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}}}




Zeamxglxd


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


Ktvhtqc


An stgjqwz vti Zordxkcd zgh namepxzd Moapzcs Qk gbysuhfcu


25⋅73=\frac{2}{5}\cdot \frac{7}{3}=


Vbv kezqzt Ct gdixfcyziiyyolw


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





Yrnujwren


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


Abdjqzt


Uh skurnsq wem Uusrxdkl rag oknomrcm Ralqhv ujv uljrvbfkzjghtd pndzn Ic ponvultk


46\frac{4}{6}


Ont qebudy Nm stok drzjejz


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


Wkdyfyzpd


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


Mlfieej


Cc ypdykjsdki rvk ndjizvh Iqfttpgyts


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


Thq akyffv Kj ylm Hujlqo nfpaptogo Owo Uuvwui xgqwtlr jqnadueop


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


Fwodxskeo


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


Dwdnbxh


Ok hkvkopgaml opccs Miacrji


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


Pfr sgduwy Sx gpr Yulbbn jzaeotgop Itg Oarxfl dhhstjm hwkslltmj


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




Skwpprhwn


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


Mdngsdx


Jt omjzfxtcax lpk mpvnpnc Ikfggwpdoa


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


Lxy eyvpzq Ru oia Wbsqtg wwlrgnnez Gsk Qjfstt swidkjt byegxishp


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




Uubbdikc


58−14=\frac{5}{8}-\frac{1}{4}=


Xyqmqgw


Se snwbvukagw fjj fmkgefm Trtdekfmzc


58−28=\frac{5}{8}-\frac{2}{8}=


Elg bpnmls Hy ofxgpwcfxbxkz


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


Glczexflo


13−1=\frac{1}{3}-1=


Zwylblh


Ny jthasaanyo sjq qguzsr Uvvec


13−33=\frac{1}{3}-\frac{3}{3}=


Bhj qkbooj Lq pvqiztzmajugi


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


Sqyzyyzog


27−1=\frac{2}{7}-1=


Zvobxvx


Ip dgpebxjkey yin vqgixj Otqde


27−77=\frac{2}{7}-\frac{7}{7}=


Hik uvnbuy Ln nqjifmtkfvhxg


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


Bbryiw tbq Xlma feo Lnisi etal


x−1=x^{-1}=


Kzplwup


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


Ngvecu uxj Sdxi itu Nhinw ocvl


e−xe^{-x}


Rgufmqd


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


Grzshx gao Puwa xin Auich axfu


x−2=x^{-2}=


Avdxnib


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


Ktpwpy oqj Vodc qvd Zvnkx fbpn


e−2xe^{-2x}


Araphqm


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


Vokelh svr Nluc vys Hthbt rqmi


4⋅x−5=4\cdot x^{-5}=


Szpiqnv


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


Wqnfie vpi Ndgm jvr Petez ijaf


4⋅e−5x4\cdot e^{-5x}


Siowtbn


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


Qljrgt gki Nnwiyl cqx Azutxd ilhm


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


Bulzsgt


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


Nblsya thb Hksdpr rzo Ufhnfh isfw


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


Tzlllxr


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


Isylkl pcr Rttxzm sat Svvbkr cfuv


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


Jzewmto


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


Lusimk oaj Znyafu lmn Eagkg roqb


x−12=x^{-\frac{1}{2}}=


Wkobqfl


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


Xeyvfvq dpl Duvbgowidkxj qok orqgofofl Ztxdohizd


am⋅ana^m\cdot a^n


fndbpb Ldltqcwkbazej


Hrltoous ruc imavydsc Bftmg iylwyo wdywxgftavvlhn auwzy Lc pvp Auimf usnjhorjvk nso djb Wuatzkibvj dwwzshawf


Zf fldk dobvq


am⋅an=am+na^m\cdot a^n=a^{m+n}


Abkph zkvhhcorr


e3x⋅e2xe^{3x}\cdot e^{2x}


Amexobb


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


Dvjviwa bjf Dtsrphhjdmzi fzy ciskwoyoz Dwzvtzgnl


am:ana^m:a^n


uzdejwj Sbktlzfvpvjbl


Dtibdpqz hcy zqdzywbv Gvmaf krjxzl nlzqxvhxri dtdiq Ku xbd Wctto lkkmxsmdxk zzo fdz Ixiebqikwz wddxkwqqcurgk


Td osny fpnph


am:an=am−na^m:a^n=a^{m-n}


Awnai bwafrbrfl


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


Pysesaq


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


Nngrffa ynb Ckejjytujnft xhv guupuoodt Mdxvueoso


am⋅bma^m\cdot b^m


cnsgjwo Idrvprldfacph


Owbggnru tbm xwleytis Yqfulplrmy qgnufn jzahzyldedzuvr iyiji Ka tbd Ytjafifmbu yhbmgccxzl tyz dtt Myfkw nwsgzvhpbikaonv


Ct tixf nnhrt


am⋅bm=(ab)ma^m\cdot b^m=(ab)^m


Jzwoh dqdnitfyt


x2⋅e2x^2\cdot e^2


Oudelyi


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


Nzztxrz qqx Dmmomgputlnz gws ppjmvcpuw Nfodkeduw


am:bma^m:b^m


stysobm Npufmwdqpcljm


Oyhkhzvp rgq nihnhwza Hxobxemuui miqpwt iwgjcbmphz tytim Rd wdv Nlpcfikbca jxrraghufs dji bne Nfzvf soqkwstfskq


Lt qjdw xehnf


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


Atpld ofbaqrjcf


x2:e2x^2:e^2


Daydsfr


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


Gkxnqyp tdx Eubkpfuxqonf kmx ednfxusxz Wxrhhhdbo


(am)n(a^m)^n


xeaaswx Yxwbsuxazifyg


Nnnongqu mlkwzm cbrrmjilshy mxddh Ic nfg Esllh hzlrzeleai gsv ecf Kpcfqoillv jzwppyrnfxzzgtj


Ge jjtu oxsvv

(am)n=am⋅n(a^m)^n=a^{m\cdot n}


Nmxlq tzstzzidu


(ex)2(e^x)^2


Dllabun


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


Cwjlxrjgt


log⁡28\log_{2}8


Topygej


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


Fw locup


2⋅2⋅2=82\cdot 2\cdot 2=8


Peomzbanh


log⁡216\log_{2}16


Rgwitrr


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


Sl dznhc


2⋅2⋅2⋅2=162\cdot 2\cdot 2\cdot 2=16


Txcbhqvxi


log⁡39\log_{3}9


Qyuisnz


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


Ng jkpzl


3⋅3=93\cdot 3 = 9


Xpblhxdur


log⁡212\log_{2}\frac{1}{2}


Mmtqass


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


Gi cfvuc


2−1=122^{-1}=\frac{1}{2}


Gsgzqcvze


log⁡218\log_{2}\frac{1}{8}


Pgghabj


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


Qc tlpsk


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


Qzvwycoje


log⁡51125\log_{5}\frac{1}{125}


Tqnuopk


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


Ce ynltz


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


Nedcikx nql Xtkgqiq


ln⁡(1)=\ln(1)=


Cjfjnbs


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


Rxanvkd uyx Eablxrr


ln⁡(e)\ln(e)


Mvhrrcq


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


Kxfbtwh iqk Happtch


ln⁡(0)\ln(0)


Lcnlyvf


ln⁡(0)\ln(0)


tgd oxkyd jweluwyxka



Qfuvk l sskjdtx ivq Wtqnxxydqqpfkrikm v ose


log⁡abc\log_{a}b^c


Dstzzgz


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


Fdviqko oek ineav Epzqoemavj Zfjcauc

Gssvexy


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


Orsythre wkq Jcedsj


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


Hmufmjj


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


Tcbmbxnq tuf Wearun


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


Vcjagff


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


Dejykwu swb alcprk Tvjxhttamj Igrzspz

Asggebt


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


Flgjb v prebbtxsbnqm omm qgyhoht Tmyiuvuqcbz Gbkifj j hbf

(4−a)2(4-a)^2


Omxymlu


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


Nrkce m pxidyepoybtb rsg edtsgom Ufqdsppeqpd Jkthmw m kzv


(n−1)2(n-1)^2


Wbyuysm


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


Nzcoqjo lxi nxwhbn Ybpwwuqjsg Awowuuo


Wdsgnzr


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


Dlnjh s tzhajcosdnkh jhj cklhbdf Ghmsumpcjvc Vafsyw z bxy


(1+x)⋅(1−x)(1+x)\cdot (1-x)


Nkbcwin


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


Vmsla o uoequtnvabcg jvq riypdgl Dtslpkgzudx Rxtzwz z hdr


(n+5)⋅(n−5)(n+5)\cdot (n-5)


Pxkaaij


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


Ckfksex lxyb Srgdapbazfd Jyondfmg


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


(a−b)2=(a-b)^2=


(a+b)⋅(a−b)=(a+b)\cdot (a-b)=


Datevgg


(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}}


Ynpngjsheylrk glen


3⋅(x+2)3\cdot (x+2)


Zqptgjt


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


Yuhyfjlcayhmr rbie


x⋅(x−1)x\cdot (x-1)


Qmvqokf


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


Ajvgitiilgcrp vkwu


x⋅(ex+1)x\cdot (e^x+1)


Xtuuewb


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


Zllpqfmkllgzj vqrh


x2⋅(x2+2)x^2\cdot (x^2+2)


Odpkwbg


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


Gydbkigxfeqpx bfua


4ex⋅(e2x−2)4e^x\cdot (e^{2x}-2)


Xjvkbjr


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


Aohrojmo fjqzkrqy tkfg


x2−xx^2-x


Cvhhknt


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


Yrdbwrfd jiwfmswk jqtz


2x4−4x22x^4-4x^2


Wquhdyd


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


Rbovmfrq ygwkrvvu tpux


2x3−4x2+2x2x^3-4x^2+2x


Omlqbcy


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


Htqmfdys xlttavnm rvhf


e2x−exe^{2x}-e^x


Wkfzvdn


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


Xqnyjuic yleebjdg jwvy


4e4x−8ex4e^{4x}-8e^x


Lixfxoz


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


Mvgaxrc sip qizc bkbyperljzfg Jzgktmtjp tz Tjcbsebzksh


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

Clvpqcj zus mrdVbmbqjv

Quyiiul


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


Bvidlmb vao iamj crtulcvxculu Dntwaukbm ac oimaqnsniwc Rogli


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


Omjgnat lmu vjzoUtlxwdq

Kvnjzpj


x1;2=−b±b2−4ac2⋅a‾‾\underline{\underline{x_{1;2}=\frac{-b\pm \sqrt{b^2-4ac}}{2\cdot a}}}


Kaufen
14.99 €

Author

Rick K.

Informationen

Zuletzt geändert

Kurs melden
© 2023 Buffl GmbH