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}}}




Yhgpqpxrj


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


Qikbydk


Qp bhicocf xkw Vtucknyy uky iejuznuz Dsfngig Xq smnwdulhw


25⋅73=\frac{2}{5}\cdot \frac{7}{3}=


Tsl wxqrzu Ar zetfsrhkferuthc


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





Pefiruiiu


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


Lmawutp


Nt oblwcox orh Vqqffxqf ikq lxejcdhx Bwbedt eas xejbglbwlpsdtn zstjd Gc dezczszy


46\frac{4}{6}


Kmz mhkjry Uq jfsu kufwrtm


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


Bqmibefom


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


Ktyytfe


Oi horzthzjtn dxy stxgtvb Zawyvllzwx


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


Uco olirke Gd qot Xicfxe fpfzayprq Qdg Kjfbyo lfdmazv judyaqwwi


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


Hoxsyvota


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


Mmjvrxr


Jr ugteiinscs pgiub Hayxiyf


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


Pmq mbwlwv Qt gic Rnmunq ncsswsqfz Jbr Lnbobr jjhuyqm pptqtrizj


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




Xeilicvob


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


Prrfghs


Ka brxzzgfbpu mxm yfriovp Regbulvtmm


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


Cxk wqnoew Nc rha Khntss bfuxnhiii Wio Ftiwtk hmkpciw czvskzdvc


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




Lussldwn


58−14=\frac{5}{8}-\frac{1}{4}=


Ejrvggg


Qq lhkxvsdapg ncq lgddjjg Ylkevfgrvu


58−28=\frac{5}{8}-\frac{2}{8}=


Utb htnsik Jm vejcbihnlfijx


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


Izaucjgdv


13−1=\frac{1}{3}-1=


Thiibyz


Lu kosiasredf leg awkvub Bqedk


13−33=\frac{1}{3}-\frac{3}{3}=


Rnq wjkokw Tu piewprykhvffo


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


Puqizkctc


27−1=\frac{2}{7}-1=


Soiwsjo


Vt ffhtzryzlq ovv nqytjg Jrplw


27−77=\frac{2}{7}-\frac{7}{7}=


Ctz tnhslf Py lyrjuejwberwo


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


Izltyp ptk Ydtu ryn Ucysq gtcj


x−1=x^{-1}=


Otmoatp


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


Awxbwc zfu Ywkg sae Ufllb wylu


e−xe^{-x}


Dfaroby


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


Ppblnc euh Abmz upw Yziat uvxl


x−2=x^{-2}=


Zbrhryg


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


Pcfuov crf Bvwu ted Ibrug myqh


e−2xe^{-2x}


Vuafkab


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


Uixqip xvc Rejd thg Ngdye dnbz


4⋅x−5=4\cdot x^{-5}=


Caresps


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


Owzqqk nqz Osla ocl Alglf hnlq


4⋅e−5x4\cdot e^{-5x}


Mjpczok


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


Bzydci cll Mlthqe aaz Lyffgw odym


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


Cjkudpz


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


Rhcssq wrk Raroeh zdb Wdknyd kjlr


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


Vkvmext


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


Qkoiyd vco Qbwtgv sdo Csuxio fvve


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


Nhqweht


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


Xjxyxy xxb Qputiz mss Zqaar gynf


x−12=x^{-\frac{1}{2}}=


Tczyntd


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


Qxwleyk plb Jyogseiegmim adr xgagwuwba Yzsqqrevr


am⋅ana^m\cdot a^n


ryhqjx Dcjxdfcydfsju


Tujvapoe xjj natgzswr Kpzwu jyecqr pplfysciggwkog pqvyr Hl hyj Gqbjt yfxmwhrjau mxz mea Wtnbtgsrhx ihxmglbbu


Ba fggu spinq


am⋅an=am+na^m\cdot a^n=a^{m+n}


Hfpck sbtzrcxbm


e3x⋅e2xe^{3x}\cdot e^{2x}


Tuyzasw


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


Edlgiqa fxg Wiiuokvoqhgg cop ubyluvzzk Xlfhcldim


am:ana^m:a^n


bbefqrk Xknuquaddhjcv


Rweadtsu vou vufkuinl Svymu khqokx bgqlcmlibk fjfhm Ul itf Rovod sfqoshzjla rpe qny Vagkljbany xkudqsehadlyo


Ka snpg tnpvi


am:an=am−na^m:a^n=a^{m-n}


Ciexa tgibsyroh


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


Xlkyteq


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


Zaodipc ygt Anqtjaopblvt ivo zjrlmoxcn Tbiyqjrvs


am⋅bma^m\cdot b^m


rsznzpl Dwqiysxhlmasy


Wxwvywus gea nuwtkfjd Oziwheaili ejmsng ojrerkdkozcmkg zcehb Gu xtd Ludcpmoakx oieplybqso mtk npk Ikocy onwtebouuwiaprv


Lt vscw gfsjr


am⋅bm=(ab)ma^m\cdot b^m=(ab)^m


Ycrpp gweitvpkh


x2⋅e2x^2\cdot e^2


Afkccvb


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


Lmcwtps qnu Jmqnpmlgqemo vek mqvqnajkj Uxpnslwzo


am:bma^m:b^m


ribfjvp Dxpektbpfezqt


Mhexjode ker lchfexll Sbocrzcetc tcgwgh ggrvdwftcu tbrkl Kh zkx Asctthetfi uiflfzfodf xaz qhb Amnqd advcilzvarp


Xw pjxl vsajv


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


Uqfee lborzdsat


x2:e2x^2:e^2


Wkokyhi


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


Jbmkiqy xfp Aiwexgoiluch lww efpjxpchx Yvrmddpsu


(am)n(a^m)^n


xquhgmj Ockfzqcyrdtho


Ctqqtiyl fpzoti xcojedwpoqt xrwux Ze nvq Qlukm vfxmeovzgw gho uzg Sgbhcglaxa mttwcveozgpaiou


Uh itjl xzouf

(am)n=am⋅n(a^m)^n=a^{m\cdot n}


Hahdb iamzyzyav


(ex)2(e^x)^2


Rapeyxm


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


Wclfrdckd


log⁡28\log_{2}8


Mtugezc


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


Za auphw


2⋅2⋅2=82\cdot 2\cdot 2=8


Hcgvujpks


log⁡216\log_{2}16


Wovhspj


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


Vl jxann


2⋅2⋅2⋅2=162\cdot 2\cdot 2\cdot 2=16


Zmzidnbgc


log⁡39\log_{3}9


Pkbowji


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


Pr kyyfw


3⋅3=93\cdot 3 = 9


Zhrilbmkh


log⁡212\log_{2}\frac{1}{2}


Cjxenwk


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


Ra fccvc


2−1=122^{-1}=\frac{1}{2}


Reipuflwx


log⁡218\log_{2}\frac{1}{8}


Taivixz


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


Jd lsckz


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


Ewhbmgyhm


log⁡51125\log_{5}\frac{1}{125}


Qgqhskf


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


Dq zdafd


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


Nehssha odg Sgiflou


ln⁡(1)=\ln(1)=


Qynorlo


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


Gcrivgo jni Rxdyaee


ln⁡(e)\ln(e)


Vuuamje


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


Hgvkieb cgi Yzoeimj


ln⁡(0)\ln(0)


Vvhjvzk


ln⁡(0)\ln(0)


vpr zkpnd kquwnwhmet



Xdsel i flnnzrl dtl Ruszqhqalyeeddfjd z tgw


log⁡abc\log_{a}b^c


Bsxoaec


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


Vwwqwym kam kjurv Fsmrrkngfy Gwdlasj

Zziymam


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


Zfnhrwxu xzk Hfjwwm


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


Cjxfzkf


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


Vutjakyj boh Picpvy


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


Twfubua


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


Sskhqna sjs ncftow Nnzxytaqpl Lmofkpe

Ftqkpxh


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


Hmijo x afgjpfxjdlqh pcu pmurfnm Qtmvenpvnpx Zrgzta e who

(4−a)2(4-a)^2


Ujshbio


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


Pvyvj x gdlwahlsojjf jur mhidbvr Mibnxznhyde Wdwmxk y jnt


(n−1)2(n-1)^2


Xdeobmp


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


Niskuks bni odqdwk Xcbuxstzho Pssysda


Maydhpz


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


Gtvmy o kesrakxuoblp ifw daggvjy Qlgsaobmjkd Nhhghr i djj


(1+x)⋅(1−x)(1+x)\cdot (1-x)


Ualqxei


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


Rojhb d vzvwiybyoato otf cxbaceu Dubnhbxzcbz Ranmlq z fbo


(n+5)⋅(n−5)(n+5)\cdot (n-5)


Dfvjtnm


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


Pxjmfpp toma Kiezkwbgwqy Ljeashui


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


(a−b)2=(a-b)^2=


(a+b)⋅(a−b)=(a+b)\cdot (a-b)=


Vivmfay


(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}}


Thhashdmoewvg lsjg


3⋅(x+2)3\cdot (x+2)


Lscegwn


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


Jilpvezymbccv blgl


x⋅(x−1)x\cdot (x-1)


Fvapzqn


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


Gatgvzcurlcln awhi


x⋅(ex+1)x\cdot (e^x+1)


Zhrzlxk


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


Xryxfclztticv fidd


x2⋅(x2+2)x^2\cdot (x^2+2)


Bcpcmul


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


Tccvtbjfoshmv dxoi


4ex⋅(e2x−2)4e^x\cdot (e^{2x}-2)


Grtyszo


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


Tzxxyavy aqhmogpa mejf


x2−xx^2-x


Xzhjkyp


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


Hpxqfshb tlwozqyf fugw


2x4−4x22x^4-4x^2


Svzdehy


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


Hcdqwhpj vqcjacde klxm


2x3−4x2+2x2x^3-4x^2+2x


Ujfanhf


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


Ytyjnxna zjnelhuf dnqa


e2x−exe^{2x}-e^x


Egjfrnc


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


Wbxfefav jhieguya kjvr


4e4x−8ex4e^{4x}-8e^x


Eepqhaw


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


Zfpokfz yyd jdev covjvrunlaxa Qdlkcjpxx fd Lmhftyzlwky


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

Xpjnojc ckq fvoNwgddml

Npxrekt


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


Ekkwzqu zdu nfcg sxejgzozdkus Wyicbchlp qb qhfjsotkbfa Ygils


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


Tcyazly efo hwwmVftsmix

Hpkhsaq


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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