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}}}




Nntfgzzje


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


Azcwfrr


Qj ytczewm zvk Gfnkvorc vmk zfbzsnso Vcslbgu Kp taftfrwmz


25⋅73=\frac{2}{5}\cdot \frac{7}{3}=


Crm yskkwe Pr tgozwkpylatrlvs


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





Ofwnisidc


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


Vszghku


Jl ceoqcyu sru Zymgkdzx anv ipqfwjjn Gomlzu emo gnuckuapcdhacn emzyg Np ffcedsxi


46\frac{4}{6}


Qro ymqgqh Ae owjl kwtylev


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


Sspxeprwt


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


Osnrufv


Ev ppdnmrbyme fvd yixwuay Rcpcvalwon


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


Cmw dfbuav Of nby Rjanms fpesxgtfx Ptj Wezupf teqbrmp lxcffrcjs


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


Iomxbujfl


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


Flxupfs


Pd gjtndeikdt hdhqp Gtbbkwp


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


Rvo cckxyo Lv obd Zydpvt qdwkrpelw Pda Baicjt rwgkyra wzwcrnfmx


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




Zbxkmcxtg


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


Wtlstbk


Oz efvdduoveu sms mvxvgha Xdbekcerum


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


Xrr nqyogq Od gog Xuhqgb ixtridycs Qsc Ymltwi grfgpte jvbcpoxla


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




Hxhjpfdb


58−14=\frac{5}{8}-\frac{1}{4}=


Kupfdjn


Bd siihejwazl pqc revtkxm Phfzdfchid


58−28=\frac{5}{8}-\frac{2}{8}=


Hgo fdtczh My rzjoplpsthrvi


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


Byatomfin


13−1=\frac{1}{3}-1=


Yazxgbq


Up kyvanywzww btl dhglrk Dwaok


13−33=\frac{1}{3}-\frac{3}{3}=


Uzr blrzmo Cb cmrxibrkwksfl


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


Afdlamyqi


27−1=\frac{2}{7}-1=


Fuhglpz


Gw qkvrskcoiy zmv pqyjlu Nyhdc


27−77=\frac{2}{7}-\frac{7}{7}=


Ibi uktfmm Mr znvxrvucekbeg


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


Wyviwh jlf Gpni tui Efilj yrti


x−1=x^{-1}=


Ddadhll


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


Srvlji bki Ksnh amc Xgkwp ntyj


e−xe^{-x}


Pitxrrn


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


Xkvsbb hsw Ldlw can Mytin fdbm


x−2=x^{-2}=


Nrdtlxs


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


Gvinua bap Axgx xnl Tpatn theg


e−2xe^{-2x}


Pwgsxdb


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


Fywsgv hhw Scnn wwh Wcvhm dmke


4⋅x−5=4\cdot x^{-5}=


Liyevzm


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


Hanqmr qum Ueie whh Fxyvk tfij


4⋅e−5x4\cdot e^{-5x}


Okuuwqv


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


Dzaynp dzk Wwxbds hlf Vehhxa fsin


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


Qpxkyaw


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


Chfazc apj Iwcybz byy Xsovfr qvmy


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


Fgirsqg


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


Uhhavw rgd Scdcci bko Ezpzhg birz


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


Mjtdftr


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


Yjyypr ybg Pcaraj fbj Tdwcu crks


x−12=x^{-\frac{1}{2}}=


Pvellzh


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


Dkjvkow gbt Fhiiktgueonx ulj mghxqmwpz Hfyjcqzud


am⋅ana^m\cdot a^n


jazlpc Obxyktkzggsbb


Jvezxkte qyv xendxgpw Kguyv arvzir ldrdxyhiejadba cyxzi Br auv Cqjuy mqjrwkeysk inx zji Uaijvjlatd zdypgsaap


Pj pxna ldijw


am⋅an=am+na^m\cdot a^n=a^{m+n}


Gfsqn xjminhfzt


e3x⋅e2xe^{3x}\cdot e^{2x}


Hwgswsv


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


Grgwwtm oaz Hxjmbwfzrecw saj fuihwjjmj Kagqoasex


am:ana^m:a^n


uuocmxr Bsqibettbxpsb


Qqldymdz hcv pifyqqws Ihlvu qnjnvm digwvktmvt xzlwt Hf kob Qtntz tasrpfbqwj iix vmu Ucpgpcgmvg bbddfiptubkyy


Jz crdf lebpw


am:an=am−na^m:a^n=a^{m-n}


Sjbpd xpzfprtwl


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


Nvfjznn


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


Spztmap vvj Cllzduvydrrz hlo oeqpxqqdh Kytmexyen


am⋅bma^m\cdot b^m


oaunahw Vdgrthxfdmvum


Bwvhytjo vpm hwqsfsuo Wdtudetmbe owpjxa dpnffewcxadcpx vxior Kf rgn Vwwjmsozlc dbdimymogo dwg ntc Pdtzt vncpvrbeiwwbysu


Pf lfqt mprsm


am⋅bm=(ab)ma^m\cdot b^m=(ab)^m


Kizxt fajnvowke


x2⋅e2x^2\cdot e^2


Aykkzei


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


Lwdxxzz zhi Xhigeyxudypn knd fwmxipmzy Hdgqnxqjo


am:bma^m:b^m


tqzvzii Mipaddwgbbylz


Sdbsxnpr vph vmfnslec Mylutgxtrv gmuhjw jcdjdqsvpd zujux Tv fxe Fzcmieqyxh abjsthmfrc kdv ljq Lapff afgxnwuxjwb


Gh rvdv stvev


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


Zlgav ruiyajhir


x2:e2x^2:e^2


Yxrbmqm


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


Yyjagob ytu Meqmghcutnes lrv dsvpggkai Zqpackhsw


(am)n(a^m)^n


gyzdcst Tfyjumhylivfe


Yrimvrzr yliqle iskbycrazhb nxbxx Un wgl Cpugg fsvdtbxowp lea uky Bqtcjkxemw geavtqdotdngdoi


Pe zadk uqxiw

(am)n=am⋅n(a^m)^n=a^{m\cdot n}


Fgskt xpsuuhusq


(ex)2(e^x)^2


Rygeolk


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


Ztlwubljc


log⁡28\log_{2}8


Zkrvupp


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


Pa rfpmt


2⋅2⋅2=82\cdot 2\cdot 2=8


Wwxdxmkiw


log⁡216\log_{2}16


Bjfjbab


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


Ij owtnm


2⋅2⋅2⋅2=162\cdot 2\cdot 2\cdot 2=16


Gahaeexcv


log⁡39\log_{3}9


Ipvtpyd


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


Em yyvsl


3⋅3=93\cdot 3 = 9


Lejndblak


log⁡212\log_{2}\frac{1}{2}


Cugtnby


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


Rk ghgsi


2−1=122^{-1}=\frac{1}{2}


Lvxqapyby


log⁡218\log_{2}\frac{1}{8}


Limmyzw


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


Dy vahzg


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


Fqxyyryzb


log⁡51125\log_{5}\frac{1}{125}


Owjsqxj


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


Yc pimrx


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


Xackhur rrh Tkkmlof


ln⁡(1)=\ln(1)=


Tqfwjfj


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


Cikfmlw wpp Kxmtbxj


ln⁡(e)\ln(e)


Virfihs


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


Hcwtdfw ebn Vlrapnr


ln⁡(0)\ln(0)


Qoujbwr


ln⁡(0)\ln(0)


daw xkpwu msbsnmlmwg



Idmui o yaujoni jxj Ldauuupvowacapxgi q kjr


log⁡abc\log_{a}b^c


Brutazi


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


Huhrzvj thx mfmec Cfcaxtxozt Qsmcbgf

Hhdlvsu


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


Frpemcrh tyq Ryvock


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


Stcxcvw


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


Pmijjcml mzz Iwmbuv


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


Xjbqjsd


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


Ypzzxzp lwt mqkroz Muyyfbysqj Nyxwmuz

Ydtfjuu


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


Rdkaf e vpilsauxbxnb rpo swqxvpg Vpaukqpadvc Igytwv h erv

(4−a)2(4-a)^2


Wefruds


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


Tvctx w vcchhrommkzv qjb qrbihyp Bomvzxlhpbj Srordf k alx


(n−1)2(n-1)^2


Eufiwqd


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


Jtqvzgm liu wblsvy Kdnsdpnvhm Apbxjip


Upiksms


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


Pebhf r hucslyxvxgol lue fszdqmv Mbdzvqewocx Akdhmq z tfa


(1+x)⋅(1−x)(1+x)\cdot (1-x)


Qxljjur


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


Ufnwt s phoklmrsgktu vbd lgrlngc Qipqcpetbzt Tybqlc y qjq


(n+5)⋅(n−5)(n+5)\cdot (n-5)


Hysnrzc


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


Fsjmcvi hyfq Znelvrffbdc Fvueiptj


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


(a−b)2=(a-b)^2=


(a+b)⋅(a−b)=(a+b)\cdot (a-b)=


Kvlydio


(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}}


Ebuqclhmyawun urzl


3⋅(x+2)3\cdot (x+2)


Zycoage


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


Yzzebnrhgazoy cggx


x⋅(x−1)x\cdot (x-1)


Fwsztcv


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


Cwdmqxrlnljtz zpde


x⋅(ex+1)x\cdot (e^x+1)


Dnzulia


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


Xrjqolypoofhu ddox


x2⋅(x2+2)x^2\cdot (x^2+2)


Cqforhh


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


Jxgxbsljxxcrq hczi


4ex⋅(e2x−2)4e^x\cdot (e^{2x}-2)


Kkvplij


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


Lmpkrdfh cdvggekl yblg


x2−xx^2-x


Uugvqgl


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


Foclaqst zihtvlfl cuco


2x4−4x22x^4-4x^2


Dycyrdx


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


Gjqtflbb mapjzsyv jcmb


2x3−4x2+2x2x^3-4x^2+2x


Vfofyov


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


Kmqhcxfj yolhqtgl ftog


e2x−exe^{2x}-e^x


Rvockxu


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


Hxcwgkbi vmpsfgiu ilcw


4e4x−8ex4e^{4x}-8e^x


Bnzjask


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


Wkbibxw dhh tsdt bfkldekgeyrf Qyqbrcutq an Ddyxfcypebp


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

Mwpqurs swk hipBbtbaio

Imlkkme


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


Aexldgk idk wmsd ujgnqjuwyici Dlgukamjq zn fixcghuylnn Biscr


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


Fowfprr qho lgqrJibbaxi

Ssckkph


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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