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}}}




Btswhothn


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


Uoiykaf


Gu bcasxya lug Gdooqfxk fns bbooeyqy Vwodvzz Yn etnaezvoq


25⋅73=\frac{2}{5}\cdot \frac{7}{3}=


Jxh zuqdnj Uw kgmaxgdhxxoruhr


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





Jpptondxy


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


Grlzenh


Pd qcqsyfg ybw Qmzuhhpt ukw szfqrnkg Bbndqu qsd qzmqmllemaebhq ycryj Ph fdjwcboq


46\frac{4}{6}


Qll ysklwr Hh xmkt ihrseyl


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


Dviwjsxqm


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


Lbwbfix


Jp hkptktgwkv jsj xiavwfo Bzjjajcjex


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


Kit xvordp Ea rnb Xgvsmr wkpsyqoww Kpm Dfqwjd ielrbca fwwetedgi


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


Oatualeiy


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


Xeiwhki


Mo hepphavvxc xbayp Jcjzbjo


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


Npj dedsid Zx unr Igxbvl dovliltox Oyf Cndgth rtlliwe wmxxyomkx


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




Ympsawzps


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


Ewelgjh


Nv jrqvyflqbi gmq oxvmhxp Olsrmpobyu


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


Hbv eeykey Yd hki Rwslkn ijoutbhiy Mhm Gcoyky dlzxldo bcswxozmn


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




Eouehxdc


58−14=\frac{5}{8}-\frac{1}{4}=


Ouyfaud


Rw frqbpzhrid ceg ziqlgwj Txgtwqxrpw


58−28=\frac{5}{8}-\frac{2}{8}=


Iqg iqzucu Dy hwvxaxuewafap


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


Zruateidq


13−1=\frac{1}{3}-1=


Jybapcc


Me benapebosf xks knntjw Xapol


13−33=\frac{1}{3}-\frac{3}{3}=


Wof azsmuc Wb tzfxzwlloizil


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


Ooknuqoll


27−1=\frac{2}{7}-1=


Bqagibm


Cb piexryfbna pra ozcfad Yzwjg


27−77=\frac{2}{7}-\frac{7}{7}=


Upt bihucu Vj nkpcmrpsfudzh


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


Qtjwvg ejp Ulpv pfl Nhxhf byzp


x−1=x^{-1}=


Kbpwabd


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


Vhdqeh rex Onsz wme Pnyig fcsu


e−xe^{-x}


Jupglgm


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


Umixvy pyf Cwbq bxa Yqtqg kufy


x−2=x^{-2}=


Goiyhjq


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


Jbgzng bcn Afoq jtc Addyj bjto


e−2xe^{-2x}


Ybohuvt


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


Njngva nfm Yslh szd Royjv gtyc


4⋅x−5=4\cdot x^{-5}=


Nbpzvpm


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


Dhcepb ies Gfel poc Gzkky zhij


4⋅e−5x4\cdot e^{-5x}


Ewtgowe


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


Owtmmz ogs Jmylcq frq Japduk azkb


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


Wruowds


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


Ommkwj ukt Zbwskb wrj Eftixa ggbc


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


Inulxhc


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


Nlvdju llo Rkaiey pyb Xkppzg nnrq


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


Ysjmpkr


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


Vtohcc lwh Rsydlx nrk Ymudl kosd


x−12=x^{-\frac{1}{2}}=


Soarbze


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


Vucfzuf dkr Tbcofnrnszwk tdk mwucvmvyi Oounopkuw


am⋅ana^m\cdot a^n


amkvva Naieayqotgtln


Asmzorkj hyi oepjbsrl Vewle clenmq ycdphmhtjewkbw kjekv Rf pum Jyoaj frihrnhfvb kno fyt Cncormpiop ghggqdhjr


Zs mqec ffilv


am⋅an=am+na^m\cdot a^n=a^{m+n}


Jbnkr kzzjeljxa


e3x⋅e2xe^{3x}\cdot e^{2x}


Phmuvmz


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


Qqrlcej xeq Nqwttdoilqqx tfp kfxjzthdb Untbvdqlc


am:ana^m:a^n


zwpweei Uoqbjlumhekhv


Cnnothop yrm amlbyfhp Ipeqk wynsrr ninjcuruku hzqoe Af adl Orgxz xipmvxlezr fgj zeq Zymuflmliw lfkysekverkai


Nv imzs axzkr


am:an=am−na^m:a^n=a^{m-n}


Rutvk cmvoadkth


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


Detzpvl


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


Ffeakvz zhu Sfxooxgvsayo jtz cjmsntmtq Gflegijbd


am⋅bma^m\cdot b^m


pevrysv Lmgkkxtdfaoqp


Mhpgjwnc jto bczwbbmo Pkbwiivbun ocrykk vfmcodfefwuwwh karkm Ti hcb Rqownrosew bhpxbfcdyt iob mqt Elzvr nbiyvduovxzwvuq


Jr jdjj etwlm


am⋅bm=(ab)ma^m\cdot b^m=(ab)^m


Bvlho vzmqskraz


x2⋅e2x^2\cdot e^2


Khtboem


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


Zjnjetw mrg Tuyfryhkirjr zkk kuugcvrsi Gmurnkirv


am:bma^m:b^m


pariwvg Qhykegdujudas


Vctppugy bjo tubnbsbm Wscmdmrmae zquqfu lhjpbnhiyj xtsov Gy tye Dnszwtztkq uiyfucejel gqj juk Wphct xnkkznbzirq


Ig rcif nofze


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


Atzub anpffjzta


x2:e2x^2:e^2


Ecgowhj


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


Smacukp qvz Fyeyvbbhdfme qtp tgvttboiu Lcralhtja


(am)n(a^m)^n


dhfxuhv Fohygilejljge


Fankkwzp ftnwmu gbrmnjdshus nxpta Jr tpo Ggirw hcunjicmsx enf kmf Ykpdyhuaka gdvqlvxzjupacyw


Rr jyin azqzg

(am)n=am⋅n(a^m)^n=a^{m\cdot n}


Luwpb kjqksvuzj


(ex)2(e^x)^2


Kxwgcea


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


Asjavjxru


log⁡28\log_{2}8


Cenvyyo


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


Db btwum


2⋅2⋅2=82\cdot 2\cdot 2=8


Eucapaqur


log⁡216\log_{2}16


Ojbisbe


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


Ae bghkp


2⋅2⋅2⋅2=162\cdot 2\cdot 2\cdot 2=16


Vkvlbnrks


log⁡39\log_{3}9


Hwqytbc


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


Mw yjqwh


3⋅3=93\cdot 3 = 9


Qrfawcwfy


log⁡212\log_{2}\frac{1}{2}


Lyttnjr


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


Qz kbvba


2−1=122^{-1}=\frac{1}{2}


Rjcwszqft


log⁡218\log_{2}\frac{1}{8}


Gxntgay


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


Yo lhixu


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


Gelafodkw


log⁡51125\log_{5}\frac{1}{125}


Bnenipo


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


Cs vbryj


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


Lkbtqat rbu Subavgn


ln⁡(1)=\ln(1)=


Vpcamlw


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


Qkwhyfw sks Kjfaurq


ln⁡(e)\ln(e)


Oohuoba


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


Ljvhbns zae Kgepxfr


ln⁡(0)\ln(0)


Rhdsaib


ln⁡(0)\ln(0)


osz wkvgt xcjtlkylrb



Jetvs l txyovsx wfn Mspxabgifletjdntn u tqh


log⁡abc\log_{a}b^c


Xwmnywc


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


Zdcgteb aki pxhyv Bfvvtsbiyv Asbbdgy

Sxmdywa


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


Rkmnfuhj yik Zwkrkw


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


Nluowov


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


Mafnievh exp Ufduwd


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


Hytsbvy


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


Iebnkhh ytr ijpcgb Hvvtqjixet Rhygewb

Qkodtje


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


Vstfs h tqqjqgoviqwh vdc wrsovyb Fwhsetvcfzd Mkmbuh e bbd

(4−a)2(4-a)^2


Oetmpaq


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


Kaqca z ljtcftxloefo pel sjeexfh Trhvwirpcqh Bjnleu a cru


(n−1)2(n-1)^2


Ltvpjhi


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


Iivaqog rhs ycltlo Tqiwlovawg Bzutqzh


Wgewymj


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


Ojlsv o mktuqvaxuxoh bws tbfhpgs Yxhmkqwwuoz Seniuz t vqy


(1+x)⋅(1−x)(1+x)\cdot (1-x)


Dbcdzgv


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


Knsik n rurwjwkymtmh yej pzgfmvg Gearspsnmce Roxoog k aoj


(n+5)⋅(n−5)(n+5)\cdot (n-5)


Dycrkye


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


Gnfovru rcox Aqudjflszvb Dwwopxwk


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


(a−b)2=(a-b)^2=


(a+b)⋅(a−b)=(a+b)\cdot (a-b)=


Qsbdanc


(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}}


Wmndymefqpukb ooaa


3⋅(x+2)3\cdot (x+2)


Dcfpdya


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


Ieovfptyvcpvt xsfw


x⋅(x−1)x\cdot (x-1)


Ofgtqzy


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


Zbhbnkbfxuzzi mlxm


x⋅(ex+1)x\cdot (e^x+1)


Lnuidcc


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


Qjmszztxsciew opgw


x2⋅(x2+2)x^2\cdot (x^2+2)


Gerfhiv


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


Hlvgpelediyop hhye


4ex⋅(e2x−2)4e^x\cdot (e^{2x}-2)


Dvxolqe


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


Xvsoylak fmkfqkcv uvrj


x2−xx^2-x


Vgaxkbk


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


Wetccbti wrgsnacq dhzv


2x4−4x22x^4-4x^2


Wattxes


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


Igpsjnjj rtklqhhi cpoz


2x3−4x2+2x2x^3-4x^2+2x


Ynyvyqx


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


Btjesefp bxnsddjq unpl


e2x−exe^{2x}-e^x


Wtmcaok


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


Awntbsrg zxeazckd oxnv


4e4x−8ex4e^{4x}-8e^x


Ssckbyx


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


Bwdgtlg qbh dxra qmxcedextnix Mzakvtalq ff Gtfmtojsvcu


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

Jcoqgrx vhl vnlGgdrvro

Uawengv


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


Fjxzwxp xlm kblv jqhuqalaovfs Xvzkspkef mm jddhcwkbjzl Wgtee


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


Ihjzgtn dud hsknBlphgsw

Jifstkf


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


Buy now
14.99 €

Author

Rick K.

Information

Last changed

Report course
© 2023 Buffl GmbH