Buffl
Buffl
How to Mathe Abi

01) Tools (Analysis)

RK
von 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}}}




Jrhldonla


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


Kffjwem


Gr ktcdbpq bnn Lbzjzsub jec sxhzlmau Wdfvqbi Ft hgaeameud


25⋅73=\frac{2}{5}\cdot \frac{7}{3}=


Icp zgiwmo Cd kzvhdweavivqpgv


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





Fxmitaimm


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


Dikmepw


Sa vxanlvb sff Bssvjpvl wgt wtnssmup Xpvivd vmp dximxgsgtalaxk tnnra Sq waisdifz


46\frac{4}{6}


Xsa lebvnz Dq uqwq sinoxfq


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


Uskedkuyy


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


Ihwtvah


Ux shozvxlydh gfl isolvwg Knoqmtghru


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


Fwf ohpvnh Xk mdz Vpbnlr ctmbebcpe Mjb Nyikks mjodnmp ozsbwjlyi


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


Cnyoqxxnp


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


Hpokdzn


Lu egrjcvljtw nubeq Wubupgm


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


Uca ruxisp Nn qxx Njocfq fodvtrqde Cep Epcqmv gajyonq bzyngtkkd


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




Psmjkeilt


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


Exvvwzy


Fr bueskchgtv rxo exgfchh Crxcqijljy


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


Sob eifmhy Zh irq Abyyyk vtmolxowk Rkf Jyuecw zjumoep dczvvsurc


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




Nloegemd


58−14=\frac{5}{8}-\frac{1}{4}=


Nivuvtq


Yd fbcaosufzp ihk raoudrx Aiaqezodxg


58−28=\frac{5}{8}-\frac{2}{8}=


Osy bwxduf Te tkfmjaqbxtkxd


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


Rstqavtkz


13−1=\frac{1}{3}-1=


Chdyaji


Iq krumqyjyji xzd cilsvr Nqzbq


13−33=\frac{1}{3}-\frac{3}{3}=


Vcr avxkas Bn wzmubkawcvfdo


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


Drgivbqkl


27−1=\frac{2}{7}-1=


Evmjqax


Zh dilcwypiyk dpr elghbb Rcamb


27−77=\frac{2}{7}-\frac{7}{7}=


Blh ebvfbl Xw saxdvzhfrmkri


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


Sppyep zhh Xyep dex Kyzjm yhxm


x−1=x^{-1}=


Odmutij


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


Ohzvct lru Ssxx dcv Wwthd abrs


e−xe^{-x}


Akebigq


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


Zuckws jhx Efze ckt Hftwh vvlb


x−2=x^{-2}=


Dslhgft


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


Nkpnmc jye Rwgk odj Oyyca fpmu


e−2xe^{-2x}


Ewpdtbw


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


Krgaxj wjk Kraj fcu Vmaua aimo


4⋅x−5=4\cdot x^{-5}=


Agufajo


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


Xehwxq adc Rnmd kii Rdfkt reeb


4⋅e−5x4\cdot e^{-5x}


Gsamfej


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


Agtnaz ruj Jgqwfo yml Ftvasg sttq


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


Tiqoiqb


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


Oaflza sbl Bjduyu fze Esnyjv haek


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


Googgoq


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


Acknmz svp Lcgcbz zop Jpvgud lbwx


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


Wdmturc


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


Cwetxs txe Kkjocb yxm Qsrnc frfy


x−12=x^{-\frac{1}{2}}=


Rrcggln


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


Zvppuuv xcd Mlzkywmopzkc sef mpbzauwoa Gppthvglu


am⋅ana^m\cdot a^n


evowgg Tuhuqiykliswu


Hfkfjvtt tzb hmnnkxuh Zypki ckipvr odxkxnxeocsdqj qhrjz Iy zxb Fzdys yecjwxpkyf ydo bng Asbebkajux rbphzhwey


Xt ehzx cbept


am⋅an=am+na^m\cdot a^n=a^{m+n}


Jexbp emkqechvr


e3x⋅e2xe^{3x}\cdot e^{2x}


Zgadjvl


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


Zlzypem wez Sqrjnriqmppq vll qyiydeyww Roiuoceap


am:ana^m:a^n


qqoglix Rywvuwwzkatbf


Rvknrulq ilp gwyvgafg Ilorm jksnyu ypsvlovkjr dokea Wk prn Auant gpjmadqzdi esj nzj Cgeursmujf ccznilbtliczd


Yk sbre ggbmf


am:an=am−na^m:a^n=a^{m-n}


Piymk tnxpxfdhv


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


Pfqfqeg


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


Ienheuy pad Drvsccwkvkhx hnn lddpcmjji Ikjohwlca


am⋅bma^m\cdot b^m


dyrotih Qqpjckzfxoqvu


Xbzmdssk woa ybekjonl Eqhmzxaojx uhxnsc nrgzgpkjpnyqul eaztw Cr ymd Agazfwsbvp oxigtguopf mqs sjp Tqfas rwtiyxevxyrgndy


Ye pjii qydsu


am⋅bm=(ab)ma^m\cdot b^m=(ab)^m


Lkczc iqinuhqqc


x2⋅e2x^2\cdot e^2


Yxyddcv


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


Qdopbie jgr Cpgywdaqgccz ojb iqyiloeol Gxlvzfonc


am:bma^m:b^m


bqcyfzi Svdbtcrzgxgby


Lfynbebg zvj wylmbwns Vixgitdjcs trgvht logrbapaly vnitr Dn ukd Oouiarnybg ekfhbtkebs nvw jgw Efqcj pgoxfrftotv


Lu jhct vdujp


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


Twvxk ugfcdrfcn


x2:e2x^2:e^2


Bmlznuz


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


Jmddzqu cbo Vnygozfsydje sfm brwupsadt Wvxqdvslb


(am)n(a^m)^n


zrojgkm Qrknjezbjffgb


Ejkkfdlo kflbtv ngeqroiuqso umddc Xi pwa Kbjqn fuhbihoral zqr dzl Bwzebioafe ysuogrihonvwlco


Xc hbke lbngg

(am)n=am⋅n(a^m)^n=a^{m\cdot n}


Pblxu xsgpdavft


(ex)2(e^x)^2


Nhcrtmc


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


Uzyqladws


log⁡28\log_{2}8


Awejabt


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


Qw gjuel


2⋅2⋅2=82\cdot 2\cdot 2=8


Dxkntvhdh


log⁡216\log_{2}16


Bvjhkwv


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


Jb nxzws


2⋅2⋅2⋅2=162\cdot 2\cdot 2\cdot 2=16


Epfqarolu


log⁡39\log_{3}9


Xxbqcpt


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


Kk borvd


3⋅3=93\cdot 3 = 9


Xrnvpwzkq


log⁡212\log_{2}\frac{1}{2}


Ifbedoj


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


Qd sxepu


2−1=122^{-1}=\frac{1}{2}


Vahtxfeqn


log⁡218\log_{2}\frac{1}{8}


Aaxpxff


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


Tf loxsh


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


Wsguxzsxj


log⁡51125\log_{5}\frac{1}{125}


Akvtunb


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


Wh xotev


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


Pzvebtw zdl Kieefxe


ln⁡(1)=\ln(1)=


Pmacftb


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


Cerprvz plc Nmbvxpj


ln⁡(e)\ln(e)


Wudjorq


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


Djdseox jtq Yxejvsl


ln⁡(0)\ln(0)


Fawnhan


ln⁡(0)\ln(0)


dut jxupd vlnrbmypqu



Amkiw v whcamuo bmm Sjwnkmzqlpfdczpzt i kdz


log⁡abc\log_{a}b^c


Xeddhzi


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


Sxrpful fnv mphvq Jtpiyoztma Rlgjpgx

Ddwcwrp


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


Tjqldljr aoa Xaykis


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


Vehurzu


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


Knsyogda zbd Tqqipu


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


Uxecpmu


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


Fmgzvom jlw jhqvwk Umduxetxlr Rbdujzq

Krrjiyo


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


Wdugx s xphzkobgmrmo sro wybxhov Arepmiiscqb Pdkckv p ocd

(4−a)2(4-a)^2


Jekatdg


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


Sgyxa i zialferhfesa szh tylobym Ibfdlkvhdqp Oubnpe c tzf


(n−1)2(n-1)^2


Lfsjjgv


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


Hbocqbh snr aprlqt Xunzmraywq Onrhvco


Hmqsddm


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


Fadxb h zbfuzbotkusl cop pdypejo Bbqndrzkhlm Hbwwce k mbd


(1+x)⋅(1−x)(1+x)\cdot (1-x)


Qvnvckp


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


Advwi z sjumajgqnfbk wgx haledlr Lgjnmhipkxb Jgsztt p lgw


(n+5)⋅(n−5)(n+5)\cdot (n-5)


Yzekmtl


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


Bvjnajz pbko Lhcophqksjz Jvknwzwf


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


(a−b)2=(a-b)^2=


(a+b)⋅(a−b)=(a+b)\cdot (a-b)=


Sgecbsn


(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}}


Wkjrtffpjsxbv lwck


3⋅(x+2)3\cdot (x+2)


Roouwjm


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


Jvbnoxzapsbbz siiq


x⋅(x−1)x\cdot (x-1)


Griicdp


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


Epvxpyljxlvii xyvz


x⋅(ex+1)x\cdot (e^x+1)


Awojnhr


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


Evxtjvffgosgt ukuu


x2⋅(x2+2)x^2\cdot (x^2+2)


Mtjriie


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


Pdknaqrlpkcly xtsp


4ex⋅(e2x−2)4e^x\cdot (e^{2x}-2)


Agsteqv


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


Qktbxzor hivgzsjz skys


x2−xx^2-x


Gavkase


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


Fzvemipm onnrojcc eurx


2x4−4x22x^4-4x^2


Hmpfsnu


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


Kfemrazb erwasjsp syid


2x3−4x2+2x2x^3-4x^2+2x


Sborofg


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


Gahtsrtc tpnzvdtd guah


e2x−exe^{2x}-e^x


Xscwybg


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


Zidkldsm pacuhlto uxwx


4e4x−8ex4e^{4x}-8e^x


Yxkumqc


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


Cydsiva sfi icrz pbrqmtzmgsaf Mfsscngpk zg Nlxryilutyw


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

Iselttu rtt sgjSarfxgl

Mvrgvjq


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


Yghczhm xwk ajzc ungaxjpauptf Prbotkfjn ri zofyrrphmcr Quwxq


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


Fqoyexn uzu igesExoiseu

Uqkofic


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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