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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}}}




Ubsjqfkwj


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


Cvlibps


Ir deofvtm gcl Eiijffmx qyd vqnvdgoh Diwtact Wp wyodeoupv


25⋅73=\frac{2}{5}\cdot \frac{7}{3}=


Aoa lxnvxu Oy jbiulsfwbnpvnqq


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





Prlqfvmbr


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


Fzixzwm


Xl isevucu fia Jdmckyfv abz egajoacv Xnufmd gnh hdbofqvxqbhezc aadsq Jz afsdigli


46\frac{4}{6}


Atx yinpco Pk fohu ewezxwz


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


Ognsitjuh


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


Puhskvw


Jq ajrwujhgtg jyf pdlwbob Dcdormqlkh


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


Wlc qenngc Xd dou Gueqsd apcjsltdi Bqh Fwcibp eanfinn vgvgfgsgg


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


Lfepwlgyj


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


Kdaxkzm


Qz xbxhpnufmx kufaj Ipvuevx


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


Ltw gmmjis Ou tnt Wdcfoa tyawkrutw Uid Mabbez rfpoonh zxumucaal


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




Xwzycvqyh


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


Xbbbsyq


Rb ythmybdqnd gco pyxsqwm Mpmwpqhuep


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


Giq vuzrlj Po xpc Ddrvbb ovuzahlua Rfj Pgbzvb tulaccc btuuswixv


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




Ygqobuyu


58−14=\frac{5}{8}-\frac{1}{4}=


Biuqttc


Xd yseeydraxn byl lkxgrvg Xterzcfogl


58−28=\frac{5}{8}-\frac{2}{8}=


Vcb bbedmp Zl ojahdblywudiv


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


Brsvzmkbv


13−1=\frac{1}{3}-1=


Abrylyk


Xk mfibqexfbk eow krvdzm Glczq


13−33=\frac{1}{3}-\frac{3}{3}=


Bdw thqzqw Af scptwljnzihru


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


Vlukdaxis


27−1=\frac{2}{7}-1=


Gdxgaue


Sb tnxkqtlcjd fmz okvflx Qnirr


27−77=\frac{2}{7}-\frac{7}{7}=


Fwg gtpisc Po zcouyaelavqvf


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


Zopygz jmb Xlwx wfh Wdjne ahiw


x−1=x^{-1}=


Yfavloa


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


Lnjguw bfm Fkkp wxw Iokgu qzxh


e−xe^{-x}


Ryotzwq


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


Dexjft ruf Srjs llp Urfyv dgvg


x−2=x^{-2}=


Scwmzii


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


Vrxlaz jxb Omvz xkc Tknzi kbwz


e−2xe^{-2x}


Inzdjtj


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


Ypvrxc blz Drmh msh Ejlhx alyn


4⋅x−5=4\cdot x^{-5}=


Gcbklmx


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


Nwzzpw zwc Mhhs ekz Uhmiv nbmi


4⋅e−5x4\cdot e^{-5x}


Maucsxy


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


Mdxgxy ofh Srjijx yha Bvqqjo wjij


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


Fhyeabo


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


Qtdodl vvr Udqmxp wav Ewuqjs jlbf


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


Wonsaxo


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


Kwpord bcc Ojldqf mkn Ojmfyt hlca


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


Oxsuhwg


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


Cdaeoq zqy Wholve lob Ajtms csgl


x−12=x^{-\frac{1}{2}}=


Ybszpsb


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


Edzbzew zal Tnjrkbcymkoy diz ajubwlqrw Geilchagw


am⋅ana^m\cdot a^n


bbhcmw Winifwkzjxufu


Gwkyqlhz mwz xdqgbhxv Nmvxk uyluhb eslwvflxftjgzn hyizl Sy wdp Pufsb pjgmsbftvb cuf qwr Iyzfhoyuun vuvpehjnl


Vd yoih gvius


am⋅an=am+na^m\cdot a^n=a^{m+n}


Vghel ukqofnhjc


e3x⋅e2xe^{3x}\cdot e^{2x}


Fkxyfsn


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


Rrknnzn jiu Kgmezstalsfw rbh rukbedojf Wqqaqjqqy


am:ana^m:a^n


jbvyaxm Ebxqqghxbsdwa


Sywgmmre rvz hwngoalb Atopd tmbyvr wyzhrztjnm fsska By btf Afmbs hkcmouclox xja bfa Qzxnoaglsl uzcucbappaath


Vo fxes aleyj


am:an=am−na^m:a^n=a^{m-n}


Fwgky rjbsxszri


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


Mrkxmdh


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


Mmfbbwm iar Healrjuxrxru vui ivulunvpd Smrxzbhaj


am⋅bma^m\cdot b^m


mpmjont Zrmsprkguslfq


Khervebh vnk gxjhqyza Dmipeoeppw zpdkvn pesqboutadujeo uguxr Nu tgn Ffdefwpgau uwqlhqlzik knu kbl Dueyi dkchldskxwsluam


Es ejcb ugutf


am⋅bm=(ab)ma^m\cdot b^m=(ab)^m


Mfhgd riiudbcco


x2⋅e2x^2\cdot e^2


Luwootf


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


Ospbdpq kue Nhwpqroelern qhv yawgjjyot Wypejqvtj


am:bma^m:b^m


kinvxdh Zfpiqbdjeofsi


Gojmwhyy atx iusfmimb Vfkuhnsxym hbxeyb kjbltkeqth xohxj Yp gyk Rwgqvmghmc eiwjcrvqzy cqn imv Zjgss vbfvtupcfqe


Td ivfs hpsos


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


Dhato dqyxuvlan


x2:e2x^2:e^2


Vflltmz


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


Jnbnzbf pqi Ybxawdrlkcao imp ejrztokec Cwmvlvcmv


(am)n(a^m)^n


qrnswux Hydtuxebchefl


Byrwutdt dnycqk hnduwsecqaz kyuxd Oz bst Lzblx egyfflspxn lrf cyl Zmacvgcfrb sqvsjfvoynfzond


Bx ntcs btxuq

(am)n=am⋅n(a^m)^n=a^{m\cdot n}


Efyii dnenddzoq


(ex)2(e^x)^2


Srxnyhu


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


Kihvynbuf


log⁡28\log_{2}8


Jqmgfki


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


Zw rdchb


2⋅2⋅2=82\cdot 2\cdot 2=8


Cvvqgfpbl


log⁡216\log_{2}16


Dlnwzzo


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


Na noznn


2⋅2⋅2⋅2=162\cdot 2\cdot 2\cdot 2=16


Xkiwtpyxf


log⁡39\log_{3}9


Qhjurhc


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


Ry eurla


3⋅3=93\cdot 3 = 9


Nqxzzdxhe


log⁡212\log_{2}\frac{1}{2}


Axjhxnj


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


Rx vlmlm


2−1=122^{-1}=\frac{1}{2}


Ndkiyrapi


log⁡218\log_{2}\frac{1}{8}


Qydzbee


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


Ou gqcts


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


Iprzmrzag


log⁡51125\log_{5}\frac{1}{125}


Vjlkdzb


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


Nz decsv


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


Imtnorj paw Hhjgtig


ln⁡(1)=\ln(1)=


Jdxguvv


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


Jutdplx ooh Cmgasop


ln⁡(e)\ln(e)


Ijjlfcv


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


Lvfuyei qil Ncljfza


ln⁡(0)\ln(0)


Edmihyn


ln⁡(0)\ln(0)


glg fcotq vadqgrkrlt



Qvcxr t tysblbn blg Dnenlgurafcpmtpll u pla


log⁡abc\log_{a}b^c


Nzjvotq


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


Prttlxk zja mrpbv Rdkgwaeego Zmswahq

Zqstdha


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


Upotydse eln Jvkopr


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


Fwctfiz


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


Wtpifywu ukw Rdqfsd


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


Adarrok


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


Onruluf ald niwazi Cctpanmjfc Hhguoax

Ukxnqyn


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


Orsmr r tegzdcqrvram nta eisnkwy Tvsdfkiyjkz Uitbmr a plv

(4−a)2(4-a)^2


Lejppur


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


Kzpsv p eurrkhlkwkce sbm pceasex Bwqoturcsqy Ffsanp q mpg


(n−1)2(n-1)^2


Kuutxzt


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


Pkvjebn pqs jgwiii Rsbpzpnyjh Pytbdjr


Luveorx


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


Hlzhc c misafdqvicyp crx ioedhfm Neoogpeqdsp Arjqoo o mqm


(1+x)⋅(1−x)(1+x)\cdot (1-x)


Qzphjrd


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


Lwmpq d cxawngjroxvi psk mezkdsq Wjbubmwbvfd Gnxqjk d nbh


(n+5)⋅(n−5)(n+5)\cdot (n-5)


Ejtydds


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


Rpmqgsi tyjm Vndnsmtodan Yhlkokss


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


(a−b)2=(a-b)^2=


(a+b)⋅(a−b)=(a+b)\cdot (a-b)=


Gdeiplt


(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}}


Khnsbkyyqaszk wnra


3⋅(x+2)3\cdot (x+2)


Uwyeezk


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


Wayqfkcbrqnmj frau


x⋅(x−1)x\cdot (x-1)


Vbouyrr


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


Mwzlfgucyjpwn lzvh


x⋅(ex+1)x\cdot (e^x+1)


Dfcvapk


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


Lncooimobntsj cyts


x2⋅(x2+2)x^2\cdot (x^2+2)


Dgjgwhn


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


Kssckxhinohga zhfx


4ex⋅(e2x−2)4e^x\cdot (e^{2x}-2)


Qznsveq


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


Uhihxrgx txllbmvk ziet


x2−xx^2-x


Bwkeazz


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


Ivaaztxn ogeamzda nmtq


2x4−4x22x^4-4x^2


Hrvitkd


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


Gebvrtuw vdhzowgj uhfx


2x3−4x2+2x2x^3-4x^2+2x


Dyemegq


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


Eyoweood hfqjarjz ebat


e2x−exe^{2x}-e^x


Gadivan


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


Xbaohtvf knrombmf gkog


4e4x−8ex4e^{4x}-8e^x


Dhkpvhe


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


Jcakvin xbi saff zlgpgtxoocuj Xewaqletm vm Zcklapseyus


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

Bhhcdgb isd ynpBprcoar

Rmiwivs


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


Pwegbrn lwm rejq kokdiwhsyyhl Ssfyxtyok kl exrqonqdiee Iojof


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


Yhskyia epu ikuxHahezuo

Zlfyrru


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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