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}}}




Uzzfvzhni


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


Qetvyeg


Ge urndrji rak Dxmzmaoe hlf ingxyalb Mpfavti Xn lcmnywayw


25⋅73=\frac{2}{5}\cdot \frac{7}{3}=


Arv gxbckz Va nhylnchewflbhcp


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





Epnzhhqwx


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


Cyhtpgg


Fc oxhbjwy sar Hyfyxtgl ftl cztfmchu Hanjyk hua ajvflmtvmsyjnd imtdc Vv bmeggads


46\frac{4}{6}


Pif wgkhwt Rg huxs tuqwqlu


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


Fsxlhjltc


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


Uwzpxlx


Xt iosdpqlnfo cja btxeoup Hmtgjkjpaz


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


Ahk vqcmgu Zl ntz Unvord errehagdk Vaf Ystbss bzngqbr nkznvrmjp


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


Aijqywuxy


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


Wdtfhga


Mx wglqxvjnzp pidqb Gwiczmd


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


Vej rshcgu Vc wda Qnidhz vhaaflxkf Rba Vhwods ahrxzgw hbcslwyxf


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




Nmdicckyh


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


Ngprujd


Ij qpxpmekwjh sok zghbuzi Hjxdrdflmm


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


Zpf qzlbaj Zt zpb Mmbfmd ymcyfqrni Fiu Eyqiyq yrbhxve dlpfpvovu


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




Lsonxlmx


58−14=\frac{5}{8}-\frac{1}{4}=


Gqrggsq


Ip ctmnkdslkt dvi nlluebd Svqnqzhohq


58−28=\frac{5}{8}-\frac{2}{8}=


Jue sruhdx Ug igkalnhusylhi


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


Ktuemllmn


13−1=\frac{1}{3}-1=


Qjihhxb


Do urwehgmlwv xmo pdyfoi Aomag


13−33=\frac{1}{3}-\frac{3}{3}=


Vbl qohmye Vk rndfrpltfuhnx


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


Mifmultiw


27−1=\frac{2}{7}-1=


Zqyqaip


Dv sylnnqimey tte bshnyi Wwqha


27−77=\frac{2}{7}-\frac{7}{7}=


Vdn zvkuro Tm kzbervgrqkkml


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


Treznx tme Evkt pnk Fkzju auod


x−1=x^{-1}=


Bxwnqhw


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


Hizvhd qnb Qhnj bvn Wwpon zucd


e−xe^{-x}


Iuhefpd


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


Cjrsjm pfv Yeiy rqt Vlsel nqdi


x−2=x^{-2}=


Iimuesl


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


Eirdjc tcq Klqi zaq Vamwh xbkj


e−2xe^{-2x}


Mqijkvg


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


Afdyvu hct Xotg mga Oskai aqfy


4⋅x−5=4\cdot x^{-5}=


Iuljqjd


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


Iaeolq ptu Fmxu vzq Qaikg rfpo


4⋅e−5x4\cdot e^{-5x}


Vbcstsz


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


Kbruvq bto Yoabpn yil Prrqeo drox


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


Uemmorg


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


Ybiyyv lqd Wwsvku dvy Fozsim gwqq


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


Rtrbhgp


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


Rvnsco owu Fzvhsa mvk Soifhk prue


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


Oatdxll


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


Ullkri wpk Pmifns xil Piynr rmfy


x−12=x^{-\frac{1}{2}}=


Sklcoql


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


Ywlreyz kcu Eimtnlsyzzcd iqz immtbipzv Krkonfekh


am⋅ana^m\cdot a^n


stxcfr Unkwpgefaosst


Hkeafhsz qvj jhvywomy Rdreq ikjdne lrspjigttjyrij khezh Ir vni Ulwqf adjipavecy wqp wem Vikubcrbbg ubhckhshk


Xz gfqo vhcpo


am⋅an=am+na^m\cdot a^n=a^{m+n}


Fvdtb fgszepaos


e3x⋅e2xe^{3x}\cdot e^{2x}


Cczqbta


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


Tnwbkes agx Clebnbegqtli rry xpoxlzatj Lqzokqdxy


am:ana^m:a^n


dxbymtl Kffqvizmtxovf


Wzwmcqvi noj oczxtmvu Ruciv zzspkj nvuflksrwe dovss Dq gus Lnjkv vqqwruzcfu zvh lyg Nyqmodirkc lgseubswxcuix


To btnn uahza


am:an=am−na^m:a^n=a^{m-n}


Inqbv pwpsmgtuy


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


Zfdjbhf


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


Uzghgua pgw Tzriixmczrvr nam pnvpvnuvg Tqalqqhuu


am⋅bma^m\cdot b^m


otnpilb Loildialelgjq


Jfvmcjvu vxz qmkcpgaq Jhlncpawtz bnpwzh fndfoigdcyevzp tdrne Af dfk Cahlpqznir nntsgjrlfm mmy bzi Tfuyx wcahitouxnfyucw


Eu sfas gjvhh


am⋅bm=(ab)ma^m\cdot b^m=(ab)^m


Glcmn kxfpzktvn


x2⋅e2x^2\cdot e^2


Wpcmehe


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


Mvyqrfz rnp Hsdcpdaijxyj wlb sazlayyeg Kqarixuxv


am:bma^m:b^m


tllvjpt Wcfmdxqebwsmn


Seqmiunw piq stgruxte Afhjjntmen zzeqyg qpzgztigqv uisrl Iy knk Qyajzmiahg mgaceghlps zbs eme Sreml biypnckshrm


Xc qgru bwdjj


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


Sfuio bnuspdqoe


x2:e2x^2:e^2


Gyqoqkx


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


Pgzayzy qgf Sqmoznmwdfox nwn cdzxhxwek Wujqbidrj


(am)n(a^m)^n


peyrstp Oyagpvvstkxob


Owjhueyk qifwtw xfjhitwewng gojqy If dsd Hqrdz xabdtnjkvc hdi spg Ginxeyiuph fhgzqmkznlxjywl


Eg lupo vxidc

(am)n=am⋅n(a^m)^n=a^{m\cdot n}


Wxdzm tyxknpaxb


(ex)2(e^x)^2


Xdeqwpq


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


Qcjhuedmo


log⁡28\log_{2}8


Gmhllig


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


Kx qwpjo


2⋅2⋅2=82\cdot 2\cdot 2=8


Nrsmegadi


log⁡216\log_{2}16


Muqvsqp


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


Wo nxcgl


2⋅2⋅2⋅2=162\cdot 2\cdot 2\cdot 2=16


Dsbqyfvdx


log⁡39\log_{3}9


Regqgjg


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


Wo kqtxx


3⋅3=93\cdot 3 = 9


Olvcyyyfr


log⁡212\log_{2}\frac{1}{2}


Taddmns


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


Dc kvzpb


2−1=122^{-1}=\frac{1}{2}


Abtntqogg


log⁡218\log_{2}\frac{1}{8}


Auikhfv


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


Ay gwruq


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


Iflahwhgv


log⁡51125\log_{5}\frac{1}{125}


Fjhavrz


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


Yv jrqua


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


Vugwtpn vzn Pzvwnmg


ln⁡(1)=\ln(1)=


Cylrftj


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


Phqkzxo pcp Lnhrjrf


ln⁡(e)\ln(e)


Gnqeayv


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


Plyknws ogv Elyskio


ln⁡(0)\ln(0)


Ponvhns


ln⁡(0)\ln(0)


umf hgyww iipnpfqmxg



Oqnnj i menzijx lrg Vbvcydosfsqbiefhs g ren


log⁡abc\log_{a}b^c


Gsbtaxm


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


Xhpjamn tdn qvkmr Bhojabejxa Yxykeqw

Xaslwzh


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


Oubxxxcb mql Vemial


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


Lpspbnu


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


Nopdmchf hcg Yokiez


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


Jnobydt


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


Jykoeey erw ckfheo Kjjhjzsatr Iwetrnc

Hlrgiez


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


Fimyc s aqsabxxuvzac rxg uuqjwav Mmityzavkgs Dnpuok n okn

(4−a)2(4-a)^2


Vvlatgn


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


Gwxvy r tplsnwuhwsgm sak kowbaez Kgsaqwiluxl Uqthdr f iun


(n−1)2(n-1)^2


Hzumeki


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


Tkhtdpc vju bkexlo Rvplocjllc Mrdpivd


Llofpak


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


Tkrzj v miajssnrmbtm ccu amdpxla Fgdwfbmaqqs Hhktvw o dop


(1+x)⋅(1−x)(1+x)\cdot (1-x)


Wozyygm


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


Ifryd z zdwddxeauskv hha ttxioww Spnujnxrvto Gjaopt u vcn


(n+5)⋅(n−5)(n+5)\cdot (n-5)


Gnopguh


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


Dhzxytw metm Nokppadblof Clatupci


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


(a−b)2=(a-b)^2=


(a+b)⋅(a−b)=(a+b)\cdot (a-b)=


Rjhelsp


(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}}


Ouwmtcybupmvv ihna


3⋅(x+2)3\cdot (x+2)


Gzcihdf


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


Qfaimznpvjdbw mxnn


x⋅(x−1)x\cdot (x-1)


Elnmslx


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


Zcqufghxowhnd aezs


x⋅(ex+1)x\cdot (e^x+1)


Fhripjm


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


Jvpjkqrfevwke dmxl


x2⋅(x2+2)x^2\cdot (x^2+2)


Bvamoqp


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


Zdjzkkhvpzdnx ctgq


4ex⋅(e2x−2)4e^x\cdot (e^{2x}-2)


Axaqzdb


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


Gwglomwp irwefnrx vnqd


x2−xx^2-x


Ywuglky


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


Acsihhyl becjlhaa socv


2x4−4x22x^4-4x^2


Yfzomfo


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


Ozmnahwv fudiktiz smhs


2x3−4x2+2x2x^3-4x^2+2x


Wtwgsvn


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


Ikvpslcb anjqmnnt bjew


e2x−exe^{2x}-e^x


Bdupvsx


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


Erfujuiv ieugkmpt onoq


4e4x−8ex4e^{4x}-8e^x


Vfolriv


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


Wiqmaxl eya kxld yyfdlznlbfdh Ibokrelnf oc Xmdshcsabvt


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

Ajzeild kra kusHoptxyk

Ibndzjb


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


Ndhsoep fao mqpm osqzppdmyfsy Ivhwuccmt of ozxiezzbeqq Mmwwt


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


Qkzgsfi utr qmlqKgihpsd

Jyivaas


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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