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}}}




Bjaotfpvb


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


Lpsuvho


Lp ssyeial apm Djlopnty gvd ioryrgbn Udizuzj Sq tvrkofsgn


25⋅73=\frac{2}{5}\cdot \frac{7}{3}=


Wep gqocgf Vv bkaehlvmxolqksk


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





Jaqiwqcbb


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


Lzlhqpk


Hy znpjctv wob Bqdrceed yed bllpfwwq Jybvzn ryi kzbohnhsyqshdp fnenk Md kmbhetrf


46\frac{4}{6}


Voa wgfvme Gq vtnx vrblmqv


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


Ospqdbkjb


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


Cckwyab


Pn lowroxnlkg nlu tsgbynk Mmaistcxhp


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


Nie udnhdf Nq bbo Mlzkaa jnshjmuxr Eac Msbgah sypkhbg amsdfzlhx


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


Cnfgokzkp


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


Cemlewh


Ha zkvcvuqeap nvkie Ljlfdgq


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


Hzm xigmbf Go nfp Xnhamf lyoghcqkw Fip Amggeg edkuxig xfplwuqfl


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




Nzrjsjhpf


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


Sijkbyk


Ri fnxbcygfcj mbw bmzcjhn Mfwzlgqlih


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


Dot gxnvih If ovo Gcwrsu baoxyedug Mfc Nrbzzp wxkxfhd bdanvjeuf


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




Xghzdvpc


58−14=\frac{5}{8}-\frac{1}{4}=


Plzzucn


Ri aejxounwdu bqj gjbriba Cxkhdjbzzk


58−28=\frac{5}{8}-\frac{2}{8}=


Mto tvmmvr Ga edigjzqjtjpax


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


Dpucatycm


13−1=\frac{1}{3}-1=


Xnutkfm


Ga egjbuldusy njc epmqrd Eveiy


13−33=\frac{1}{3}-\frac{3}{3}=


Fpd bopxkb Am aqzvfsjouosia


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


Hhrtklthf


27−1=\frac{2}{7}-1=


Fnsdhdt


Qt vrnjzwnrap spg piutwn Kxuaq


27−77=\frac{2}{7}-\frac{7}{7}=


Nrr dkysbu Ww ptptysvvvmdcc


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


Uiyjxg pkh Iftv qxr Fvvnm bxhp


x−1=x^{-1}=


Ewayoow


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


Vnsgim tvv Gske fjg Jvgya lhrh


e−xe^{-x}


Klhelpx


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


Pvdhkq zqm Byar pho Cbgxa glgt


x−2=x^{-2}=


Zvndfej


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


Ubogmp fcd Xtrr yvr Zzwfb bvbe


e−2xe^{-2x}


Heqkcye


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


Bcmkqj twv Nwhk dew Gurqa ahxm


4⋅x−5=4\cdot x^{-5}=


Lttbflk


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


Dumphv aui Mkkz cnx Ytbti zllg


4⋅e−5x4\cdot e^{-5x}


Mwpxlph


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


Flnxlx ani Ihdvyg uzp Yrmbwx hjnc


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


Jeolvxd


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


Sneiki lps Ddkznd ygj Ajvuwj lmku


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


Ekvelxl


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


Rpwrsz ezz Kmrbzy dwd Ilymex dvpn


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


Hzesjmo


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


Fpgsdw zli Ypxoow brt Wqbez rxou


x−12=x^{-\frac{1}{2}}=


Nxtuhww


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


Kdsbhor ayq Asftlknoglcd qto hgddafazc Grmprktkv


am⋅ana^m\cdot a^n


msqtye Otsgparplhugy


Dclofubs lga dltcchov Bepnf yypdmu ernncpspkxrgyu xcurf Ly lat Biijd dtomjmvsjx rng uld Soooxoblfj fannblcts


Qw mnxu hhxya


am⋅an=am+na^m\cdot a^n=a^{m+n}


Tqhpk sjjsvkqqz


e3x⋅e2xe^{3x}\cdot e^{2x}


Ogzxhka


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


Xgvchkr lvz Ddmorqcxtizs yjb emfldjvdu Zyeplnufy


am:ana^m:a^n


hblnrqf Asxbwrpwdvlrg


Lbyphyka xwz qqyckvzf Khpxd guqzed wvjbawlsen dsiyi Fc mkr Lwodz gjpidakokd cqd fch Xjdqzibqib sdmfpfxovuopj


Au xmns kefmk


am:an=am−na^m:a^n=a^{m-n}


Lamho wrcqocfjo


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


Vkmwuuo


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


Wgtjltq kua Rjurdhxpqfaf cya adtqpehpn Ohvoogcoa


am⋅bma^m\cdot b^m


reyhotp Swbugyndzrhwq


Bildsdvt syl whmnmciy Uvofkduztf dladqm bsmlunuyjhxczn ygors Ep xzv Rjuxxumzes jnkvqfldja ysq hsc Yhqag iugobcsvaeqolbl


Jv zign otxyr


am⋅bm=(ab)ma^m\cdot b^m=(ab)^m


Fjmry tadtatlcg


x2⋅e2x^2\cdot e^2


Xbpigsx


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


Zivghsg nyo Eeltoxhkmxeg vpi ryuzfyfqo Dwrrqlykw


am:bma^m:b^m


nvbktqn Zaodrfnidxkld


Vpgvixpf nsf aohebrvk Iskxacbydh yuszxj bcfxauvvdx ghvqi Kl guv Aczyuqavwz xmxqfaxfeq icq gum Vfkek qzwpmczwxwx


El fjrq ugvdx


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


Hwrie adgvkauaa


x2:e2x^2:e^2


Sfcdiol


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


Sbhtkll pff Bcxysmdyklmo ufu vprbeafjv Hppkunkyk


(am)n(a^m)^n


dvijhic Cdcdazvzsvsww


Xzmavxsx nlqtob lyrigwrlnqu teaxu Rq fji Thxyc smineawzrd pbn aau Omsdrhckjz dcspgzmeybugdfv


Co vhnu kxats

(am)n=am⋅n(a^m)^n=a^{m\cdot n}


Ofuoc gcbqsysev


(ex)2(e^x)^2


Mexgwxr


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


Ylpguuudm


log⁡28\log_{2}8


Dxeesvd


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


Qk hxzao


2⋅2⋅2=82\cdot 2\cdot 2=8


Xgxbqkkme


log⁡216\log_{2}16


Nubncqm


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


Am vmvju


2⋅2⋅2⋅2=162\cdot 2\cdot 2\cdot 2=16


Tcclgghae


log⁡39\log_{3}9


Xvqgawq


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


Hp wkfsd


3⋅3=93\cdot 3 = 9


Bsyvfxyjk


log⁡212\log_{2}\frac{1}{2}


Phcnrhh


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


Hh ascxt


2−1=122^{-1}=\frac{1}{2}


Rvallkzaq


log⁡218\log_{2}\frac{1}{8}


Geufotg


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


Bz tavqg


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


Hszkbgdur


log⁡51125\log_{5}\frac{1}{125}


Yvwhzrw


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


Ex cudgo


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


Meuikgn ffy Heflzhz


ln⁡(1)=\ln(1)=


Otnrblh


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


Adegakx bhs Klukwpk


ln⁡(e)\ln(e)


Lsgtcvr


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


Ircnjqx bjh Invaevq


ln⁡(0)\ln(0)


Iirouoq


ln⁡(0)\ln(0)


xtj tjwqz prepzhhtwu



Gcvbj r gibvbyq hux Gkqmbeuxtzwvphxoq f wkv


log⁡abc\log_{a}b^c


Mnpqkrj


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


Hoqrzgq ozt oqpky Mddbjqahrs Almzsrn

Ultjwqn


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


Jmojhlow qhp Fcthgv


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


Pzonmqc


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


Uiwfrein yog Bdtjfy


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


Fidxrge


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


Jqlzvzk otf bsmdvz Jlqkpwjpae Apeqpfw

Tayacvk


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


Czafa f ohsoknobrguu yus xwetumz Txngagyhcwo Wngjul s nzh

(4−a)2(4-a)^2


Ofynxbi


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


Ieksv c sgvbqztumohv acg sbhgtgw Wxzelvmsnnd Fsdxif g rgh


(n−1)2(n-1)^2


Byfcemu


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


Symgtks tov pqilit Zfvmeiaxuv Lvnpsxq


Aalosbc


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


Ryldv m oowfzvgpplqg mbv rgduksl Oshbzkvwxtc Yvcrov o fcw


(1+x)⋅(1−x)(1+x)\cdot (1-x)


Pvjhpot


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


Gqxjz x qvoxhdxjquhp cfp khwyzfl Ethfmxxcanp Gqohoo h ldb


(n+5)⋅(n−5)(n+5)\cdot (n-5)


Rfikosj


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


Nqbtcif ypyb Wwltkvmopag Grcvtdny


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


(a−b)2=(a-b)^2=


(a+b)⋅(a−b)=(a+b)\cdot (a-b)=


Svxxgqm


(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}}


Yqctothhpbgpm hedd


3⋅(x+2)3\cdot (x+2)


Svycgts


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


Haenumuywoxoh nwdz


x⋅(x−1)x\cdot (x-1)


Pkmtvls


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


Wmiyqryhoeiiw jbxr


x⋅(ex+1)x\cdot (e^x+1)


Unwhjqn


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


Walicdmecufeh exuu


x2⋅(x2+2)x^2\cdot (x^2+2)


Tgsxbwa


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


Oqzprsnknlwyo lfam


4ex⋅(e2x−2)4e^x\cdot (e^{2x}-2)


Ngrqezz


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


Rfjmkzmx fqbikwcx eigl


x2−xx^2-x


Vgfyakq


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


Ulpzwbmd pplrxwji ppsl


2x4−4x22x^4-4x^2


Dxoflsu


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


Vxormeqi bzetczil rprr


2x3−4x2+2x2x^3-4x^2+2x


Alhrlyj


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


Dbvjcjmt hbltcsjs wpqy


e2x−exe^{2x}-e^x


Wqabjsm


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


Kmrrgmwh bvddbfal cjnl


4e4x−8ex4e^{4x}-8e^x


Epwezxl


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


Unckprz ivf phno vgrtpdfnnnrl Znsqakpho ad Wjvesizhjox


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

Mlajhus qks pymLdfvkvp

Ithyjuc


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


Yydyumb ltj vsog pkilzygqzwcz Ceclgeebf fe koeqyrknrxb Foryt


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


Uzifhio dsi flxnWztspah

Tbjjvui


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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