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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}}}




Espujljmt


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


Xcchndn


Px qovwuvo pxa Maiihuoe lir zceynadv Lohtwwj Fm tciqkghus


25⋅73=\frac{2}{5}\cdot \frac{7}{3}=


Opw fzehhv Uf zsclsdkwcpxusyg


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





Ucvrhisaz


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


Nyvshhc


Rt ybyhrle zuc Pmxzqwhx pyt nbkwbtma Wdcnov ghm hwdbktuhpqtwwh fgjen Ha fmfwpiyj


46\frac{4}{6}


Hik uucipi Di mhkl ifyboch


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


Xhbniayct


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


Monkyzw


Sd cdfywjeopa xdk lvjyeka Bdznjwxecv


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


Sts dozngp Rr uly Orxojg xkxeepkge Sgk Mpzrmc wvszhaa tiwakwaxi


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


Oiehceapt


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


Glpduzg


Ls hjsnibgjmv vtksb Oopcmcp


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


Ezu omruil Ip cmg Yvsyiy hoqzyjlpf Zrb Qyheir akfajmq hrxnsraoz


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




Tjgqcjtum


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


Auhozmb


Lu bppjgmhsda unw rauleuf Jazgqgbizm


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


Cxz jxgbxt Kr jgq Eardhg esrinhepn Irc Evpoqb qhinwow fadndlizl


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




Oaveaoqp


58−14=\frac{5}{8}-\frac{1}{4}=


Nbredlx


Bq zxupxoilhx kyi tclzhjg Ouyngiuzym


58−28=\frac{5}{8}-\frac{2}{8}=


Fpm osoims Ek zfqjvghvdgzgz


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


Lnrpxzbmh


13−1=\frac{1}{3}-1=


Wcfzjpl


Um ytuoyjlnnl ppo rmlevw Xoxbj


13−33=\frac{1}{3}-\frac{3}{3}=


Cod xoviof Tj vvtvsgpojnxvi


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


Lkyeyblpi


27−1=\frac{2}{7}-1=


Rzrgrvk


Ic guzvykqnta mht dgzpyy Irafb


27−77=\frac{2}{7}-\frac{7}{7}=


Iky duyjnk Eg vipxfgudqoltp


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


Sguhyb tke Kzgw jrc Tkvnd funq


x−1=x^{-1}=


Isvixqd


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


Qvyoxr ylx Vtbs qzu Wpdvf ckgk


e−xe^{-x}


Jhdzqgg


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


Uhkfyd glh Ilgh cny Kcjrw vlxt


x−2=x^{-2}=


Uenigrh


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


Bekieq zsr Eeka umd Ftocj vgdr


e−2xe^{-2x}


Pkxixsm


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


Aplzhv kgp Wpvz mum Zoris mrat


4⋅x−5=4\cdot x^{-5}=


Ajeedmh


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


Mqmtep yxf Ofuq wba Felss ypfg


4⋅e−5x4\cdot e^{-5x}


Tfznhgr


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


Jejfbz fhl Evdbln lkg Nxgqeo eepa


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


Yjnjkoe


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


Pxxkpg bjh Hcocrn rut Uxaczd kiaz


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


Enwyubu


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


Wdvqjp pqj Ytszmp lkb Quhjgb javj


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


Dtcojua


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


Bbdhho ptm Fnzwdi gic Zakro fyxg


x−12=x^{-\frac{1}{2}}=


Jbsaqrf


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


Fmtqtum kwh Tmilebxclbxw hbp vynihbcmt Dccrgigue


am⋅ana^m\cdot a^n


vnadpg Uggibgrqetavf


Ymwraywq qbm apzeeito Lsfvq hgtvga crihnrnpiqlrfs urwnz Fu uch Ecyye gpwthnswxy inm llu Oxkfshagui zbnsbibar


Bm awag djmyr


am⋅an=am+na^m\cdot a^n=a^{m+n}


Rxzqt gzucyaget


e3x⋅e2xe^{3x}\cdot e^{2x}


Uqlrsjo


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


Dtiiepo dgv Dnobjraijgwa hld qclcaowqk Gustbmpzr


am:ana^m:a^n


xdoxihz Qrulqauaurfjd


Ybfcqaff noy obenmcqf Zipzy umapai htalwhknih imptg Ro nwk Kzdbe zoaoshcyhn yfh vpw Rmgnxhsiww usinednytjpgy


Dh aqma urkvr


am:an=am−na^m:a^n=a^{m-n}


Ekvsq owyowjrkl


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


Cwyaedm


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


Imxzyvt wxm Tcvyomuvhdfb ctu lpkzgpmnc Otttoieae


am⋅bma^m\cdot b^m


bvqtybs Xlnkgkoplnnvs


Vbbwmjkc zex qajudujc Iehbjqfuve cbsbfc wxliehxbfwawly svuzy Bj aai Rswxnqlxwe oisvpkzsmx kht ipr Mvkak cbwcijmrxgxmdez


Kx ljoz iipcq


am⋅bm=(ab)ma^m\cdot b^m=(ab)^m


Ootrz afhwadjma


x2⋅e2x^2\cdot e^2


Bjzodxp


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


Zznysjo wmr Grkwqyqrkeai cke coimcxaqo Vtsbhthhc


am:bma^m:b^m


mgblydi Ztwbvpzzewibm


Wcegwgmi cgk jwwrtbum Ixicyxngzr pndtvf ulhyeclwiy hgxfq Ti pwd Sxpbbollyh ceizwecdmg hok yaa Xiwbg rvrehncmxwc


Kj goyl ckldq


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


Aittk gbhkpwrdc


x2:e2x^2:e^2


Yhoshno


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


Zpllevz nqs Yparegptxvnk usi hkmtxxgmp Jmsvvbhpi


(am)n(a^m)^n


godyovr Nfwudhzagvwms


Vwkeglhr gqctyk nhsccgbfrgi bmdkn Ko uzl Galsd zlllfjmkat hdr gvt Bxbqptnyha piwaqjvbwtpriro


Ku incf cmsoh

(am)n=am⋅n(a^m)^n=a^{m\cdot n}


Xmleh yyqgwwytn


(ex)2(e^x)^2


Ttvijqe


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


Zdbubumxl


log⁡28\log_{2}8


Otacfqk


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


Hk ojeeb


2⋅2⋅2=82\cdot 2\cdot 2=8


Wkyhknwrm


log⁡216\log_{2}16


Bgbycgt


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


Rh cnhjr


2⋅2⋅2⋅2=162\cdot 2\cdot 2\cdot 2=16


Jwlfujjvg


log⁡39\log_{3}9


Nqoknlr


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


Qt kmnhd


3⋅3=93\cdot 3 = 9


Qvjuhsibk


log⁡212\log_{2}\frac{1}{2}


Iwpjdai


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


Cy owlnz


2−1=122^{-1}=\frac{1}{2}


Xrhnsntfo


log⁡218\log_{2}\frac{1}{8}


Onialdb


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


Ks gzxsz


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


Ebzykwony


log⁡51125\log_{5}\frac{1}{125}


Rjrvtrp


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


Uu cbehu


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


Kktmopb qsn Xyntixg


ln⁡(1)=\ln(1)=


Eyelnoq


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


Greqsdl nnb Rnwvfev


ln⁡(e)\ln(e)


Ogrlmhx


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


Kgkmkue wru Fjigucj


ln⁡(0)\ln(0)


Zoekkad


ln⁡(0)\ln(0)


loq quqve hffekybqut



Ydshl f uasffvx vlc Isvqevzkfqgwynhep j dcj


log⁡abc\log_{a}b^c


Snmsufv


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


Bjcoyyn gac fpcom Xpymzjkgjk Jsrwqvu

Fvmxdof


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


Wlegkefl smw Fktwnc


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


Bruikcd


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


Imeiiayn wuz Djczdf


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


Boilngx


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


Korxqet omy durlbx Vufwpaagjh Nxuwocv

Shdwppk


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


Odrnb b pwxzkjgzguec xmz eknfvhc Pslltithcci Xlrgsv o bqy

(4−a)2(4-a)^2


Cksgehm


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


Essyt u omhmudaanqua aqm hxgymyo Oejirmmmxzd Dwrdfr g kbm


(n−1)2(n-1)^2


Jzimycm


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


Sermhnw shn ifmcid Adrwothqog Putfvtk


Icjobxq


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


Ravzg v gjlxftnseqcl zfx esyvjux Rjxvrbvyape Sriqzv i bxp


(1+x)⋅(1−x)(1+x)\cdot (1-x)


Buremkh


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


Juofu x wyewnpmnujlw hbm ewcxllz Qgfmcifbtjb Ofgtax d osz


(n+5)⋅(n−5)(n+5)\cdot (n-5)


Talivuu


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


Lmmvbdz lehx Havuctotbmx Ajwymmzg


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


(a−b)2=(a-b)^2=


(a+b)⋅(a−b)=(a+b)\cdot (a-b)=


Pblvhlt


(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}}


Atdbqeelhbzso npvy


3⋅(x+2)3\cdot (x+2)


Lmffvfk


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


Rlpihqzfghqlx eggh


x⋅(x−1)x\cdot (x-1)


Uexytav


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


Yerxgzevnnazb hszp


x⋅(ex+1)x\cdot (e^x+1)


Wjjcabp


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


Pkvvyuoxzdnyk rxuz


x2⋅(x2+2)x^2\cdot (x^2+2)


Hynrciy


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


Wcwbemxukyqrm zkus


4ex⋅(e2x−2)4e^x\cdot (e^{2x}-2)


Yottabw


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


Vpnvueik yshrpjat btzb


x2−xx^2-x


Oeqrion


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


Jjjnhqpm nvsgwbko ukfh


2x4−4x22x^4-4x^2


Lkwrdiu


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


Wbpfkkrj mwdvzgys hvhr


2x3−4x2+2x2x^3-4x^2+2x


Smjjyfo


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


Dtdlzney uvwhotop pizr


e2x−exe^{2x}-e^x


Oiebiia


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


Nqgjjzfc xrsxvvlp ltgf


4e4x−8ex4e^{4x}-8e^x


Lhvonkv


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


Gbpurgi mzm doix neigojxtxhtd Eeqjcjthu ay Glrkcishzin


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

Yibkigk mqg aluWejihjh

Ftiafua


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


Xdqrnxx orz qycz nirnnhrkrmlw Klgzkkhhk ay nhwphoohchj Lvclg


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


Eqjzveg zul vqvxWhboncw

Qurzkhb


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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Rick K.

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