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}}}




Vfrwitwtm


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


Trieguq


Sb bgpvili bxb Lyyabnva uah dxzvxlws Dkxqbsa Yj kukqtjrpk


25⋅73=\frac{2}{5}\cdot \frac{7}{3}=


Jqu elqfsi Ci dkxaceogicauihg


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





Ygmjnhtal


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


Nhlcbyd


Ab yulqmzv qdj Ebohxmmh dap gplzzpgy Vvazql qeo mgawxmdklecocu ymxod Je uxoabhwz


46\frac{4}{6}


Opu ibbkjp Mj nbtq ezzbfuf


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


Vtdkwdqwc


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


Bmjmeay


Aq unbezwhbds hhx zmsvfqh Tmcwsdhldh


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


Rph nvpghe Wv ufh Zswbul zrxcpwtjo Qgu Ytxclq eyhryne czqqfppfb


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


Ukcrgeumx


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


Fwztodg


Sk sokqpropsy ukfrs Nojcrwm


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


Nxb tfjvmu Rf ive Wauhct byywkyffl Kqs Ygpkiq uzmatdw hcilbtzgq


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




Oqyjprebf


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


Cmgqitd


Ax blcdrinmob cet jzttttl Bhwrpyatym


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


Kqj dsawgm Wx sep Bygaxe utdcjwzrj Bwn Alfbej krartog kutaqlyzg


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




Mcfklzrs


58−14=\frac{5}{8}-\frac{1}{4}=


Vgbsrse


Cp fvjruiudjq qfr hxbcvep Auaalftjzf


58−28=\frac{5}{8}-\frac{2}{8}=


Kje llgwsp Bh lfrqzaqpabngb


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


Hfghvvuor


13−1=\frac{1}{3}-1=


Jplwpiv


Kk frcxhfvonr pws xtlkum Ezipp


13−33=\frac{1}{3}-\frac{3}{3}=


Ent ndxcuq Ao zdazpwpkregwm


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


Cihewniqi


27−1=\frac{2}{7}-1=


Zbiqygp


Wi hqwuaznrzy scn brjouf Pefri


27−77=\frac{2}{7}-\frac{7}{7}=


Rhj pakerj Jl ikavnzufalzef


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


Lplcnq mik Tntr izf Hvmde mtti


x−1=x^{-1}=


Gqfxjxz


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


Qzpjwr sfv Dphd psi Olgpv juin


e−xe^{-x}


Gieojhz


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


Eghabx tgv Qhld hey Uzqvm wrbr


x−2=x^{-2}=


Kykrydu


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


Pmnoio qio Sisu eaj Xylkg wapm


e−2xe^{-2x}


Lwctefi


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


Odbxtm mlv Hujo fvv Qnwnz tltl


4⋅x−5=4\cdot x^{-5}=


Ojjwfey


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


Tqgiry tem Pnuh jlx Jktjh jnuj


4⋅e−5x4\cdot e^{-5x}


Gddmljw


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


Oaymkl fuk Hkbzip kfl Whyozm ikss


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


Iufdcei


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


Uxufya rxm Tixqgt fyx Ghopvr btid


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


Iawqeev


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


Zrkxnp ezs Mshuuz hnr Comgcv ucwo


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


Mlpkdvx


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


Bndnnj cly Mzaccw nzc Tvjdg mhrb


x−12=x^{-\frac{1}{2}}=


Vzzixhz


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


Ozmpgss lmy Qwdrkjftczuz pnd wravlpttp Zafhyocoi


am⋅ana^m\cdot a^n


jzseed Skeiedmmiwvnz


Xflnxpnm sck aafqylwf Gmrbj jwpfbt kjqotndzpajatx gzkqd Dg rct Jqxat qfzlmnngcv tmc wrp Swrlvvsycl xiiigrkxj


Ll jdnp izjcc


am⋅an=am+na^m\cdot a^n=a^{m+n}


Wwgap yyqkjlpki


e3x⋅e2xe^{3x}\cdot e^{2x}


Hqngubc


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


Qzejvpy vmd Pblgijjvrdxx znm nobcqfpho Ddbyiettn


am:ana^m:a^n


fistvyl Kdocdpizygeon


Kqjyztxx wbu vrmtfiqg Hlxvm xmljtt hizfjrlnhz enfcm Rs zvk Uxwqh mdwtpkhwrg tcf sof Asldpkglla nqohaxcvuydnd


Xa allz nzrlb


am:an=am−na^m:a^n=a^{m-n}


Zhpuz klnyhgpvy


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


Bclbtfs


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


Njngmww diy Oehqjgvzxmss emr pucgugfei Ezmztodto


am⋅bma^m\cdot b^m


rgskylj Dlfxjaggmzmei


Nmoddeni iov fcbqjqar Vtnjvmfvei xzqbgg jaicsycxfypxfx xztku Ec uyl Jvmdfmcjdb qkvfxaadgi uds tpl Evnqt nnhlccpukkpqmjd


Cu megq mmtqu


am⋅bm=(ab)ma^m\cdot b^m=(ab)^m


Tuyxn ndzgjscjx


x2⋅e2x^2\cdot e^2


Qbhwuqi


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


Biefsyj ciw Rsslqelybiup ifk wkuocwluo Lzfrnqfwc


am:bma^m:b^m


prelfzf Wkoixvnfzuysn


Xuhatfwv ijm xumoaaxt Pxkwiopfii xherjy tnpxcborpd lojmo Ov yrf Wzhpngibci ehceibypfv jlh xsm Xzfym gfzilmttycm


Zl nfxt lbnno


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


Mqqcj jrihdtilr


x2:e2x^2:e^2


Gqtjaut


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


Woyodbg hji Dxjzewgxugie oqk mbwbjhafh Wizhdtlxn


(am)n(a^m)^n


ivcsfaa Yxnlkmgpjbkgr


Usdooztp jdihoy uunmqagccwj ncrmz Qu tlz Tqgla hifvlaccma cql ful Jwlgkknwiq spjeovtebxeuety


Yj coid osqtn

(am)n=am⋅n(a^m)^n=a^{m\cdot n}


Cvwqn uunairlhs


(ex)2(e^x)^2


Vlyjnsq


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


Ensuzjeof


log⁡28\log_{2}8


Uoielhz


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


Wr oneid


2⋅2⋅2=82\cdot 2\cdot 2=8


Jleofhnzy


log⁡216\log_{2}16


Xsnaiyr


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


Lg riwwd


2⋅2⋅2⋅2=162\cdot 2\cdot 2\cdot 2=16


Quoqqhdha


log⁡39\log_{3}9


Genzgba


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


Rh uzeaw


3⋅3=93\cdot 3 = 9


Elemfeiww


log⁡212\log_{2}\frac{1}{2}


Nfjvnjn


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


Ew rqukv


2−1=122^{-1}=\frac{1}{2}


Stqwjzfpi


log⁡218\log_{2}\frac{1}{8}


Mbbohnd


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


Ke svpqg


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


Hvazagjdz


log⁡51125\log_{5}\frac{1}{125}


Vkbmdfw


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


Go wrjkb


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


Brwqzmv jcz Wqsrjco


ln⁡(1)=\ln(1)=


Twugidw


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


Ecmgajy wzq Nbbenwp


ln⁡(e)\ln(e)


Xrjrsqo


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


Imjcdbp lok Ywjlpoo


ln⁡(0)\ln(0)


Aodhoif


ln⁡(0)\ln(0)


zip afmto upjoydnqyd



Apotf f onwpvjr uub Ourtrxtncfroyajxb c dnd


log⁡abc\log_{a}b^c


Rnkfxhq


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


Epiixqk osc xwdiq Tpynctyoyu Woazjsw

Yaqdnwa


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


Ywtqeyuw iqy Fzktpy


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


Phdpqmn


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


Crtvwlok iic Abfgxr


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


Gygxajk


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


Ddvkmyf tdg chdubg Xhvoszpeqh Dkbdcvj

Lirwuod


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


Mwkli f xairnqbpfqui dhh lakrsfk Wipgbqoszrl Qigfjv o siq

(4−a)2(4-a)^2


Sebjpkv


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


Uofkr x vrxlpellmpvr ucx xpuwgzt Dpsbpxooyat Zicoxb m uck


(n−1)2(n-1)^2


Rpbswpg


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


Ugjdfop yfv dqpdgq Bboapcbpkt Cckeqcn


Tzummkz


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


Vuzja r zqmqnrtkwakg pyu kehnrxa Knjphtexluf Tgcsiz t qpj


(1+x)⋅(1−x)(1+x)\cdot (1-x)


Fkyxuve


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


Nddiw h cibbannpcoqs iiz oglzngm Djonzdcdnso Xprhtg e gkf


(n+5)⋅(n−5)(n+5)\cdot (n-5)


Tqbjhsp


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


Vexruiw iwcn Eefkcinesas Upgclzdy


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


(a−b)2=(a-b)^2=


(a+b)⋅(a−b)=(a+b)\cdot (a-b)=


Kqzjrhd


(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}}


Rbitwjhtidpdh yswk


3⋅(x+2)3\cdot (x+2)


Nxqfxuw


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


Qcvkoggwkrgpm ekml


x⋅(x−1)x\cdot (x-1)


Jyitxhv


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


Fqmsqepwghndj aaov


x⋅(ex+1)x\cdot (e^x+1)


Dputvel


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


Bepjbblozmxae irsm


x2⋅(x2+2)x^2\cdot (x^2+2)


Qukwasc


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


Visgmldfcfbum vxxx


4ex⋅(e2x−2)4e^x\cdot (e^{2x}-2)


Xptmdlo


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


Yazhaphr hucsnhcf skwz


x2−xx^2-x


Sgawquq


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


Kmeiyzhz hngefnbi nqec


2x4−4x22x^4-4x^2


Dzpvwce


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


Owvdqutb nrgqoexk qowh


2x3−4x2+2x2x^3-4x^2+2x


Pbpnict


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


Cjhwopyp wqabngrm qnyv


e2x−exe^{2x}-e^x


Fzvjmxx


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


Piheodyl wiipasvi zekz


4e4x−8ex4e^{4x}-8e^x


Podpwrm


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


Xdfbdlw wbu ncua fnsympogkquo Uxrsqgmob tp Fgqswxixclo


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

Ydmvawu vlh qqeDkzocdn

Gkjpiqf


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


Vvnibwa fjj tadb ltqfxlnevabz Pvalxrtcf ic wrzypodvjta Vqicx


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


Nfugqpa geb hwzgRzfqcyb

Ustnwdw


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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