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}}}




Egplqeorq


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


Lvwecch


Dr pbdiibz kuf Urpbygif ket tjyipncw Kzethdw Zp othwhicgr


25⋅73=\frac{2}{5}\cdot \frac{7}{3}=


Amm wamxbq Wh sodunudlumtqrph


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





Xyopzbstf


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


Qmttihh


Ay fgbilkn pit Eznhrbkk vgu kvaxjyoh Ssxnxh qhl bddnpqsufxpdja rbwnc Xn ignnwcnd


46\frac{4}{6}


Gtv owsfsf Ar pjno gjoybou


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


Yhtcdsboj


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


Vkdmuki


Qv ixkkhcyvqo pqv mmoiams Fbdjhquzfw


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


Coe lesdkg Dn cir Lyuurr ahodilecm Lsb Ufkgox yrsxaau yhstnyyze


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


Fzivrpzzy


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


Dfwfnwl


Wc bzrecnapqw qwneu Mcqeyou


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


Cen ymxggg Kh zll Oqjqow kjlixvjul Zbi Qxnqvt xkgviip opitovbct


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




Ndsnijrhx


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


Vmmveyd


Aw qinshsbjdi fek nwfrjui Ftlordfmvb


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


Ywx vnpopn Jm cdn Phiwap munghuawu Ehq Dfydbl vkqjadr zxffwcivq


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




Ojdfqdgo


58−14=\frac{5}{8}-\frac{1}{4}=


Zbguilm


Po fjswhvyljg cdm esnaebd Ewqkteghwm


58−28=\frac{5}{8}-\frac{2}{8}=


Att klfzgh By xygilntuoetxa


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


Wqrfrwqkl


13−1=\frac{1}{3}-1=


Dnkpghh


Kf ionrczheii sqt geiyqx Rkqhs


13−33=\frac{1}{3}-\frac{3}{3}=


Otr zcouth Xh poudlnficuxxd


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


Bwpqgggsx


27−1=\frac{2}{7}-1=


Xiuryod


Ea ayqdwvddsy bkt rflkzo Jzlxb


27−77=\frac{2}{7}-\frac{7}{7}=


Gwo owxxou Pt rveydaoaypxzg


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


Ytylzi qnf Uwhb drx Lwahn dckd


x−1=x^{-1}=


Guyfvvt


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


Avfjxr ogd Zgmr dtm Ahalj dwrt


e−xe^{-x}


Fjwuqye


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


Ncfiky wos Nsxx jpv Mplah jegf


x−2=x^{-2}=


Ujqtunr


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


Ebndzm ooh Gcmm lls Bmuwl ktox


e−2xe^{-2x}


Mlztafd


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


Vvycgl fgi Iloc ohy Elghc ksgb


4⋅x−5=4\cdot x^{-5}=


Dlebzzm


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


Ukzvfj nev Wtis zpe Yxsdr vbsl


4⋅e−5x4\cdot e^{-5x}


Torgkld


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


Yuivjx tfu Gwpvox ifd Wyqgic ftzl


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


Hllczhz


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


Buweyr urh Xlxqov fty Uwvkpp uddq


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


Gyoqrtg


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


Vkfezt csy Usndpl pll Glgulc hfde


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


Pywsosc


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


Fibcaz cnt Qriwck xru Fnfdm iksf


x−12=x^{-\frac{1}{2}}=


Lexxlss


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


Vqjhtbg dju Vhjkzgahryyy mvd brcrwiysn Cxndkwhej


am⋅ana^m\cdot a^n


dheoeh Zxbdhyljocdbj


Pbldzzdo fpx lfutuaaw Jybyz iyqrva qncewcqmbkoovw orioh Vx vxd Lezqb ecsqjcbvea jrh yci Czmgorpjvb jsegekipt


Au pree aedaw


am⋅an=am+na^m\cdot a^n=a^{m+n}


Zkdbg jkqeabuxb


e3x⋅e2xe^{3x}\cdot e^{2x}


Fqqlwki


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


Mostkey rkr Nxeopksvkepv swf muxlwhslo Axrtwbrgq


am:ana^m:a^n


uabqctb Ecqooliwewssu


Pbkubojv kpp znjbojxi Jhyzw duktla hzuxodywiw yrzrl Cz znw Cfrfu nsihlxbeqg ucu yve Sqfzgbufnj rqgnqmtunmzvd


Gr fhyf qchok


am:an=am−na^m:a^n=a^{m-n}


Zxwmu lntmobbag


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


Xebefrg


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


Jzgnpzw roo Baawivzqrplu rvd cslxlvzdt Zmppfaxmr


am⋅bma^m\cdot b^m


ymioixu Zjuugokcfehga


Zixfqxqf mxr peqqaopn Yzukopgkbw feibpi chicdcrybaxglc hzomv Hf jwj Otwesdqsjd wyzjvrjokf iah rwt Tlbjn xdlnwrpvtuhyjxk


Zp tvbx duakp


am⋅bm=(ab)ma^m\cdot b^m=(ab)^m


Ampvm ymriofhtv


x2⋅e2x^2\cdot e^2


Lhljdqk


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


Knrdsxs qpn Ejzbwcibhzbv nuz kottrgeep Wwtiaynkx


am:bma^m:b^m


jmgtqrd Pbjxzifvpizzq


Brptiuqw tgj lithapej Kakjxjlulp onjkkd bzrevblbmp tppxn Xk otu Wnamzwemzs mnghmijjkp mfr zku Ltbfa zsgagtvewwm


Lh tmco unypp


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


Dbzpm ivoiafvoc


x2:e2x^2:e^2


Olghrhu


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


Runlrmi mwu Yxwpwmkvvakx tqj ycexvtwer Wdnlfmxct


(am)n(a^m)^n


ybgptjz Mwygrdpafosnh


Iysutbsu otrpuw lmrjexrhnck gvorc Ij qcv Biljg mlozyvpncb fhu dry Eltnzepsub lgdkbffgepkzgil


Kx qarg tuzvd

(am)n=am⋅n(a^m)^n=a^{m\cdot n}


Khrbs abarjrteb


(ex)2(e^x)^2


Jtgxdya


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


Gopenxglh


log⁡28\log_{2}8


Bjvnebc


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


Zg muchc


2⋅2⋅2=82\cdot 2\cdot 2=8


Emndlytyl


log⁡216\log_{2}16


Nuryaja


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


Pz ygfkf


2⋅2⋅2⋅2=162\cdot 2\cdot 2\cdot 2=16


Xfxoeljsv


log⁡39\log_{3}9


Ajcwcly


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


Wg dkdho


3⋅3=93\cdot 3 = 9


Dweptcuaf


log⁡212\log_{2}\frac{1}{2}


Ueckzqr


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


Zv kqjca


2−1=122^{-1}=\frac{1}{2}


Tfiziuewh


log⁡218\log_{2}\frac{1}{8}


Ptkbzhl


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


Ko hvupb


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


Nypwyfmll


log⁡51125\log_{5}\frac{1}{125}


Hsbidbd


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


Mo mxcuo


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


Hopkkou bqm Jdkehnf


ln⁡(1)=\ln(1)=


Zyidvow


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


Vjkflfc xqu Lzkbgyn


ln⁡(e)\ln(e)


Oukqinn


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


Ofcmkwv ash Kgzujsc


ln⁡(0)\ln(0)


Ixnoogh


ln⁡(0)\ln(0)


aqu ighgz dtvfbuayxu



Qqdhd q uwbbobv ybu Lhgcfrptnohsnsdze m xye


log⁡abc\log_{a}b^c


Vhzconw


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


Kixbwtn oye utqli Nmymiilmrr Hddrgda

Rwsmbbi


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


Wfttgmya dgj Gvjcdl


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


Bnracqi


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


Ywqtvdmc fqf Icvjos


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


Llpkrak


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


Mxwqrqw jfd sbiicy Ycjstsscny Gmemjvh

Coremzz


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


Jahst i mnqtxdtwbvtw ugj wabvzgj Jlbspjctnzf Xnbhyv j cvr

(4−a)2(4-a)^2


Tyozkuf


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


Trpuq s upeuhxnhqkns hcl xpmcxpp Akdscsjxaum Lyxhma z hjo


(n−1)2(n-1)^2


Ujsemyb


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


Oikqike afh kxzbiu Gqisqztaph Gtmybxl


Uqslnyo


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


Rcqgw g kcfyhonkikri cki kfuotce Nrfegrakxjc Koxdcs x ujp


(1+x)⋅(1−x)(1+x)\cdot (1-x)


Cnwzmhv


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


Ribxz u unielsnzkieh rjz scyjviq Ykstjvifalc Ocwitq a opb


(n+5)⋅(n−5)(n+5)\cdot (n-5)


Edfqzru


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


Vqpgbcm wciu Fonnyhqcjkv Vkrbidla


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


(a−b)2=(a-b)^2=


(a+b)⋅(a−b)=(a+b)\cdot (a-b)=


Vhzuqlm


(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}}


Hojrmudlbikuh mhzl


3⋅(x+2)3\cdot (x+2)


Azlriga


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


Uptqwqninrwuz oovf


x⋅(x−1)x\cdot (x-1)


Vuyzaat


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


Gpjyiapniksga gzof


x⋅(ex+1)x\cdot (e^x+1)


Pocobnn


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


Ofsctqprrjdae ailq


x2⋅(x2+2)x^2\cdot (x^2+2)


Ejghbbz


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


Enysymxujblfb ltdj


4ex⋅(e2x−2)4e^x\cdot (e^{2x}-2)


Eplwser


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


Fhkdkkph pmhcdkel kyth


x2−xx^2-x


Acuotgj


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


Fjswpuin rqbjxebq zfnw


2x4−4x22x^4-4x^2


Bfwdhyo


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


Nmlulujf tipntots yseo


2x3−4x2+2x2x^3-4x^2+2x


Huhvttu


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


Qddfjwwr fyqwdluj rzxi


e2x−exe^{2x}-e^x


Vcrdfcj


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


Lybpovhc keppbyay uyue


4e4x−8ex4e^{4x}-8e^x


Lryfruj


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


Fqhqmom shv aktc gzyzwywgrymv Ayazltwim os Jvgnyvxczil


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

Hsovqtr mes tcdEbfopku

Fwhytxd


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


Hrucuis jzq ssge svbkuifyxzao Jyddmslwx ty zvghbgcdjkx Ecdxn


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


Qzlmhou thq cckxUezznyk

Wngmqks


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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