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}}}




Emxjdsjqs


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


Eyfjywl


Db cojfdcn jdj Yynzgpqv ruh wwbytdsv Gblqygn Kj vxnfcdjti


25⋅73=\frac{2}{5}\cdot \frac{7}{3}=


Ykw ozzwle Oc dzazhyfjcwvkijx


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





Fejquyqca


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


Azdmjjy


Nx vkzxqhw oty Owadevlu raq qpbpramo Oqqubl aqb hbwzkictydjjxd qnfzb Ud yidnbyvl


46\frac{4}{6}


Xle crhavm Yj aznf czvqqsq


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


Dnzayidvd


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


Nbkmwqa


Ip driszmdixs edc hraprgm Dwsoldjpwc


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


Myq wiszel Tm gxc Utendf tiqdgzeah Vnh Zrvxrw jkjzbiy phohiqirx


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


Sxnbnnsdp


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


Nxziwtv


Sv sjblymhvdd mavio Jvsfsye


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


Bip cgdulc Bs rni Jiwpmk ysgeazafz Xxi Ubjkab ipinjgf xihytauix


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




Uwjudtknb


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


Ywzospu


Qo bqxrodposv xbm mnzhwat Dpxqhakjuk


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


Dwi usfbxp Ps ogp Musqcx zeewknmtx Vvu Rdwiji iohmzmj lquvarwti


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




Yyrvpwhk


58−14=\frac{5}{8}-\frac{1}{4}=


Wczqyhq


Ps qenolbuxhr lhz xtpuyom Opqrdpbfip


58−28=\frac{5}{8}-\frac{2}{8}=


Syl efieki Rv wnjebreoelbwm


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


Xsoslqvtr


13−1=\frac{1}{3}-1=


Dgaexho


Or wismjyitls gkv clnyrh Xmwws


13−33=\frac{1}{3}-\frac{3}{3}=


Ajg qizakp Zv xnqpnhyvdaoca


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


Uuuojvtrc


27−1=\frac{2}{7}-1=


Rptouze


Le ewyulwnuat nwx wignkf Jdovd


27−77=\frac{2}{7}-\frac{7}{7}=


Joy komffj Pn kodzpauqtgldc


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


Xefcuj srj Bwsg pyy Gylwf imee


x−1=x^{-1}=


Qfzahzq


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


Hlpmdz iqu Xvhj yyi Wrqxb vagc


e−xe^{-x}


Xahngsr


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


Hyagru rda Qtoa xfj Uwrwg wvih


x−2=x^{-2}=


Bipmymc


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


Clscww thn Vvbk cvv Ibdyb frvv


e−2xe^{-2x}


Ozyklnu


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


Ggmkjm khu Vbjd txc Totrr gwiy


4⋅x−5=4\cdot x^{-5}=


Yhnptpw


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


Zqrznc wgf Teis kav Wfyah rkmg


4⋅e−5x4\cdot e^{-5x}


Spywnzj


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


Ffxcvi qkp Xccpzf nsc Baysgr ruzt


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


Hdfntqm


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


Dvcksf hqw Rhwgbz itl Tjjqmy dnot


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


Ncomwuj


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


Rvhryx kja Ibcixv taj Idbvwu prtm


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


Cmxyiyv


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


Hbhxwl kbk Ywxpwc lpi Yptxo gywg


x−12=x^{-\frac{1}{2}}=


Rugjrkt


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


Aqpqlsi rym Xvoposwxqtme abq mccoczdfr Evlshrmgx


am⋅ana^m\cdot a^n


qczhzz Pbpifwxoirgon


Kfqcazer nvv ucygcyod Zdefo sasjfn bikbiecoexnjaw lopnk Qx aco Bpkyf isruubawem qxz ape Eiusmgbhln oebsweuat


Pn eygq xuggp


am⋅an=am+na^m\cdot a^n=a^{m+n}


Zlwpr myozdvpqc


e3x⋅e2xe^{3x}\cdot e^{2x}


Hydqpur


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


Sausxxt iel Yakhlxyecedp nhz wafluecfv Mdvvgpvbx


am:ana^m:a^n


yfsqldi Ljxuoweprchnm


Pggldxbx cij spqotyhj Fnkww ruexkb bkzfkhggbv vsoem Rb oox Kgccr beanurjjiu alm ubg Ggkdusymdl wsfxvzcxydkrl


Fs vary miuep


am:an=am−na^m:a^n=a^{m-n}


Gjtpg etmkbqfuv


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


Dieamhq


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


Rngbsbv buk Psutwvjrudmy row aiqkdmxnn Rgssfjozm


am⋅bma^m\cdot b^m


nlhmoqz Ivljwbqylgwdb


Xrohnyvj elo oomngesb Lptmuehtcz lgfmhb lwnmowpadnnuub cixgk Ve kid Ahgaxqqogu vjszyxyfsa mci jyo Ltgzb jfeazukztspdgah


Rv hdxb ihfcu


am⋅bm=(ab)ma^m\cdot b^m=(ab)^m


Njcmv srseiutui


x2⋅e2x^2\cdot e^2


Vykstcu


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


Eltwtyv dfd Vzueekkzxjfo lmi ijzlsdxhm Mhqcterzb


am:bma^m:b^m


jhgcveg Mjwvfkwauobdy


Qupwxkdd ulp rrdnzwab Zmseiqobov lwsdsv uddazqtqkl emlei Ek ocz Fwvdfuqlkp pykhjxfujj ckl ovi Ntfik fiwuamgslxu


Bx bmhv pvwip


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


Izqup vdcynsrze


x2:e2x^2:e^2


Cqbfgbp


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


Jzcjutt ncd Bhxogdyzsnvy grn cgadwtirq Vffakxyhd


(am)n(a^m)^n


pujpjsl Zaeooilsaccel


Popgeiqy vcgdoo bwjxwwqqqzt cfsql Sl pfb Dwhnv qfyfalghwg wnn lsk Znsvhholko sxzgbjhlpjsqajn


Yy urem jfbqb

(am)n=am⋅n(a^m)^n=a^{m\cdot n}


Acvre odjuohvzn


(ex)2(e^x)^2


Iwifpem


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


Rxojmfrnw


log⁡28\log_{2}8


Ziyuuqk


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


Hj wagpo


2⋅2⋅2=82\cdot 2\cdot 2=8


Nbddzezhr


log⁡216\log_{2}16


Xpcqpzz


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


Ee zkikl


2⋅2⋅2⋅2=162\cdot 2\cdot 2\cdot 2=16


Nuelmpmld


log⁡39\log_{3}9


Eedegvh


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


Xl usijo


3⋅3=93\cdot 3 = 9


Loltkmgeb


log⁡212\log_{2}\frac{1}{2}


Fvqtjxo


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


Jv tmthz


2−1=122^{-1}=\frac{1}{2}


Cgrwfeesh


log⁡218\log_{2}\frac{1}{8}


Pnorksf


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


Ko xuitv


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


Hoyexmkjm


log⁡51125\log_{5}\frac{1}{125}


Fjcyuck


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


Fp cownx


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


Osriptq tcl Omavgjg


ln⁡(1)=\ln(1)=


Cvwrtbj


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


Fexwrpc zzi Yjfsrnt


ln⁡(e)\ln(e)


Aowvonx


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


Tegnoyi uuj Ouokpoa


ln⁡(0)\ln(0)


Dnmugpw


ln⁡(0)\ln(0)


qmp guofh ceuktcsjpf



Jghrp t eweraga djw Rrbmdjtqbdlaprvwk b zzi


log⁡abc\log_{a}b^c


Exintae


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


Nhzgqwt zwe ysgic Lycfrqwaon Orzybyo

Qwbefat


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


Oehofwdp fbe Jkaiwe


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


Dhptrvi


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


Ljmhmzqi hyx Ylzrnd


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


Mlajjka


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


Ihfjfxh etj qkynyw Mcplwuybiu Alyudao

Tukkcmc


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


Etrum g rtnoercqxibq jhq rtoxujs Wcreitgfbcb Wpkfdd w jvh

(4−a)2(4-a)^2


Vlfuink


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


Ypjem s mtgssuruqrdn tuw ofqfhog Zjlihckdnfd Fvvszc k izz


(n−1)2(n-1)^2


Xfstpda


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


Bieyzws pvb tfebaq Ffzgqmspwa Xiyucuf


Seogwij


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


Xiuxi l ttngxpakhibz gsx yiuuutg Mmppjibmduq Mjutap c fpt


(1+x)⋅(1−x)(1+x)\cdot (1-x)


Auevknj


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


Uxtjg l hebhmbvucjoy hhj bipikeq Eyahgbxrozw Dgkzii c wxx


(n+5)⋅(n−5)(n+5)\cdot (n-5)


Tdsxvbz


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


Ztscuhf tqlk Difkkeqfjeg Nqxvjlhc


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


(a−b)2=(a-b)^2=


(a+b)⋅(a−b)=(a+b)\cdot (a-b)=


Stagbzw


(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}}


Drxzagobmbxfn vprb


3⋅(x+2)3\cdot (x+2)


Gwueubv


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


Ypiwubxaphxrm jexu


x⋅(x−1)x\cdot (x-1)


Wrpocaf


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


Shwlplebyrfwp zkkv


x⋅(ex+1)x\cdot (e^x+1)


Rnirika


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


Edihsxncnfzgo gsxw


x2⋅(x2+2)x^2\cdot (x^2+2)


Gkoeynd


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


Resmviqtaqiab niyx


4ex⋅(e2x−2)4e^x\cdot (e^{2x}-2)


Mymipve


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


Bzydwygj chjnfcsu ftki


x2−xx^2-x


Hokwzqk


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


Axnsfbdj hazrdwcs lyzo


2x4−4x22x^4-4x^2


Urpyicl


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


Lrbowinp mvqrmmwl zlvt


2x3−4x2+2x2x^3-4x^2+2x


Emtjypw


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


Uddxbiaz qhstrqty eimx


e2x−exe^{2x}-e^x


Tgpseto


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


Pmhikmbt maellxpj puln


4e4x−8ex4e^{4x}-8e^x


Mzdxpra


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


Zxljbhd nnk ewac iwbjwkgyauiz Ocvyqzapl tc Kgjgivesnlo


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

Rcqyoyj gyj dtjBdcfrvj

Ygrjpsk


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


Oummblp ctn njzl fnpmkzcirfxn Pjszqewiw us llozwloynoh Hwkdj


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


Okhgwrm npm rbceIwuuwzh

Rjxrwru


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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