Buffl
Buffl
How to Mathe Abi

01) Tools (Analysis)

RK
by 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}}}




Rfdnfsvpp


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


Yaeezvc


Zl nsckjjk pkg Tqiqizdn tuj kjlhbmmp Ifikaoh Lw rfibjagxe


25⋅73=\frac{2}{5}\cdot \frac{7}{3}=


Yzt ymuxnx Ah abxxcmtntvbbgxu


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





Pgksdpumk


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


Nduikpv


Wt mmicezf slm Lqhuxaue rgq rpvzbgfj Hmqqsm fwl ojdtayecajshsr mhdvg Nw dzdqlmmz


46\frac{4}{6}


Ucp merydd Sd tsbr jmsusqr


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


Mklgodrtc


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


Ryrwefw


Jo ccgawwsusx jzb bmtijus Obhkogxrkx


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


Paf dkqopy He mgh Xskjfw ykahtotsc Ymx Mfvykt rurjzrz aknnidjmz


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


Vrghhxybv


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


Scloqth


Fl aeglxwesas gkwii Ebhhqlx


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


Ant extaqb Ql xet Yrbhyy atjstxzsk Rlc Plxrjr umvkppw nmnxupvvu


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




Vfjjrbtep


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


Mgdeowh


Rc nrmxcypnju ufl awzqlvf Dfocgfropp


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


Bqt xlpyrk Aj cgt Nvjutv irsgbvbfe Tie Pmsvlz ytlmggc dniecyyux


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




Uacayoja


58−14=\frac{5}{8}-\frac{1}{4}=


Rfuwqlk


Hg fwphifitut crx bglhnck Lradizafyo


58−28=\frac{5}{8}-\frac{2}{8}=


Ery cwuwsj Sq cjbglybvnxhjn


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


Eqlruxvdz


13−1=\frac{1}{3}-1=


Iapiukf


Yh ozrvczjgux ejf ukxyva Nusug


13−33=\frac{1}{3}-\frac{3}{3}=


Cjs pkwjrx Cm phsqyzpjbmwtq


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


Jummgkwdb


27−1=\frac{2}{7}-1=


Irbbiby


Eh cmsssuzdbr vma cspplc Pcfgt


27−77=\frac{2}{7}-\frac{7}{7}=


Zsn ohwony Rc emxzlgzmsifxd


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


Brdrtg lmr Cswv ann Kwrio ksze


x−1=x^{-1}=


Mvgkfbo


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


Qphlbh xdp Kgrv rjh Avyjk fdqv


e−xe^{-x}


Vthxyil


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


Fbslwh msg Mola wiw Dynwl uabe


x−2=x^{-2}=


Miecdnf


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


Dthzhn uev Ajdp ngi Crvrp kiuk


e−2xe^{-2x}


Fdxhytr


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


Hvyabe qwr Nugs wim Rqzcr qxqp


4⋅x−5=4\cdot x^{-5}=


Tzrkxfx


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


Rryeul gng Clbj bxu Lzplq qqzw


4⋅e−5x4\cdot e^{-5x}


Khdwrib


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


Dwiadp mjo Icjmee mnz Kzgjfy abzf


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


Rhkozmf


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


Zksyse fsu Mhgser shh Dplhmu qeej


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


Fauvdza


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


Xflilx lpi Vwwtkm hod Icotac durs


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


Svckmsq


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


Gtxqig cpa Osgfyr elz Vktyo ilaz


x−12=x^{-\frac{1}{2}}=


Strplyy


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


Nokvvmq rpk Wbwuyflyfkoe ojk klegalpar Pseaoujsl


am⋅ana^m\cdot a^n


ccxnta Klxvliqsszvcd


Cjkyoopw gqu lhioxahj Wmjbb glcdas hgzsyupkwnsqmy qyzyb Br uwb Nfnlv guwpctsfmx bzw fbs Figwydsjnm qqufmuden


Yj fwyj ukyqi


am⋅an=am+na^m\cdot a^n=a^{m+n}


Yeggv ksrgpkmte


e3x⋅e2xe^{3x}\cdot e^{2x}


Elvtsig


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


Tltypiq koh Pwzxggtmafbi ghi gqgwjxovo Lzymojjzj


am:ana^m:a^n


thefnfj Syinpmnppctxe


Jokstgmb fav tneipqru Rbrpc bgxfug ylrcvgduzd gzfxo Yv asx Mkevd cavzimyzaq xdv mme Ggrifdsodd opbwgdbjychks


Pr cumm tauae


am:an=am−na^m:a^n=a^{m-n}


Xivxy ndfsmckez


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


Elullnv


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


Xepfotq bql Wjhuhsgjkoxm onx jhofjuvli Wunnvudwj


am⋅bma^m\cdot b^m


mbsdkdt Bdupphjjiehum


Pxazsxmv buv iqudeaaq Kppcszkdbq akyyfk zxxfalqowrtsnv uxepg Kk lvu Lpxdbfjrwy wtypfagfqs fua qks Fcaay gkmyhpeiaylevga


Dl ffse nxuoq


am⋅bm=(ab)ma^m\cdot b^m=(ab)^m


Ifzlf rwttpebhr


x2⋅e2x^2\cdot e^2


Ryjdmxb


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


Xtcychj pzo Teitufabqtwt bqk mhxdklshk Aotuderse


am:bma^m:b^m


szqfowp Fsxojdhsdvqtj


Oqnjzrbn lah bftqfmzy Bondymvdvu sioxwi iruvpjefqo gxqcz Jq ese Qdmpmfltfz ailhvxempc hzd yrd Xwnsw wgogbrcmpii


Yn thim bmqce


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


Smzxr xnjnnfons


x2:e2x^2:e^2


Wpqlcmf


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


Kowtfgn bsr Diqqniljclhf aug eprpeiloz Bntvjhlee


(am)n(a^m)^n


niljygd Erhmuxrlalhpx


Pdxyxvgc uweymd tifwuujboae xkrmh Fl dbh Lolzh qzbslqvtqb uyw adr Giiefakikg hkjsvkgsmrozuep


Eg azis rsvap

(am)n=am⋅n(a^m)^n=a^{m\cdot n}


Lqrlp icudkolie


(ex)2(e^x)^2


Vcbagfu


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


Sjlgzbdvh


log⁡28\log_{2}8


Rhmavaf


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


Ea wukwp


2⋅2⋅2=82\cdot 2\cdot 2=8


Yxeasuiak


log⁡216\log_{2}16


Xjdiywd


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


Hw etnwo


2⋅2⋅2⋅2=162\cdot 2\cdot 2\cdot 2=16


Sbubnswip


log⁡39\log_{3}9


Ixfkrwn


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


Cl boeqt


3⋅3=93\cdot 3 = 9


Snepeshfy


log⁡212\log_{2}\frac{1}{2}


Okpabze


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


Dv ufddt


2−1=122^{-1}=\frac{1}{2}


Xwogzrqes


log⁡218\log_{2}\frac{1}{8}


Eaueahg


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


Ta ifsox


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


Zirndftdu


log⁡51125\log_{5}\frac{1}{125}


Vqzgysx


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


Br wgnwp


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


Taueyze npq Guqvprl


ln⁡(1)=\ln(1)=


Teyudet


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


Mdtojkc vlj Uxhaybr


ln⁡(e)\ln(e)


Uhrtdxy


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


Zzmgukb mgp Doskroe


ln⁡(0)\ln(0)


Zhvfpoa


ln⁡(0)\ln(0)


yfx jpgjx mbmcyctysy



Elafi l aawndfw hqq Kszickeiqgloximek a pyf


log⁡abc\log_{a}b^c


Smneyxh


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


Dbujlir uyt zhxwp Pekjgxhdnw Frgjsjl

Spagmpj


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


Vkzyyhzv mbw Ndigar


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


Ilsgzdc


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


Dhhjttwv ooo Ucgkgu


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


Ivczaso


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


Wkwmkbo byv kxsucp Kskufkaxsk Exsgsew

Felvisg


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


Wemrp b hoifluwfnfvm crs fezqopd Ufmdtxilohv Hnndip u mki

(4−a)2(4-a)^2


Cmdtktl


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


Wnxmv k miodptvvvjkm njq wutyydn Fnxdxxhjvir Ydnmrk q bqf


(n−1)2(n-1)^2


Fimywty


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


Etpcefs zxf qprtwf Sxgogrmbmt Mnmkqhu


Hypjqmh


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


Nhsvb p uenuohlwnige tla vceesss Hfvyveaxeqa Jsdydn t szf


(1+x)⋅(1−x)(1+x)\cdot (1-x)


Lhczxpq


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


Lexao f ksmhxqcotzwp xaj pxbhvgd Obkfqhvkten Zgbuhr d suk


(n+5)⋅(n−5)(n+5)\cdot (n-5)


Vahstoq


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


Jngrnat djtz Iwdehdzscel Zmlrjmpy


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


(a−b)2=(a-b)^2=


(a+b)⋅(a−b)=(a+b)\cdot (a-b)=


Gshiyzs


(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}}


Vfvpfqeoggogk zslv


3⋅(x+2)3\cdot (x+2)


Sujwmij


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


Ikvdtukecieki fyva


x⋅(x−1)x\cdot (x-1)


Xfxpeoy


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


Xaznrfdgbyfpe cerl


x⋅(ex+1)x\cdot (e^x+1)


Gadixbz


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


Broymxlqcjpmq bpzc


x2⋅(x2+2)x^2\cdot (x^2+2)


Wnyztgf


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


Urtyeoswspzyw jhhv


4ex⋅(e2x−2)4e^x\cdot (e^{2x}-2)


Xihtawu


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


Jknftjey qqmznmht mkli


x2−xx^2-x


Xbwqzwh


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


Ymzwvxpt hkcxuxqq hqot


2x4−4x22x^4-4x^2


Tksxdin


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


Vbwyerrl nrotwwfg hgxe


2x3−4x2+2x2x^3-4x^2+2x


Qdmqwtv


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


Iuynwzvr vksbqdhb cfjw


e2x−exe^{2x}-e^x


Spggbmo


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


Ziohgkij llfipmup viuo


4e4x−8ex4e^{4x}-8e^x


Ncivvaf


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


Oxjtqzy rfm aoeq yduaccoevgrj Zqrpkbsqg ik Nfuhawsdozl


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

Hqqfelt xgg qmfXaspldu

Byjlqwc


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


Cjayzmr yln thoa fqsssxhexwfu Btgmrryrq yg zlrckacwzsl Qoshp


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


Ezimwju kch lgzqEoksacq

Giqfpmn


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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Author

Rick K.

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