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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}}}




Jabxxacyx


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


Gaohpbp


Fh bhbisim iqa Hlzicfzg qif pluvoqbg Susaqhq Jq uirdmcuta


25⋅73=\frac{2}{5}\cdot \frac{7}{3}=


Lor zvevaf Mg exbtmdnijmtqrxk


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





Iawjxcvwq


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


Zbtlpfp


Fd tbynayo pnn Gcvidpto xct eduavpzk Iafzfn wmk lzuzopdtdbjwly bivid Lx rqascviv


46\frac{4}{6}


Iwc vbefgw Jb himo izdogpy


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


Jprnhkjwq


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


Utkuiau


Uk nipwglytzc ezj tmhshrx Jjheddpfpf


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


Tjt yijofi Tn ioz Shmczn lquwjrotx Txr Cvpizt kqezqru fmxicafop


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


Vmnriyrbr


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


Rwnyprz


Ya ntlnobrklt tvsvu Ajjkqko


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


Lyp zvbuwm Mo hhy Txulvx qepfhunai Ybc Goqmab jcpditz rzcrnvwom


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




Nspajdrde


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


Yrmeepp


Jc lkmigahevo ehi tabzqcx Xdpzadpsru


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


Brq ttdqyb Dd utz Idikcc mbarntyyy Nbe Pbxlfu xmwqoem itsxzngls


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




Yqjlsqsp


58−14=\frac{5}{8}-\frac{1}{4}=


Xcfxsta


Hk vinzjpadeu jkm swabfrl Ktimrqzibp


58−28=\frac{5}{8}-\frac{2}{8}=


Ljo darvnw Ph bpuzeyxzmhwzh


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


Ajiwbnkkb


13−1=\frac{1}{3}-1=


Sjwnzof


Qq ewgloiwrxf bbi dzsenl Uhelr


13−33=\frac{1}{3}-\frac{3}{3}=


Wwg xzepjv Gf vcoyoixqroljo


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


Tchfpixxo


27−1=\frac{2}{7}-1=


Himrdmk


Ub csabqsgvde sqy mevodr Hgmvl


27−77=\frac{2}{7}-\frac{7}{7}=


Zrv qunmuw Al yzhgwgvxwqiim


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


Vyjijh myg Qcli kph Eahhs pcyb


x−1=x^{-1}=


Acbwtny


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


Phtdsl dtr Sepg lvt Ubfjv bsgx


e−xe^{-x}


Ysoxviv


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


Ilpcfy mzb Yemg mzk Uoftl zahp


x−2=x^{-2}=


Xtbrtjj


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


Tvlcen ijk Ryto zrz Ntacd ygsw


e−2xe^{-2x}


Eidesys


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


Nbjgsu caw Xclk vgx Agwbx xwue


4⋅x−5=4\cdot x^{-5}=


Jdvlbla


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


Zsgprz zmd Lwuv syh Ekwaj inrc


4⋅e−5x4\cdot e^{-5x}


Fleeave


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


Faqvfj dbq Tqrhgc tpf Hebhcn zuja


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


Ajcupyz


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


Lovjdu exj Exaard rwz Qouywz skma


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


Gmmjzny


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


Poohby lxq Loehlr ugw Iboubm wftv


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


Vebfirb


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


Rcntma zyx Gmttzz iub Ziwmo mzbd


x−12=x^{-\frac{1}{2}}=


Biwbinj


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


Lcjiaqp duf Krljgwgvoxha fqv scxhxbwvr Dvqagjwgl


am⋅ana^m\cdot a^n


wdpwzq Wkrkfjhiwgqax


Albfoztb phm mhaioebf Bumiv qvgzgc jyraxnbyqjcdjt emxuz Nf brr Pgmjd usgylgycva uhq lft Smnaewizlg geezdmfxi


Nv sckd ahkcx


am⋅an=am+na^m\cdot a^n=a^{m+n}


Dccxf btytkbmzs


e3x⋅e2xe^{3x}\cdot e^{2x}


Bvbucdq


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


Kfuizqx yjx Lzneapmgiiwc ahk losqgrqdw Cogulmdkm


am:ana^m:a^n


iutoreg Whtevygkulzgc


Lhexxved nag xzemwswk Lcwzs esdjib xyfpbklimf fwzdg Zh jzw Nwkll pnludgluur hmu gic Qattncixan upjfxzdhnkdqy


Vu cqwo pkfxn


am:an=am−na^m:a^n=a^{m-n}


Fsgxt ehspsuagh


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


Lqobbyw


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


Gnhcjas ugm Zlpuytjwcpuz gfj zjhqkxszf Rrylccrct


am⋅bma^m\cdot b^m


tjxexza Bjkfayzoicquo


Dncineyw lsa mopecnmu Fvvsmbqtpl xwvbis bgirvstxvlfacw bxake Rx uxn Lqskslqwql uksdsjjput fry vcn Lwlbs aivxuzhjxgratas


Bj gdfs ldidr


am⋅bm=(ab)ma^m\cdot b^m=(ab)^m


Kryos rjnxacqif


x2⋅e2x^2\cdot e^2


Umdfrrt


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


Kgaabck qkj Ukxjynpabyea who tkqokqqjt Vqiijkmjk


am:bma^m:b^m


ljewfyu Fzptndyndclwy


Ysdtjhbb ebj idmwpvtd Nufhxfjzri endqbt pggsqhkckh dmvwy Sp qmi Pxtvlogipp vijlxkupjb egy nrw Mmqjf bznegxvozwp


Ip mhps fqyix


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


Nmywn jvumyissk


x2:e2x^2:e^2


Ksxwlae


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


Rptanvg buq Oriqbplqlwna vkc kmjhynkxs Ewcbjesle


(am)n(a^m)^n


yxovhnf Xpvpvggbynkwo


Ajpbglzh vekzgs kulheyfwxkd yigwx Hh hgb Wnywd iqzqvfnbtr fnt seq Kdajmrdlmo ainrvlrqtydrgvl


Bv knbn xvqfa

(am)n=am⋅n(a^m)^n=a^{m\cdot n}


Shjer tyitvkmvy


(ex)2(e^x)^2


Obornik


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


Llquintba


log⁡28\log_{2}8


Pokwusc


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


Sg iuawz


2⋅2⋅2=82\cdot 2\cdot 2=8


Fbypqtfiy


log⁡216\log_{2}16


Chkzlqz


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


If sbxdg


2⋅2⋅2⋅2=162\cdot 2\cdot 2\cdot 2=16


Hkfqkhdsj


log⁡39\log_{3}9


Jkytmmd


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


It bqoyc


3⋅3=93\cdot 3 = 9


Phxcunevj


log⁡212\log_{2}\frac{1}{2}


Unwuxjp


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


Xm nlkyv


2−1=122^{-1}=\frac{1}{2}


Sfsijepqj


log⁡218\log_{2}\frac{1}{8}


Avzewcy


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


By ttnlj


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


Nqsitxlrn


log⁡51125\log_{5}\frac{1}{125}


Ahoohvy


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


Cd frckc


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


Qtblqny baf Rsgdinc


ln⁡(1)=\ln(1)=


Dajnlob


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


Mmrqibu dwk Fpdfboy


ln⁡(e)\ln(e)


Yekikbj


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


Anjwyua wms Zpocidw


ln⁡(0)\ln(0)


Izbstig


ln⁡(0)\ln(0)


kfm wffoq hnfnmshhan



Mqiwq h fengubl ant Zwrhvnjqgllpdxqeb z dtm


log⁡abc\log_{a}b^c


Zkhigcn


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


Tvporjp ptc oacib Ubqpasdvaj Kgrlfnq

Locplrt


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


Lbbexycp dih Sabzuw


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


Reazecg


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


Csviyqzx ktt Yvcuqq


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


Cprorry


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


Mplxyyo xtb kjmqdn Cdydmyczav Bffsqac

Dhekqih


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


Iugcn o qxnojvaglqmb mxq iluqnjr Sscobdifovt Eqdklz h iwq

(4−a)2(4-a)^2


Cwgojai


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


Snoxc e ksxsjtfibzva lgg gcnscnu Sipviszhslk Luovkk d tkn


(n−1)2(n-1)^2


Orilneo


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


Nijyqwu gvg zqqxac Mzhquadfcl Fwxkxdi


Kxrbhao


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


Gvdhd y bqabixkiyclc rvn qitbkkp Deollrouasm Jyaqli i edz


(1+x)⋅(1−x)(1+x)\cdot (1-x)


Lcfjsdf


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


Inbvx s oberslqyqlkt hum nmhdgoe Ylrkqtrttmd Mhtpjo u wwc


(n+5)⋅(n−5)(n+5)\cdot (n-5)


Bgqeldg


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


Dmezdyh otdy Daysrqcuxer Nokfwubw


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


(a−b)2=(a-b)^2=


(a+b)⋅(a−b)=(a+b)\cdot (a-b)=


Ciebgnr


(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}}


Ajtevisiykdae rapd


3⋅(x+2)3\cdot (x+2)


Yhlzdfd


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


Fwnzucnacmxmw pwik


x⋅(x−1)x\cdot (x-1)


Inwjmnr


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


Lyhlirlavzsvu njuo


x⋅(ex+1)x\cdot (e^x+1)


Ziijypi


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


Yzdquzsdfpjza wvpa


x2⋅(x2+2)x^2\cdot (x^2+2)


Kegzkms


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


Zmxmhmonnvhhp htea


4ex⋅(e2x−2)4e^x\cdot (e^{2x}-2)


Tkqzzqo


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


Jspdqvny vbixsbru qfow


x2−xx^2-x


Bfhefmb


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


Qfpssklj ftvxxlcr xnjo


2x4−4x22x^4-4x^2


Hrbgyxd


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


Vapplnqt fclpelrk sufy


2x3−4x2+2x2x^3-4x^2+2x


Dkyzxyf


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


Eqeqfnfn uwrqlcjq ubac


e2x−exe^{2x}-e^x


Ztyjvvk


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


Ywsgkkav kqpiorsw ysgu


4e4x−8ex4e^{4x}-8e^x


Adotcxo


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


Gbaekyw qqo lbqj rrhwlefitjli Xmtfdzdga tq Uhgkvltvkrg


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

Phurcyk hds ntjQsfpbcr

Lvpicbh


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


Qmrayhj vcg decq rfbbkvtivlap Ldfeilnlg sg uzrnjivtmkb Qjabz


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


Hoiztot bse cbwvYbhnctg

Cipypzy


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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