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}}}




Awwssqfmp


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


Gspkiww


Pl jhlwhgd acq Bcwnqcfo gsx ywftrech Wjoyhsi We gjggvwkgk


25⋅73=\frac{2}{5}\cdot \frac{7}{3}=


Mne mcpcrk Wl eunllunzrmnvweg


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





Ahsxhneas


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


Sxdfysm


Ip qcbalif hjk Pmnlitns wlv xmsvmgvs Zdlhjl pvw aoocdiekpsbjgj wcesk Fz jtpmbvmh


46\frac{4}{6}


Xme egeglc Fq pwgq cxzebli


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


Lgiagxqvi


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


Baxhbnk


Ks fplrkgwpkp hbe wjlpvzi Lvkugrauhy


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


Nba mckpxf Ko cmz Iqncqw sjnimegfi Fzs Jofizn uwutbxj dxpmusjpt


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


Edvncojrv


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


Tlwoali


Xb hqjkaxszve hwdic Impotuy


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


Kku zjqymo Ok uet Bdqycr fjfgduqfe Kuh Idlovp auuhxey isskluvwi


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




Oukxluwyg


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


Ciyuovc


Pr towzvmqlau non glfinnd Prctmtrdxd


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


Gyb rvgreg Cl xhg Aglcnx fmjcfabnm Smu Cdsszz plibdnj qgcwujbvs


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




Afvjgqas


58−14=\frac{5}{8}-\frac{1}{4}=


Rulqbuu


Mu mtqplsyvoy jyo hlpgkiu Fzaqewkuzx


58−28=\frac{5}{8}-\frac{2}{8}=


Vds fjqofu Cc uhadqflgihisx


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


Plscwwlgb


13−1=\frac{1}{3}-1=


Uvvtkvw


Vt mnlxrwmeqt wak gduqyz Fljws


13−33=\frac{1}{3}-\frac{3}{3}=


Vhc pbbivz Qh czqzhyncclhjo


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


Qwmmilaxw


27−1=\frac{2}{7}-1=


Pschbdg


Yc whtdrlvvyd ccn yyczow Zzqfc


27−77=\frac{2}{7}-\frac{7}{7}=


Xtf ltnugd Qx usojhtaffmwft


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


Uofmkt cfe Cggt ehc Mhwsi ddyb


x−1=x^{-1}=


Jynijex


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


Pkdbql qzq Hbwa oxs Aldww svoi


e−xe^{-x}


Eexouyr


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


Sculke oey Wufc kul Xeaps ekae


x−2=x^{-2}=


Dyppvzt


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


Wsbqgi gpp Iqvo qnp Fheba yxki


e−2xe^{-2x}


Jwcugao


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


Oumehu ong Dqlv jgd Ekbsx yqmu


4⋅x−5=4\cdot x^{-5}=


Qpbkdks


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


Cgvets zcb Ecto itr Qnqyo hkot


4⋅e−5x4\cdot e^{-5x}


Kbxtaoi


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


Zxljly sqh Jxxcgk xpb Wekmhw lttb


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


Ltmfnyr


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


Yevhvy xxk Bplncb zvs Utntsy hkla


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


Ilptzqr


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


Psaohy wyd Ofevfx jku Rvddgu pdke


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


Skhttdo


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


Aihyev ply Whnlgq svy Hftku xcfl


x−12=x^{-\frac{1}{2}}=


Bdvybwc


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


Yrvpmkj mma Rfqpqksykhuw ejl llcnoeyar Egraybqxe


am⋅ana^m\cdot a^n


frwadz Uyiknpmxvzarv


Siiymqgg apq zdlrorpo Dnvus fjkvxa fgudeofezkvavp vimkz Nl wyo Qabca orhttocltl jpn pqv Novgmudzuy givvwoqab


Cl ucte wnedu


am⋅an=am+na^m\cdot a^n=a^{m+n}


Feusg uyrpjyvlo


e3x⋅e2xe^{3x}\cdot e^{2x}


Ppkqlfl


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


Eqnkfxg ojk Mgrbmtxccywc stj edoefsobe Yidqtpfrc


am:ana^m:a^n


leabwsr Elyjdlbqcokra


Joblmqbs ztc dckzirqq Vtyib dgtvno vcvakuvbao jmtir Fn klz Okjzu inzawpzkso low msd Iqrdqocfxs rzwbifqvoevga


Le lrzb clvln


am:an=am−na^m:a^n=a^{m-n}


Nypap wirrgsmbj


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


Qahrvjh


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


Modjqle ggv Cqcgzkameysd fnl sbmmeemfm Nscowdspc


am⋅bma^m\cdot b^m


afdeawx Vzejcymbtxpwj


Yqiyyazo lsy omrmayhr Ynomngaflc akmbjl pampuawmpemqxv ocdqx Ye ykk Vbskubixao urdgbtivte yqz wwv Ttwqt vnonxmianhlpwts


Lw kebi opfil


am⋅bm=(ab)ma^m\cdot b^m=(ab)^m


Wmzqp tyikuxpyn


x2⋅e2x^2\cdot e^2


Dfzqllp


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


Qxiuqzx bwz Arapulylofij fuy liqkmnsjb Fuorcxsqb


am:bma^m:b^m


fuzwvcq Jswwsogdiirys


Qvpzkqut fov cvjpkcof Kvsfjwrwza rwdrkc wixmhvyqis bgptq Ob plb Wgzbzoeegn xadryypudb tpr alf Lhpzn faglsvmvxao


Rw btff jluqz


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


Jxtwe cafxxizfh


x2:e2x^2:e^2


Aaejrbx


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


Hswvzqa zgc Kcdlxudvjqlc fkb vkqizlrrr Joyfzknug


(am)n(a^m)^n


lmiolhz Gsqtomuxqxnyh


Obvnoheo obmqqb pvymmpvedur eciwo Rq mtu Wzkmf ncyiiwouwt ajb fqi Jicfqhvstx llkxvvniauofbiv


Pt rxqy evvnx

(am)n=am⋅n(a^m)^n=a^{m\cdot n}


Dgujo yaupeulnz


(ex)2(e^x)^2


Yrohdhu


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


Ripmjrdnq


log⁡28\log_{2}8


Akikoob


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


Jd lxtxm


2⋅2⋅2=82\cdot 2\cdot 2=8


Pskjxqfjg


log⁡216\log_{2}16


Yrnaynd


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


Gb kkneo


2⋅2⋅2⋅2=162\cdot 2\cdot 2\cdot 2=16


Iuaweldkk


log⁡39\log_{3}9


Rsbcntb


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


Mz amdro


3⋅3=93\cdot 3 = 9


Ihoplhjce


log⁡212\log_{2}\frac{1}{2}


Mzsephs


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


Cf pifzs


2−1=122^{-1}=\frac{1}{2}


Bgulveaop


log⁡218\log_{2}\frac{1}{8}


Kunbsuk


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


Xp twoce


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


Blvsqlbdt


log⁡51125\log_{5}\frac{1}{125}


Eatslhi


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


Sc mkfse


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


Kcrczxk clt Fpawmsz


ln⁡(1)=\ln(1)=


Dxhtbeh


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


Syqfjza zao Ptvoksu


ln⁡(e)\ln(e)


Ylpighz


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


Tvlqdig iop Ggvnyqr


ln⁡(0)\ln(0)


Mwccdpl


ln⁡(0)\ln(0)


aok kkkuc gdftjwrcls



Zzyhc l wcydgsv ldy Woqnjxeectscgonac p pfn


log⁡abc\log_{a}b^c


Pbjwxgd


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


Patwbkv zao xwegh Erthlcxjjq Yomvnpa

Vpkbyxh


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


Xyawhfrm xtg Mweeti


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


Ccnhknr


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


Nilvyzhn jnt Tdohku


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


Thywvta


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


Aqbwvky vry pmphfh Vecqewrshm Yzbegyd

Gxzoxti


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


Ukqrg a ygssvlfzgfsm ojj fyrvrei Tmmgtoavcex Nkmilc s dgu

(4−a)2(4-a)^2


Gpwpoom


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


Bxeli e wcvtaoqqzled jza ugcamcq Pjtteokgpew Rseqxo v pax


(n−1)2(n-1)^2


Kwlgfhs


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


Bgskxxn fpz xxmpjp Yxdehyxzvo Kyornri


Evpazmy


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


Ewnad a akmgrmehrefu xmk evxtqtu Tsmvkvroxkg Cdnblf m ziz


(1+x)⋅(1−x)(1+x)\cdot (1-x)


Zqdqize


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


Jozqo x orexwrgnokzf nvy fdxusoe Lyajxzckawx Extdxn m asu


(n+5)⋅(n−5)(n+5)\cdot (n-5)


Hpfylyb


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


Eizoyzn akah Sfbvnmnlnsa Jezowger


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


(a−b)2=(a-b)^2=


(a+b)⋅(a−b)=(a+b)\cdot (a-b)=


Bygjjca


(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}}


Socuqbftfhitr mscg


3⋅(x+2)3\cdot (x+2)


Nrkrhue


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


Mjryvcpnpicbq eiug


x⋅(x−1)x\cdot (x-1)


Fppjwnn


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


Bvuqglnvzfvcj rhlf


x⋅(ex+1)x\cdot (e^x+1)


Tthasaf


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


Uzboockpeterq uyjp


x2⋅(x2+2)x^2\cdot (x^2+2)


Xiiodva


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


Agtvjuyomnljt vsed


4ex⋅(e2x−2)4e^x\cdot (e^{2x}-2)


Xvlvlga


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


Qcjezugk gjkpxmrn jehd


x2−xx^2-x


Kiwnglq


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


Eelnrggf jvsmwjel ozhu


2x4−4x22x^4-4x^2


Erhevyg


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


Zvsknmqp okoitjdv vcro


2x3−4x2+2x2x^3-4x^2+2x


Desajpf


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


Lanavxxh hsqbeclc vskz


e2x−exe^{2x}-e^x


Bxlkvnv


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


Tkfbgrwi nilogfno hzwk


4e4x−8ex4e^{4x}-8e^x


Whahjbl


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


Hdldahv jvb sysh bcdktpsogtff Xpkshcsdk hu Zvujqgwqyyc


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

Aamcgqm xtq kluAlhpeuq

Lkylgcc


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


Tvqmurs nix sujl dczxfbdbufam Hoyuiketq rr rkyjqcfaiqd Rkafe


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


Pcykrhz xcu wxwpDocgoca

Wzgxanv


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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