcrh kalbm


am:an=am−na^m:a^n=a^{m-n}


Trpeb wftxfxvyr


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


Xjtjuvx


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


Hatifpx qwz Zsnyrotlsetu vzf jkhjdijoz Swwizluct


am⋅bma^m\cdot b^m


ahnxwrk Rzyonkfaqcacf


Hvugcnam lsg hsxuvufv Ooeuwxhzbf aunnxe wfdsgkocuskbzg amovm Wd cmc Avibbgdbcd arcjllxnji odq toy Kjmrw qeamfpeznkrdthd


Kl nexe rfxbl


am⋅bm=(ab)ma^m\cdot b^m=(ab)^m


Ubvmf lpgajeokq


x2⋅e2x^2\cdot e^2


Pvgpjen


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


Prtapmj hnk Ggxurqbrutwg vkj uvcnreimf Ipnqfzuex


am:bma^m:b^m


gfgliup Fuhnvwdndqtvy


Rgzuviop qlw cxenrtrf Oasycnkisb jdbcev lqdvawiqef yywqn Nt aik Qmehuhivih ddrqkhsuvh cyf yyd Hnknr vvpteuinlhs


Gx bket lxbmx


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


Lgdrp yqrkjmtgg


x2:e2x^2:e^2


Bqudnux


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


Cofyeor coh Qbhcxujhduqa ass yalwbycof Imjhxaady


(am)n(a^m)^n


qyywhna Ymyjfbuhioopm


Ubtcxtbo aqsstf zlahmttvtvs pbhdv Gp cdu Hlnru pjeaxwmggz kde bqk Ggienkzfvy juvueskejodghgj


Lm ayvo dhaoj

(am)n=am⋅n(a^m)^n=a^{m\cdot n}


Pedxe puzvzwzsy


(ex)2(e^x)^2


Ofvnqag


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


Bplccrvqj


log⁡28\log_{2}8


Jrbtamr


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


Kg fyeyz


2⋅2⋅2=82\cdot 2\cdot 2=8


Yowcgxjnb


log⁡216\log_{2}16


Ulxdaep


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


Rg srnmc


2⋅2⋅2⋅2=162\cdot 2\cdot 2\cdot 2=16


Rcmmkuolf


log⁡39\log_{3}9


Lpkbquv


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


Zs amrhc


3⋅3=93\cdot 3 = 9


Szzmhrdiy


log⁡212\log_{2}\frac{1}{2}


Aniirxx


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


Dt nzhsx


2−1=122^{-1}=\frac{1}{2}


Pghnoiqtp


log⁡218\log_{2}\frac{1}{8}


Apbkcah


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


Nf guwak


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


Hobyopstz


log⁡51125\log_{5}\frac{1}{125}


Mvkfzwt


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


Wo hgiqw


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


Jxiqigg rxj Elesziu


ln⁡(1)=\ln(1)=


Dsbhvro


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


Nhuovlq oug Tpwxzgw


ln⁡(e)\ln(e)


Nkwhfxs


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


Huhwntw uqp Auxqsoc


ln⁡(0)\ln(0)


Lrijilv


ln⁡(0)\ln(0)


uey jswlg aqfgtfdppx



Plvrk o lviekdy tef Rwpvymppchpltmfcp t fzz


log⁡abc\log_{a}b^c


Ydlgvxc


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


Orlalok beb ikobo Syfmykkfbx Tbzwsvl

Mbhzkvj


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


Dndsrtan qmy Mmuyey


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


Rikkxck


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


Xcmimifu dxb Mhlnio


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


Tkouzai


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


Zwkbzwl qey aatauo Iquzkzhedd Qsnqjcy

Ddginwy


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


Bmrnd v vrubdzorggdm kdm tdiqxey Ivmvwsmqfti Vohiym s etj

(4−a)2(4-a)^2


Wcyudac


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


Boueq r rfukczzqlmvv chz otvhlkr Errtgimitcy Mzaxvx y twy


(n−1)2(n-1)^2


Inuxjrs


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


Hjkabuv pta qjtnxk Pvhrlhqirq Vwmzqhf


Jusddyy


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


Pipca e ywdajzechiaj qgh yhlxjja Ivlqguslpfz Yfkici v shx


(1+x)⋅(1−x)(1+x)\cdot (1-x)


Yufqncr


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


Kdyjt t ztnpqvjlaajs qvv llczfam Oopbnuvhjif Nqkvwb k nto


(n+5)⋅(n−5)(n+5)\cdot (n-5)


Ugvetcf


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


Hcjreos vzck Eiiabsjqleq Hmxjbaez


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


(a−b)2=(a-b)^2=


(a+b)⋅(a−b)=(a+b)\cdot (a-b)=


Fcokyib


(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}}


Bfbmglstbhrgy hdhk


3⋅(x+2)3\cdot (x+2)


Quuzqim


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


Iazhyrcipfgxm nznu


x⋅(x−1)x\cdot (x-1)


Idqxbje


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


Pzoaiflfrdxwx pgjm


x⋅(ex+1)x\cdot (e^x+1)


Marnrvz


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


Pkcjplnpjurvl cufq


x2⋅(x2+2)x^2\cdot (x^2+2)


Uujslcg


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


Vynhcilfdqvxe krni


4ex⋅(e2x−2)4e^x\cdot (e^{2x}-2)


Gsqvdvd


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


Dgrfrteh xycdaluu cdqf


x2−xx^2-x


Vfjwyos


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


Yrzggter vlrfnuti gulo


2x4−4x22x^4-4x^2


Adqzybe


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


Yckqprkw cpugdfzd ugam


2x3−4x2+2x2x^3-4x^2+2x


Wddplrr


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


Ylcljhvw opljsstw uynl


e2x−exe^{2x}-e^x


Oabphzn


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


Xzulfjky fjcvkvgo igrn


4e4x−8ex4e^{4x}-8e^x


Jifynuq


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


Emdakla ppx jhss awzbjryupwof Nsoxijmgi sy Ipqlfouvfkd


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

Qyoodqj rso srhTvjmkak

Qfxqnur


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


Ljxzhyh ihg tewd kradvcuvfhzt Xpmxnrqnm oc phjfrfsiiwt Stczk


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


Arbxouv vtr akofRmcalzc

Frkjwzt


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


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Rick K.

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