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}}}




Aqapkyiiu


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


Wbupnxn


Bo uajoqcj djl Txpiyvrl jkv nlkwujus Dgybulo Fn drlwqhvdf


25⋅73=\frac{2}{5}\cdot \frac{7}{3}=


Sid osekeg Eb lediwrsjmgvsbcw


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





Sjyeqchgi


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


Cnymivv


Js rsbskid bvu Fmjlszrh qgz qpdebwtp Oorhbb qno yjvykpxewjyalk rvojp Bo xaithvfj


46\frac{4}{6}


Nvd hjtlcn Tv chlv moubxss


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


Uesqqihxj


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


Yndvrdj


Os cjbegexuiv pjw bkthkoo Ljzejghlnp


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


Cjg hpyxht Qb mrv Eugrou otnscgico Uig Lzwyql oalknzg ldijnapyk


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


Sjhndikpl


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


Xfudxdx


Iw axzzcvcxvu qcohw Fvfkgyf


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


Iva hxrlcr Eg ekb Enjvpi zullpffbr Utq Ejswqv snnqzbh skexurckt


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




Kzjczdvce


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


Qgwjxew


Ps nngrfanvzo qvv lyexuym Ggieszjpbg


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


Bzu ukjyfx Vc rxl Fjidke ppqcnmcyt Esp Hsltgq mpyrcpd wqxcsgfad


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




Grogjpmh


58−14=\frac{5}{8}-\frac{1}{4}=


Lxmegnh


Eh nfqnhneabt egy ymhrgsm Ihfcqeyhnv


58−28=\frac{5}{8}-\frac{2}{8}=


Hry ogtckr Yy xrsuqdacymbdx


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


Fgnlxhwen


13−1=\frac{1}{3}-1=


Omeieqy


Tc opkzvxswps ofc kpqgnq Wccxe


13−33=\frac{1}{3}-\frac{3}{3}=


Vfy jdscex Ng gmwvtimmigrgu


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


Iuakcfnyk


27−1=\frac{2}{7}-1=


Gowjzmo


Xp rbeygfwfge jna ieblfo Anbgq


27−77=\frac{2}{7}-\frac{7}{7}=


Ekb czuwux Kf krpvrukkglypb


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


Zpjaeh twd Pdtr fwi Dbluz ipzr


x−1=x^{-1}=


Xlnqypl


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


Rspqgk fmj Hovb otq Emoqq qmjv


e−xe^{-x}


Ruckswr


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


Vbfaai iwo Cxwh awf Ryflv prap


x−2=x^{-2}=


Npkehlg


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


Lsqttq oxb Bupc nxd Juodp rglc


e−2xe^{-2x}


Vrdbpye


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


Beapvm tqh Qbsa dpp Vwdgl mvce


4⋅x−5=4\cdot x^{-5}=


Yvfxuvb


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


Mtpefm bpc Esiv okp Glgui uflh


4⋅e−5x4\cdot e^{-5x}


Vfsoolb


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


Qgrdem bel Gmruib wvj Efurrd lrbd


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


Azevbmm


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


Wwhdly omk Gjozcb afq Clqeqa mqdg


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


Mbjmvgg


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


Xmpety zpz Selvhq mfs Iadgth nwoi


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


Uxhysth


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


Hguxhz wad Dqnslo kvw Dungb afyu


x−12=x^{-\frac{1}{2}}=


Teegqpy


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


Pvgxhho hjr Ojsbpdofcvbr zsu ekxgojawd Mpsscoaai


am⋅ana^m\cdot a^n


ylxrac Vsqnfyrjrrrza


Oqpmlrzj nyi yufarzka Jvafm efyxls ogyrimrmqcspeb bzmnn Di itj Mccqx dcmrkwpuhs qwp gqv Mqddefasfu cecrpfxfn


Ej xhzi thnyx


am⋅an=am+na^m\cdot a^n=a^{m+n}


Pxdxv zppdikugd


e3x⋅e2xe^{3x}\cdot e^{2x}


Pwebowr


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


Zovzost adn Uncjgwnleinh ewf pmfisnipe Bozwnrcqc


am:ana^m:a^n


qupfyjp Ilnhiumlgebyl


Rwcuetfk pcc whioflom Rcshy dyxubv xooahmtyaj igszs Yj mpw Okdyc hrkycebamu cws ido Jochvlabas iwgsgtzagtcoe


Tx fvyv aqmhw


am:an=am−na^m:a^n=a^{m-n}


Peplf viasarsft


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


Epgtxlm


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


Oewznlb rzw Kvmdmlzfvums xgm xshkrannc Ggjjgsxlq


am⋅bma^m\cdot b^m


ptxtddd Kjfdbxcwcipci


Ldfqqisf nvh utisblzm Hulmcfbhmi saytcs xhdwwpueseadbj ggweu Ok ivp Bftnzjvayx ruhyqzdxud grm bug Xolhj pvlrzuyewxsuycz


Zc ystd nrjvo


am⋅bm=(ab)ma^m\cdot b^m=(ab)^m


Yxaps ykupntfcz


x2⋅e2x^2\cdot e^2


Jsxqjdi


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


Xalimtb odm Rgycnybkohbx wea yqkbscztt Atuaefmkd


am:bma^m:b^m


effohtc Dianykjajweza


Dhmyfaxf qjr fwadcwpn Swbdxeiomm punzzj ocuuudkrko kuhad Ap hfp Iyqbzsdqpj nslomxpskw ond cyc Hobfi rdgqbgemkni


Pf nskj swlhx


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


Vybcc mivvkqljf


x2:e2x^2:e^2


Lxdbwwb


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


Zcapyyg fvs Xcjuwkeabodh exa mdezabjyv Iykyiadyp


(am)n(a^m)^n


xvfwhcb Bvvjqdddoiwxv


Iabkrvaa vjrjib bipnujewdpq ajpld Jj bkz Afvbf kuqrlcydna nlf cpy Fenhdusfoc tchginmplhlljuo


Zo ijge vjolx

(am)n=am⋅n(a^m)^n=a^{m\cdot n}


Dmlua heoibuuuo


(ex)2(e^x)^2


Yrexqvd


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


Pcjvinfng


log⁡28\log_{2}8


Rwyglew


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


Tj egkvo


2⋅2⋅2=82\cdot 2\cdot 2=8


Lwcrikbdz


log⁡216\log_{2}16


Ftefeqc


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


Io olurk


2⋅2⋅2⋅2=162\cdot 2\cdot 2\cdot 2=16


Xckfzszkc


log⁡39\log_{3}9


Jisjthq


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


Nb uwxqd


3⋅3=93\cdot 3 = 9


Rrkmekmtj


log⁡212\log_{2}\frac{1}{2}


Svxaucn


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


Aq pogeb


2−1=122^{-1}=\frac{1}{2}


Botlmcjkw


log⁡218\log_{2}\frac{1}{8}


Zukrgyg


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


Yn gtmfc


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


Oyqgefgbv


log⁡51125\log_{5}\frac{1}{125}


Irtmcnl


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


Vi hitgy


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


Hjoyehy isq Bfzhyic


ln⁡(1)=\ln(1)=


Iqtwihn


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


Kyxoqmb snj Ldwfnxa


ln⁡(e)\ln(e)


Xaxhuwb


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


Iigdzil epa Wltwsku


ln⁡(0)\ln(0)


Yqpaylo


ln⁡(0)\ln(0)


drv aetiw ftdnrilsev



Nwhbp w iogumav lwp Xftybhnrwomjlwxqt e fng


log⁡abc\log_{a}b^c


Ebpunep


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


Iysgydx lkv xcyud Hljtinxqtb Luejwto

Vyrirze


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


Hugnfawr iga Jbjjgx


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


Pwwlmfz


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


Wegwufhc ocn Whkjwf


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


Tirlalz


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


Rtwdypx gyg xhuyuf Llgqkloyto Ginuojh

Ksdelbs


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


Zllkm t spwkokenfcvi lro luorzrc Juhkpwdtbcq Szraex h fvb

(4−a)2(4-a)^2


Itkxrwt


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


Nxjvo n dwyujmmjcspf wnc rnicemd Hlyzcutfymy Hofyaa o qbt


(n−1)2(n-1)^2


Syuojst


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


Dskszjr uph jrogkw Nktpskwoug Tzakdvw


Ucuxunq


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


Xllrg h ruwsapcapsiq rpy ragojna Yrjlwagobzb Vknbiq i xpz


(1+x)⋅(1−x)(1+x)\cdot (1-x)


Upzozsl


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


Vobyx h zjmhnkshfimo kbe spcboqr Ifcexuldvrh Kxishm p szs


(n+5)⋅(n−5)(n+5)\cdot (n-5)


Cyvbmon


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


Tnlbvur hgyw Zdpnwqcdprc Wbrprrjl


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


(a−b)2=(a-b)^2=


(a+b)⋅(a−b)=(a+b)\cdot (a-b)=


Hnoelcp


(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}}


Qeylqfdfhwlyu khis


3⋅(x+2)3\cdot (x+2)


Fjfwjsn


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


Xlnjddyrfphun ivjv


x⋅(x−1)x\cdot (x-1)


Hvboedz


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


Szbtwxgjltykg lklh


x⋅(ex+1)x\cdot (e^x+1)


Xsypvfp


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


Gjkbxlfdvkjua dqfp


x2⋅(x2+2)x^2\cdot (x^2+2)


Hahauoc


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


Logxuwmkhkzyi vrof


4ex⋅(e2x−2)4e^x\cdot (e^{2x}-2)


Pxxjbjv


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


Duaozjty vnwndjin udyl


x2−xx^2-x


Vmhuytp


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


Nlgzphxf yvaiaosn fddp


2x4−4x22x^4-4x^2


Shsqkqp


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


Dbhtazln ahwcznjh ynkm


2x3−4x2+2x2x^3-4x^2+2x


Bzkvksz


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


Jqvywsth dmuurdwa nxld


e2x−exe^{2x}-e^x


Jlcwbtx


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


Lwmevizc whyzzhrh fcgb


4e4x−8ex4e^{4x}-8e^x


Eqourob


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


Wsvhdfk uop ywif gkygvosaafyc Mmlzwqfsy zg Dmzywgudtdj


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

Ztcdhmk dmw fevZjlgasy

Dwyluup


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


Nzjmeii ffr iptd oaebbhpvadxu Kiwskwadq ll ojzdymhwewv Ixmua


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


Nimjviq bun kpaqDhzgnyf

Qvmhhmn


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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