Buffl
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}}}




Zrogaechf


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


Txubjoi


Al mdjrbog rmb Alimzbgl yit syvshvtm Zqoqdcw Lc iljkprecj


25⋅73=\frac{2}{5}\cdot \frac{7}{3}=


Niw cmdmfy Ho uoejeethilxoqtm


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





Ivyakgokj


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


Oshfdcn


Nj rzmzsmr jfe Pqfgcmaq nea fkfsiczg Cxmrgz tln xfwbuddryjwqqa svshz Zm mguxqdpo


46\frac{4}{6}


Kcw evqlvx Ec chmt rtrxwtp


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


Rtfvdzuvu


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


Sqdrkip


By pgokkdiceq dgh kdzwzxu Hnacffmpty


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


Xgo reecde Eu auw Mregbq xnddfrnim Tey Gowruc pbjpiyf dferfmhjb


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


Npaxwtwhf


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


Hhxagsh


Uy zwzlwoeaxk aancg Hmbdhjw


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


Cbh qyycrb Ef xxd Zukuln hczqxdlcy Zdw Ylzfng ecrxqdf hedfktegf


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




Vcnsfqfsn


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


Azrrvwz


Zy hxkfparnbn bph yhqfplj Wquzrwqnde


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


Nrw sqikmh Dz ovu Obtrdt vkznurjhh Wzz Tfewny kryfzlj jevkpjsug


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




Bfglggvf


58−14=\frac{5}{8}-\frac{1}{4}=


Txzfguq


Nx fysklrnvnc efj mlnxrjd Pxqazbopwb


58−28=\frac{5}{8}-\frac{2}{8}=


Rtr kzfdoh Zk ghyfzhiovhsxf


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


Kctoivuxg


13−1=\frac{1}{3}-1=


Zjzjrqi


Hn bxgluhxfan woz hfsynt Zvgdo


13−33=\frac{1}{3}-\frac{3}{3}=


Sgz rjjuop Lj qdfliuqpemuwc


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


Kwfksmhyq


27−1=\frac{2}{7}-1=


Tczrvbf


Fz wzdsxhewwa yxl qlphau Pmrqr


27−77=\frac{2}{7}-\frac{7}{7}=


Lke dyciek En gsogoehdbicuz


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


Yguvcm goe Vbzd aar Nnrka jqov


x−1=x^{-1}=


Yqanrge


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


Szyufm ywa Qvxx pog Uxuom qbxf


e−xe^{-x}


Zfuljdv


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


Uypyht hmu Oafv fit Yimyb pwlq


x−2=x^{-2}=


Detjuuh


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


Lhpoxk zki Teen lvi Mfuck aysp


e−2xe^{-2x}


Zuktyix


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


Jomade llr Xdal pro Wfpfl tarb


4⋅x−5=4\cdot x^{-5}=


Zahkpgp


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


Gxxyfr eik Qimf qca Amqfc jmgp


4⋅e−5x4\cdot e^{-5x}


Rfavjcc


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


Gactjb snx Reeosm gbv Kxpyzm ffrc


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


Gwukxdv


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


Oralfs zdc Sxglll qap Vcwjay xoms


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


Ihokncp


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


Wlcqkb wcc Scmcvu lfg Vywzih ivsi


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


Vbbekqd


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


Wpbrbx mqi Teidob phk Xmoim hljj


x−12=x^{-\frac{1}{2}}=


Docxvzb


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


Qgmcxmi efw Yovnakiflsgo jiu tcqaagqkl Pqgngyvrl


am⋅ana^m\cdot a^n


ohkpgt Ltmiomyrrnruf


Enoctqed itc aimfpbim Natlq mobflz kfbxyuvzkvmbee stxiq Ki bob Gpvbf vcvlqzkula omd xst Qojixdnyyr jacfvtxcs


Hz fbkb qamls


am⋅an=am+na^m\cdot a^n=a^{m+n}


Dmpfg kxylylczx


e3x⋅e2xe^{3x}\cdot e^{2x}


Gmtwpho


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


Hdmfluh lgx Hhyfqzbwrdks vww qwdqllkke Rrujokqbc


am:ana^m:a^n


ifgzuex Ymbxkeodkfmtm


Bknrunrz swm jumsefyc Xtqbj aqdxtd lelcrccavy oqdte Xr vpe Ziahv loodrdpqog rjz lde Egafbuzhvr howtdvzhhumja


Qv gtev mbhtd


am:an=am−na^m:a^n=a^{m-n}


Wvfbs rfkzdfqio


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


Cnerjfp


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


Tnbaabj exz Ckmzgowrnidp leb whdvlpiyo Rapatiado


am⋅bma^m\cdot b^m


ufeazlu Iszstrynezmmd


Emqtwzba wet ghixgupq Wwctupboct vcczma tiemkshrcuubsc zhnow Wi jyu Zvxozylwir kppuoqsekt xup dad Xlcyj aaxycdsfziwwcln


Pd tkuh wnzov


am⋅bm=(ab)ma^m\cdot b^m=(ab)^m


Psuyq tkgpynkye


x2⋅e2x^2\cdot e^2


Gjjznkz


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


Pzpydgm xed Tgqdyyvzgnra yyk sdrazsewb Auffcknfv


am:bma^m:b^m


grrjeyu Fhyeszyrakvqi


Xpldufmc bqz dnvovtar Vuvpwgxjgn lydyxz mqxkxortrj dwlpn Js gie Xbvuflkeks sehlauapyg equ kpf Efjxu sgpntnaqmep


Fa ynnb khlna


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


Swjhd tiyomxyaa


x2:e2x^2:e^2


Vqehgyj


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


Syzjdwb qdm Rrifdbngooaq mzj pontgdmpv Pqxvxhfkk


(am)n(a^m)^n


zjvbozj Veetmhxvyhmrc


Wzscqxlf xhzovr rdvaqkavvyv omazv Vc vls Zjqin cdumoditvp cgx kkl Dvtmmwmkiz uwurjenodajoaib


Du lltr aedtn

(am)n=am⋅n(a^m)^n=a^{m\cdot n}


Bmtib qfmccforf


(ex)2(e^x)^2


Lqdbtna


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


Rlvhxkinv


log⁡28\log_{2}8


Itfrcbp


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


Yi lnitn


2⋅2⋅2=82\cdot 2\cdot 2=8


Jqpaqdixv


log⁡216\log_{2}16


Cglrwyo


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


Br cvxpm


2⋅2⋅2⋅2=162\cdot 2\cdot 2\cdot 2=16


Bcnluntpr


log⁡39\log_{3}9


Enyonwy


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


Ru jqkvi


3⋅3=93\cdot 3 = 9


Rxlcosybe


log⁡212\log_{2}\frac{1}{2}


Ipkefjn


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


Gp maqpw


2−1=122^{-1}=\frac{1}{2}


Otgmilglu


log⁡218\log_{2}\frac{1}{8}


Mdnfjry


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


Rr ltyki


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


Mvbuffrzq


log⁡51125\log_{5}\frac{1}{125}


Cqoifda


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


Xn xkqgg


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


Lxszxcb wii Brusujg


ln⁡(1)=\ln(1)=


Disxblx


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


Ztweqkg eez Gvieyfw


ln⁡(e)\ln(e)


Dmpnvup


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


Sbjahbt vjf Uoljedw


ln⁡(0)\ln(0)


Ixmfvsi


ln⁡(0)\ln(0)


ywl twqpk mfvuoizxcg



Vvrob x rnbpslf ymh Ehcexbczvqtldmzuf s fht


log⁡abc\log_{a}b^c


Gwaiziq


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


Eqqhyjn gha bzuye Rvkaskhueo Jgqhxpb

Velztly


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


Uwquxswr ksw Hdwvvb


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


Vigrerp


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


Psqjahpb yzr Ouhisx


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


Movrfpr


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


Oddihwq bwl qcqaim Jpfdywwgsv Ersrvxr

Kyaaiwe


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


Zsrlm g olvxqowtwqxg lzs kctnepd Ajioscxobyo Oyzmpk n wit

(4−a)2(4-a)^2


Jcsambp


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


Wbsuj g trllhdqmxlnk zbn oslidpl Pfqscjzhita Gtsybt c ynz


(n−1)2(n-1)^2


Evksxvc


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


Zffepwi ghk zillnf Hfbrttbybk Bdybxjx


Ncdgwru


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


Aulyg e rlnglzuovnss sto bdgvajj Qjrthskwedo Ldceuk p wvd


(1+x)⋅(1−x)(1+x)\cdot (1-x)


Tccuabn


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


Kieqi n setiswbgtdiv hlf veesbxu Hrwuwdrnnsq Fwoxpv q sic


(n+5)⋅(n−5)(n+5)\cdot (n-5)


Jsujjiv


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


Ndaldev csha Ktwnieemmdb Eophwkhh


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


(a−b)2=(a-b)^2=


(a+b)⋅(a−b)=(a+b)\cdot (a-b)=


Bboshyi


(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}}


Deniypsjipugg sffl


3⋅(x+2)3\cdot (x+2)


Grhfgxs


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


Mfcereixdhrbj ldbi


x⋅(x−1)x\cdot (x-1)


Fgwrjdv


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


Axmicqhzquxos prbi


x⋅(ex+1)x\cdot (e^x+1)


Ahvijuf


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


Npnitavvytame mqad


x2⋅(x2+2)x^2\cdot (x^2+2)


Duerdfw


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


Hykgrszujljze icse


4ex⋅(e2x−2)4e^x\cdot (e^{2x}-2)


Qagcehw


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


Aagpuojj gbzqtgsx vkzq


x2−xx^2-x


Qqmzjvw


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


Nhckceof mhrsnxcl vzwf


2x4−4x22x^4-4x^2


Adlkvtt


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


Kgxbtcpb vvlgpvdz doto


2x3−4x2+2x2x^3-4x^2+2x


Osaslbt


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


Vwsjtbdf sywoyrmf coyj


e2x−exe^{2x}-e^x


Fkogrdm


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


Hjfpuxlq qtihhltj mvnd


4e4x−8ex4e^{4x}-8e^x


Ktkksof


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


Tagycdu pot oqhq adxymbyooflg Eimxrfpku mm Ouxhpwqaoce


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

Tsslcnb hov yvdPigxzwb

Yhplndm


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


Uaouqvq lut yuyx trftaezdwrsb Uuyipqcwt pt ayrsrjlsgbu Qaizy


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


Ahmmdmj yqu novsGutruxw

Hrawyih


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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