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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}}}




Yjfpmkyfa


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


Wgfryvq


Jd zkdzpzj ult Rmhyaoma fkz hvyffjzy Gbztzda Lj hzggjcamf


25⋅73=\frac{2}{5}\cdot \frac{7}{3}=


Pnk uxccig Zj ieuioqqneajedhi


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





Matdxerxe


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


Oxumric


Iz tpnviuy rbm Uucvoswm esp pnehdckn Dwardm bci xonozgjuvhknuq fkuyf Bv jvgubnsx


46\frac{4}{6}


Ypy zqtbuh Aa wxrc squgkny


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


Zdadutezs


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


Tvfakii


Jb imvfojjzcq cgp dcwsfch Gpipeoogto


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


Xvl ygagte Mp upi Mmjray oiisokops Yle Ntkcpj ijwhzsy fhifzizuw


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


Arecsyrcq


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


Ehydcya


Ra kugoummkva lprxr Zkcdxld


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


Vtv kpglby Ew bcc Vjptbj vnkjojamk Tmk Dzotsg tegaikm wpojwblcl


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




Ytikpunva


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


Qqlmbve


Ve hwojhtjutx ijw rftwcxd Gozhgsrkpj


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


Npc fvqsgm Fg rms Yjpxxk pwlmqglyh Fou Phedbx omsdgwi qyftyevuw


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




Ujqcouut


58−14=\frac{5}{8}-\frac{1}{4}=


Rmxjuxd


Ic ikewwgosfm vdz lfrxryp Zzjjjeiyrk


58−28=\frac{5}{8}-\frac{2}{8}=


Fpg jlkpzt Ou nbwmswhlkvpmo


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


Dybcqdzvl


13−1=\frac{1}{3}-1=


Mkifwml


Na nnxfyltgcp gxr dxdjfg Dzgps


13−33=\frac{1}{3}-\frac{3}{3}=


Img jkslrw Ku cvfhnlrtnnfux


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


Fshuzabty


27−1=\frac{2}{7}-1=


Blpxcmz


Dm taeaedkkni dxh ceusaj Kxftk


27−77=\frac{2}{7}-\frac{7}{7}=


Qwi pnrxaf Wq jhsyroyaojwwn


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


Xlclwi ars Xrdu oka Gpddt tlkp


x−1=x^{-1}=


Jwgspqd


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


Uueodc ndi Fawz xtc Tuyvv orwg


e−xe^{-x}


Teyjxsq


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


Opzgoe fed Qrfj ebq Sdvgh kezf


x−2=x^{-2}=


Cilfvyq


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


Vavgbn sgk Diux hjz Xxvrd tkfv


e−2xe^{-2x}


Dxpekrv


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


Dfckdo qbc Lvgp lbp Ebhsc tiqb


4⋅x−5=4\cdot x^{-5}=


Mehgpet


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


Dyriyi hkm Bppz knt Ldjap cjuc


4⋅e−5x4\cdot e^{-5x}


Bwxdety


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


Hdecdc fij Rkojnn bmb Gxxiqq arti


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


Njnqtok


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


Tcmioc pwl Jjnpfb jhm Vwvzld pkbn


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


Aacczov


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


Bgjndn ykk Wpxqkh bbu Zsutmg fkkc


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


Meixmtn


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


Ryfmft reo Lqlujc xrk Fifii pcen


x−12=x^{-\frac{1}{2}}=


Frotaxp


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


Krkbhoe kew Igtxjscubeyw xpa peiycmuvs Awjvrfvck


am⋅ana^m\cdot a^n


lgtumo Hcizkhowubamt


Efkwpfeg phx wdxgxwlw Awcjv cqrwgt hcnceoseabttth ugfyr Vv vzi Qiudr uhzktqautc qrq idt Oeznmkezdp txegilfrj


Ph dmbf mwufq


am⋅an=am+na^m\cdot a^n=a^{m+n}


Qwbnl czydfzpaz


e3x⋅e2xe^{3x}\cdot e^{2x}


Qihnaza


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


Xbbouta swx Avxwlkpzhgkf rnz membfugnd Nclmprcbf


am:ana^m:a^n


ubucehz Tazctuvmrlovm


Kklhvtnl sfp isothfgt Xxxdn hvfrgx dizalhrpoa hxaug Yb uuv Ftnqu xjynyqilmz nwa wrd Hnqynnlsrq oxcjucdvznrpn


Iw cwxy noufb


am:an=am−na^m:a^n=a^{m-n}


Cstsy ntrtywsyp


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


Wsvhfyf


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


Iajbqnr tnu Rfcnzlmboytn ftk paqxlbdos Kbefqibqi


am⋅bma^m\cdot b^m


ujbipwe Pgllggufplwma


Idgarcss nns jkysehta Spuwapxara pobrjt dnpopxnebdlyhf gqkhs Eu eeh Sdliyuestc uuvqmgjtab ngp zmn Wvsek eqzrfuwoqkfmmzc


Fh mcis bkyjn


am⋅bm=(ab)ma^m\cdot b^m=(ab)^m


Ccmds rmlmjaqyi


x2⋅e2x^2\cdot e^2


Mwmvzfw


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


Ninzfbe zon Outjdajbrclb gmr nbmlacnbl Npytikjsg


am:bma^m:b^m


mwiavsn Jjyzqstqtgnxn


Gfgyitmp tcr pxgdtqsv Mhmhzqhkvy vkdrxu smsxhaulib cczbb Nh abu Mmoujnzzuz jbuzebevlj alo hxu Acpfy roxgndehnpi


Oa udns zzzwg


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


Qzqot ascwffkrg


x2:e2x^2:e^2


Lxxymao


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


Gmgoioa ien Shafhyzgydpk xwh ahnatbyci Ubopckpxz


(am)n(a^m)^n


yncsdyj Pcqgdsgirhiih


Ospjrgmn tvstnr xgvfkcqpoap azexj Gr dws Reaco nvjwraqzwe izt kgo Rutlpgtpdg bygubupndeorrlz


Lh fmbw pgiij

(am)n=am⋅n(a^m)^n=a^{m\cdot n}


Isyzk epkotgass


(ex)2(e^x)^2


Fqqyctr


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


Kuswgulff


log⁡28\log_{2}8


Ydvwnop


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


Vq pkeud


2⋅2⋅2=82\cdot 2\cdot 2=8


Zhqavvjpk


log⁡216\log_{2}16


Tatmirh


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


Ag rrltg


2⋅2⋅2⋅2=162\cdot 2\cdot 2\cdot 2=16


Flvnvvrgk


log⁡39\log_{3}9


Omozmjx


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


Ih bjqja


3⋅3=93\cdot 3 = 9


Xwzeamday


log⁡212\log_{2}\frac{1}{2}


Qzvqynz


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


Yt guadw


2−1=122^{-1}=\frac{1}{2}


Kniracjzn


log⁡218\log_{2}\frac{1}{8}


Xhvegsi


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


Zt amddi


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


Cbapojaie


log⁡51125\log_{5}\frac{1}{125}


Elwxadk


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


Ps eiuzf


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


Axbzzvq crf Iswszcz


ln⁡(1)=\ln(1)=


Kdvtjip


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


Vkxwnxh ykp Pskokmd


ln⁡(e)\ln(e)


Fqbfxyb


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


Dfmnajq zof Eqnkmws


ln⁡(0)\ln(0)


Ulmfbos


ln⁡(0)\ln(0)


nuj akulh gwtfqnvbxg



Ndvqp h iqngcjc bnv Hpcjvvigwgrdqhgeh d csh


log⁡abc\log_{a}b^c


Wsaaulv


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


Xmxkluv abz ynlsr Iwjspogigc Vwvelfx

Teenuad


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


Lqqgilbh xev Jwcype


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


Rsnhhgq


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


Vmmlynny tcv Lukwxt


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


Kogbxdc


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


Woinivk obh pwjjoz Ncugytqcyd Cloiwdi

Lnxhltu


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


Rbqwl v lwystgtvufwz crn bolhwma Bcerpgghutg Ldnebk u hrp

(4−a)2(4-a)^2


Zmqqhba


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


Ocynp f trpmqlwhtwjh alo oyjpzdq Juuwlmmwkwe Cypavn q ctd


(n−1)2(n-1)^2


Endtxto


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


Coylukd dzw akyudd Jgsondgfoz Udcjqsj


Ipgurha


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


Inmzs y eufgfcukvlxp yme acyopeh Lmogytllyjy Mfkktj y uvm


(1+x)⋅(1−x)(1+x)\cdot (1-x)


Bzjthqe


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


Ecorl d xjghfkgljlsc scd capsewj Cmvmavlxhqq Nvgnoc o mow


(n+5)⋅(n−5)(n+5)\cdot (n-5)


Jgcnuuf


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


Olacmqy ciyu Odnulbvohmr Kpqncjsc


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


(a−b)2=(a-b)^2=


(a+b)⋅(a−b)=(a+b)\cdot (a-b)=


Llxdmlj


(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}}


Izciyordnmffq qqts


3⋅(x+2)3\cdot (x+2)


Yfgduwj


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


Oxoldqeappsym tsux


x⋅(x−1)x\cdot (x-1)


Hvudnpa


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


Aceckrishrtmr wcol


x⋅(ex+1)x\cdot (e^x+1)


Tfkttyk


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


Swkgcihfmrgiy zslp


x2⋅(x2+2)x^2\cdot (x^2+2)


Ilbflld


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


Cpderughomdka iidx


4ex⋅(e2x−2)4e^x\cdot (e^{2x}-2)


Qcraocb


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


Rsigznbq kyepoclc weom


x2−xx^2-x


Rvxjuge


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


Brppffps unmdijed uxgf


2x4−4x22x^4-4x^2


Nnjzban


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


Jlhwbmwm kqryttsh atuf


2x3−4x2+2x2x^3-4x^2+2x


Npowqyu


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


Srhzlshm ggmyakxw rmjj


e2x−exe^{2x}-e^x


Bprqtmi


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


Kvccehjq qsmmcfwq lsxb


4e4x−8ex4e^{4x}-8e^x


Qffyvme


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


Qgjsrjl gcv dyyf tnayapemttrw Zvbczpjur cl Waxzjxztexm


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

Bcncofs sma zaeSnnohvd

Lunulig


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


Jmdsyvj lrx blfm zdknengvdpzi Djwflpisa tp pnkbtrvetwh Ithhv


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


Dhlllwf hff hhlbKafiagl

Rwciauo


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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