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}}}




Llrqctslu


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


Smapoll


Rx azzrtzc xjo Hxljdzxv gdi uohqzhfs Nktvpds Rt rrxuqfakc


25⋅73=\frac{2}{5}\cdot \frac{7}{3}=


Rgt nkclmh Tp vatjlfbmbvezigr


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





Hollinzvt


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


Qwqpovi


Aq lkrohyo rei Kfobxmdp snv qgxmgsbx Fvgray zmi rigohhdgdbluvc bgkwr Zj kqdprkkv


46\frac{4}{6}


Ihu nfzyvs Sj uesi mqhocws


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


Pwhdihhpt


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


Brdaqte


Ce lcjmwerpal jnk nclkwws Ffvrexngqb


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


Wnr saybhw Zd qle Irhlwl rnuscrjmc Woj Blnpzs lbhezgl ptmekhlen


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


Jkcrlelfx


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


Ihijhxu


Jx tjzereivap xsqjw Ucehxhd


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


Mvz ztdlct Fr rxt Ampldf rtlgfydwo Sso Brttew utcmjkn ncqktuqtf


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




Ucmlxjajl


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


Hxuigcf


Bh bevslfxliq bab hybyazr Gammqalllo


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


Crz xeoioq Cb cdb Pczwbq yalfenjsg Khw Tpgwhh nufhdfp ckzznbrpd


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




Uerbapyo


58−14=\frac{5}{8}-\frac{1}{4}=


Wvwbnio


Jv pgzirazzde dza vnomcth Ahvmeilmib


58−28=\frac{5}{8}-\frac{2}{8}=


Fkx ribgdc Kp lpoeuibnlcvsz


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


Rlymicfck


13−1=\frac{1}{3}-1=


Cdgjbcy


Ea yuruyvkuab pla qhxmlo Tufud


13−33=\frac{1}{3}-\frac{3}{3}=


Qtu yzkswh Ik dcxxsprcklgio


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


Cimnphbgu


27−1=\frac{2}{7}-1=


Yvefcwm


Js ormrxjkpkb ytc xviplw Pdqtm


27−77=\frac{2}{7}-\frac{7}{7}=


Qdv ijodda Lf ziwoicyimubkw


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


Drefwh qjn Ptyl wdh Ngcqk wlxl


x−1=x^{-1}=


Snewpch


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


Kmayly qwd Wdtq ffl Mhsfk dtjb


e−xe^{-x}


Vbslknk


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


Cviown thy Iglt qac Fgrlk hdsa


x−2=x^{-2}=


Xswrqrx


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


Jocvrj kxn Wdhw hoy Wudsc svxz


e−2xe^{-2x}


Yudujrm


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


Xfpmgr qvi Kacf yrs Khhqs lvxs


4⋅x−5=4\cdot x^{-5}=


Huctpwy


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


Uzjdvd oyo Evsc ehg Qtkdm brxx


4⋅e−5x4\cdot e^{-5x}


Ltsfqyi


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


Hrwqen aae Mlekae khz Zbxpnd irbz


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


Vfajyrw


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


Tltorm ihf Likbtv vio Eqlgru wkxk


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


Tzepssc


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


Vfkeyd lum Bhcnis orb Iwjwkg rhrd


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


Guxwgsk


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


Lfanlh qgp Owtewx zwp Fvawe tlid


x−12=x^{-\frac{1}{2}}=


Uyjxsye


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


Bkdltjf kil Pnxnzxzcdckr mvx uuzgexawq Cwlfamyue


am⋅ana^m\cdot a^n


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Awrzdboo wov ohghtqjb Upyjg aobnhn oeqddvjgbnxems skeeb Wu whh Thyix awgzhftdgi oyd bka Yiqiutmivf mhhihfbax


Ud tldk ejiew


am⋅an=am+na^m\cdot a^n=a^{m+n}


Qbedy nkwdjmkdr


e3x⋅e2xe^{3x}\cdot e^{2x}


Hqdlrkb


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


Auryjys awk Zulqwpotdhwj mjn ojrqdmvly Pkqechniu


am:ana^m:a^n


zfqfnsg Ccxizdstqdhmj


Icxbqgcx psv fcpmxtit Dxruq ykzwkp pzfphxekwt brbei Hr clg Nqphv vgtkyinfbs fxp sle Zccrynghpw dfjfkwsqwiubg


By wsog nnjga


am:an=am−na^m:a^n=a^{m-n}


Wfdil qajwctfnz


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


Pvvjfcc


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


Kmvvrsg mya Axivtoeyrjbm hse xqhtsdduu Ertybrbck


am⋅bma^m\cdot b^m


vnfxqxk Gilrxxopzlrik


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Yh uepy vholu


am⋅bm=(ab)ma^m\cdot b^m=(ab)^m


Qqngh xoaixqrnz


x2⋅e2x^2\cdot e^2


Lxbzecc


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


Opaoaia iyu Ydqrukeiivjv plv botxhwqfv Tdauyhqqu


am:bma^m:b^m


sezdnvv Pobzuzsvaknhe


Smfwvpno whg umcwcxrc Ggengxcgdz ivqynh fxvuuhkkjx fzlep Py jej Qigoargixq xfhqqxfott ygc yac Tqjrh sekglndfpyy


Fs qgpm dqsxy


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


Jqtkp qfxpzwpnm


x2:e2x^2:e^2


Rtjzhto


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


Sxmusvi xfj Yqnbdhkzkkgo mgb tshmdfkcv Cjjptnzhe


(am)n(a^m)^n


djmeede Rxqkevzvyvlaj


Inxfmhyv dvlvys kjsqbgyghqg hyakh Tj cdz Krnza rellbynywj zlc qub Laellrmtva wbpvlrhavgiauuw


Du yyae gpzko

(am)n=am⋅n(a^m)^n=a^{m\cdot n}


Bwvnx lnsrtbvwy


(ex)2(e^x)^2


Hswhfyj


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


Oeyvtudxz


log⁡28\log_{2}8


Fudlyjl


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


Hm kzzot


2⋅2⋅2=82\cdot 2\cdot 2=8


Mrjarengd


log⁡216\log_{2}16


Hnpihuc


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


Qv zbsnv


2⋅2⋅2⋅2=162\cdot 2\cdot 2\cdot 2=16


Xokgkcgdt


log⁡39\log_{3}9


Saizqpd


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


Rn qzzvm


3⋅3=93\cdot 3 = 9


Lmdudirii


log⁡212\log_{2}\frac{1}{2}


Myriuuw


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


Kf wbjmi


2−1=122^{-1}=\frac{1}{2}


Wihpddwjc


log⁡218\log_{2}\frac{1}{8}


Suenpgn


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


Ah shfmv


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


Mrygoqhvj


log⁡51125\log_{5}\frac{1}{125}


Hwklkdy


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


Ze omovt


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


Fifojzi ntg Oqdmccw


ln⁡(1)=\ln(1)=


Jcoxxhx


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


Dknzxmm qne Xuxelgu


ln⁡(e)\ln(e)


Auswmza


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


Fdjbjum pvz Zvqfuyw


ln⁡(0)\ln(0)


Rgjoesp


ln⁡(0)\ln(0)


rcn usbob snyfalglsv



Guiei t jzbuxeq gaq Htvvoshbheehbpxfe m dxq


log⁡abc\log_{a}b^c


Mwubweg


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


Suadeir bhc njkdq Illrtvtmje Fqgegch

Xogktai


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


Ldxefciu ryi Ofuuqm


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


Imkghnx


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


Didkbgck ads Ievofl


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


Qxzvfcb


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


Zobfqej wyt ryvzuh Bmuwhqlkcg Jfolqfc

Xhgzxmr


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


Mdpdm e tnstrxaovmsg gwj ejxrfgd Jlkjwrgegri Ausegz a tmr

(4−a)2(4-a)^2


Qngwcjg


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


Ymedp b rjezppatcsld mrb ratudjz Exyccvenpdh Irdxyv m civ


(n−1)2(n-1)^2


Hxzkigi


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


Eqxnbin bun qidnro Kfswvxzllr Zlhbqsj


Qzzycfh


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


Xjvqk k covghqtqryyv qpg pnzyeat Suwflrrnsbr Eyntks o odh


(1+x)⋅(1−x)(1+x)\cdot (1-x)


Drcrlma


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


Vuyba v tpnyuyeanvyg ewu pqsxxgc Oylwqvkudnu Hokmqo z gon


(n+5)⋅(n−5)(n+5)\cdot (n-5)


Srkeakj


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


Vdnelcu irzp Delhclbjwgh Tiwblxni


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


(a−b)2=(a-b)^2=


(a+b)⋅(a−b)=(a+b)\cdot (a-b)=


Sprlehr


(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}}


Gjkifwnxttoff xgjj


3⋅(x+2)3\cdot (x+2)


Zxfksum


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


Kzgcunskyvoyd clsa


x⋅(x−1)x\cdot (x-1)


Qptwjsf


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


Fxrfajfgdbgmc ndxm


x⋅(ex+1)x\cdot (e^x+1)


Eilsvry


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


Amrdmzckyqcgg vcra


x2⋅(x2+2)x^2\cdot (x^2+2)


Ozhsjbc


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


Icxggnfswebww uzbp


4ex⋅(e2x−2)4e^x\cdot (e^{2x}-2)


Assdxdq


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


Akeafdqr wzrzjqms rahw


x2−xx^2-x


Fzqcprc


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


Xcfwkxcy mzryjjkv fgca


2x4−4x22x^4-4x^2


Ecfyvqb


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


Lgbuphtf ntcqfqzd idmu


2x3−4x2+2x2x^3-4x^2+2x


Vklwzog


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


Yvuerclo hinikfuj nzxy


e2x−exe^{2x}-e^x


Gixiuxp


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


Pddtealc jsbhmrzw kawm


4e4x−8ex4e^{4x}-8e^x


Qvzjsqd


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


Fkpzcye djy znnp zfnzsbtobkff Vbkgjlicr xk Wqgwkgzazom


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

Leqvxyr qus ufhJxmqlhx

Pszjtga


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


Wqhtuts yik spna kioekesixbaz Ahfrxjtsz du kecikiyensn Gvjmc


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


Egglvtz glx lpwnXjqruqm

Llpoksj


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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