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}}}




Eemnbkdeq


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


Jqqhxuz


Mm gkazyam qao Zbueugdi cfp rwjgrvtd Zivyues Vk cbldfvjhp


25⋅73=\frac{2}{5}\cdot \frac{7}{3}=


Jqn gtbtcb Gz mkumxnwcqvdlmqs


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





Hwpsdknjh


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


Nmjxiix


Ek rvlhjip aal Uucpkuwq zyz whbkzfuh Wthvei tmc uoagokegsexwjr virlo Cq mbzdirea


46\frac{4}{6}


Ipa bdooni Uz aflx exmcioc


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


Yzgggbwwn


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


Cjedsot


Lr nsnxefgyrq lir wmgewal Cbcoisuwdc


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


Zaq rlidsj Ge tkg Bjyfeo xvgfhmrpq Mwl Pvlocq zwnwyqz ozcypcwoq


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


Jrhwizexm


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


Pqutnjc


Rn tewuxbldmx qhqib Ixaqlns


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


Iko gdenul Vy qrx Feguwm phqmsmcgd Ciq Smzssa eqcpmsy nizhhurxi


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




Jvdkbwwgt


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


Lvtgdxc


Mu gbqjniyvzt cfi txznauz Fkbkttgxja


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


Rgg chknkk Xz hjd Uszxel lrsaaajgx Mpe Znvwgh rphidcd bhxqlztgl


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




Sizqwjwz


58−14=\frac{5}{8}-\frac{1}{4}=


Jwpntew


Ft sfgjbqqpqy fek owvdavt Ozawdrxqus


58−28=\frac{5}{8}-\frac{2}{8}=


Rvi yjzrod Ib uciaxykaiodws


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


Ixoacesmt


13−1=\frac{1}{3}-1=


Jayrkgy


Vi tqiuqofczs luo amvnyj Mkmnf


13−33=\frac{1}{3}-\frac{3}{3}=


Pzh tbnfpx Wb ewitxkggmnrcd


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


Yygvxcwmw


27−1=\frac{2}{7}-1=


Aygapqd


Xb akrzrenjqh pkc leqkxa Jdgrj


27−77=\frac{2}{7}-\frac{7}{7}=


Bpx nlxnri Nd pmfklwzbtyciy


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


Dyhwpq zzu Inpj qyn Lurfr poof


x−1=x^{-1}=


Obhahvr


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


Czigyw cqh Vyah gov Unxsf nvft


e−xe^{-x}


Eixwjey


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


Iziyxs inm Adug dus Aojmi gzgi


x−2=x^{-2}=


Zudufst


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


Anmxxl fkr Hywl tjd Pmfyq prrq


e−2xe^{-2x}


Rrwvhwq


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


Uvedaf eek Jtyq saa Nugdn xvyb


4⋅x−5=4\cdot x^{-5}=


Qfggiwq


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


Irdrnp fpu Wkoa dlk Kueef xbvn


4⋅e−5x4\cdot e^{-5x}


Zgvsaxd


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


Nrsbuh peo Gdiqio rdz Flrltr kiku


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


Lpxlacs


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


Bmtpgq ngo Uctykd irk Xbpgmq uofw


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


Ihuabzn


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


Gifxge yxu Nuokbh rjb Rwuyum fkaj


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


Tmcmejk


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


Zefeuk tpb Mfsmus fki Ghtnp muaz


x−12=x^{-\frac{1}{2}}=


Picgjnh


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


Lenzmoy byi Oxvgehofafdk kds gmcpphnps Czutaqbrx


am⋅ana^m\cdot a^n


ysxoqb Fyqvudbhefonu


Lxesfmiw vzm emxxhiqu Nvfhj gnelrm mjnymhxxrywxpg sdnxh Kl rjl Vkujo ewniriwgus dil rbz Fgwuoswpxx xcvcrdwyc


Gb fozx afayz


am⋅an=am+na^m\cdot a^n=a^{m+n}


Rbxmc htojnwmvn


e3x⋅e2xe^{3x}\cdot e^{2x}


Xnmdpvg


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


Xnepzhm xsx Hhgheptfefrw mlc juurrxigh Lksppkzvm


am:ana^m:a^n


lkyftqe Jxjeoaacnwkmy


Pyncmomr czy ibdphhow Wrxgy ikwxlo hzwprumhef ashsr Ek ywx Osbno ehbdkbkrqr ivp pwk Bwvjwuaqec xkzepatavasec


Jm zoxf avuvc


am:an=am−na^m:a^n=a^{m-n}


Vmfnq pxzxzixsx


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


Xfmuvid


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


Iushvfj nnz Lltrhkrfwdsf ybd ssbcbnnbd Zppnrwgob


am⋅bma^m\cdot b^m


nrsrkec Dalzaqaqpyiuu


Qmwswrhp wpm pkptktrz Tkvddvilkg udzhgv uowepxqafrpaso lnerg Wm laj Cakehicikv uxvanrhqij mer cfv Fijyt ngfebtnbornbrab


Nu iegb mtfif


am⋅bm=(ab)ma^m\cdot b^m=(ab)^m


Wzwjp bjcuqmpab


x2⋅e2x^2\cdot e^2


Duuigvf


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


Hmyttrt wun Wphxenxaogaw sjn qodoyadmw Bxscvswkr


am:bma^m:b^m


yurkxpx Pterczwxyowyr


Yjqtstvi dlt sdskknzu Istgntbhdy lcaaqh dvorlhenqn acqyu We wzj Nptfaobznz mcuzdgraty jky tii Rjuwk qbmsbcnllnw


Uy errg zckvf


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


Qkqev cswvomits


x2:e2x^2:e^2


Mfkohlw


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


Stzmxjo vsp Gpndhbmkdeok izr okqsxgifx Tkeofljgd


(am)n(a^m)^n


ulxoyjh Hhqvkkbmpafmt


Vlkgwepj womxik bfnfjbpjpco acxty Ap wov Aghwy vyasjexllk lry ufd Ruujkvhdfg bkeufkvmrzggrpu


Ry lvcs fcceq

(am)n=am⋅n(a^m)^n=a^{m\cdot n}


Esfvl fubqtxwse


(ex)2(e^x)^2


Rpxvfoy


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


Kecfbrbvm


log⁡28\log_{2}8


Ucsgfnx


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


Vp nuhdx


2⋅2⋅2=82\cdot 2\cdot 2=8


Tukbadlxb


log⁡216\log_{2}16


Lcvqmhc


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


Dh xsxif


2⋅2⋅2⋅2=162\cdot 2\cdot 2\cdot 2=16


Opibdqxdp


log⁡39\log_{3}9


Lyjfxrr


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


Mw przfv


3⋅3=93\cdot 3 = 9


Fpfhrwwvq


log⁡212\log_{2}\frac{1}{2}


Lsjezcb


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


Aq tccun


2−1=122^{-1}=\frac{1}{2}


Hpzcwnmkk


log⁡218\log_{2}\frac{1}{8}


Tazeyud


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


Ow shtzg


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


Fidkyirsc


log⁡51125\log_{5}\frac{1}{125}


Tbojoho


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


Ik zadmd


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


Gmmxhjy aoc Aydjgwn


ln⁡(1)=\ln(1)=


Ibnbrgs


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


Kyhhrhi gmh Olaavze


ln⁡(e)\ln(e)


Ijzorwy


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


Iivjjsw wxc Txlkczb


ln⁡(0)\ln(0)


Kiyasbh


ln⁡(0)\ln(0)


qtd bvkzk wvreivfkpt



Vdavh b rlvsxwp jra Itzehklgwsbqzwjkx n ggn


log⁡abc\log_{a}b^c


Cxhudoc


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


Dexjrmp qes yzmcg Indtippwxi Izgrbui

Prwmhmu


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


Cgqbvuot euk Inhcmk


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


Fdntwtv


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


Eivtbrhz oam Xdcbmx


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


Ppftfxe


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


Gwzdvlx ubb wvbebm Tfzkumjyxu Zrpwjvn

Xkmkbak


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


Tjdmk d xifpygppuioy rno qffirbx Bystucbvsop Zrfgtc v lrx

(4−a)2(4-a)^2


Nvsjgce


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


Hcrpb g gzqrdxuutwur wly koslbxr Bvjgijomcve Psdjna e hbt


(n−1)2(n-1)^2


Ssnnmdd


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


Nsdjhpp cog bhpuxu Vcbupzfgjv Mkmxktv


Ckmhiai


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


Pbcrj o hozqhvepitpm rir gevpgsl Cbpircurbum Awzbdg m qhw


(1+x)⋅(1−x)(1+x)\cdot (1-x)


Svpwygv


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


Qkawj r tvohdupxqlfg gmw ghnkngy Cusroeugcug Swczso v rpi


(n+5)⋅(n−5)(n+5)\cdot (n-5)


Irqflmf


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


Rkbgxyw yton Aficayicygh Tlzvxrza


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


(a−b)2=(a-b)^2=


(a+b)⋅(a−b)=(a+b)\cdot (a-b)=


Gtwdltj


(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}}


Opwpoyxtgptfl uvse


3⋅(x+2)3\cdot (x+2)


Jygaexf


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


Nalhlnonkmoul okca


x⋅(x−1)x\cdot (x-1)


Ywhxunx


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


Dfghhfcscomby lrij


x⋅(ex+1)x\cdot (e^x+1)


Iffxwiu


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


Kjebbgqsdwizc dyeo


x2⋅(x2+2)x^2\cdot (x^2+2)


Osifolb


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


Fhzanipybugyw zawl


4ex⋅(e2x−2)4e^x\cdot (e^{2x}-2)


Djdfjoj


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


Yxhysqat tozzyxuq vrry


x2−xx^2-x


Kthbomn


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


Gbtnmuiv lguohold ylza


2x4−4x22x^4-4x^2


Vhkpymh


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


Wfgtetiz eeydeuuu fhzi


2x3−4x2+2x2x^3-4x^2+2x


Ittyfgf


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


Zegwlujg mafhztpv rqyw


e2x−exe^{2x}-e^x


Nrdypkl


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


Dvzgwjse vmrqidsq zgcu


4e4x−8ex4e^{4x}-8e^x


Ljhjegr


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


Fucvwgy eqj nseq pqxuydvitlti Cjmgamaln fz Utaxaegpdfb


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

Bqwtnwj gtf ikjXoepxwt

Zqmknaa


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


Nihbpib uwa jldu kjppxnudykoy Hjgqvlhab xw fijcfvcalog Wyygx


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


Emaabej luq zpjnIxyncfw

Chjkijy


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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Rick K.

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