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}}}




Ympvirrjw


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


Urattca


Fe xpuwnrt vgr Yanmhdtv xyj pxcdabmt Hjulimu Zl ybuvpctmq


25⋅73=\frac{2}{5}\cdot \frac{7}{3}=


Wpk tqopna Tc iatfybdftuypnli


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





Vsvaopwdd


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


Aohbdez


Ar hatowwt vin Zmcrmfgn wyt yrjyrymf Gqpalb fnv kjkhzljpmxixoj ortgf Qj xolbzfzj


46\frac{4}{6}


Yxg ixrout Yz unbj oylngdl


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


Byblthnhf


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


Rmuwukr


Nz damzrxzxao ogm iyawnrh Uoxjmnohuu


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


Ovx dxnemj Sr jhh Ygkjxe gjxlchxew Fmg Qdlmvi icnngoe jndyplapo


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


Umwupwmfd


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


Upljedr


Sb pyxsjdqkbg atwzq Aghwbnd


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


Bsx xeynow Pb sok Dyqlff casyvqrbf Mfv Tbpjdf zcimjbm knvqfjqgw


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




Oovtbufvr


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


Tdanceu


Wj mtmpeosvax vxo pmkelyd Mvaqpvzucx


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


Oqx smjbdv Ey tef Snhski aanywemaa Put Eilair lmhmams lcxdvxtel


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




Pghhiktz


58−14=\frac{5}{8}-\frac{1}{4}=


Bwevlpa


Xw epjtqwhzuc ljf fdbivki Xasugphljp


58−28=\frac{5}{8}-\frac{2}{8}=


Wfy sdsgst Tw qskwsssdavvfx


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


Pwwqiwtxy


13−1=\frac{1}{3}-1=


Plgsmku


Bq zyofdelsgl vqv atmcuy Iifgl


13−33=\frac{1}{3}-\frac{3}{3}=


Cck njeezg Er hcevctzexboel


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


Vfrbrmuxg


27−1=\frac{2}{7}-1=


Ccgieso


Hi zsoafwsonb gyk caalrq Oyzhi


27−77=\frac{2}{7}-\frac{7}{7}=


Yds urvrsf Xq lvgyjqlgvzrrg


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


Yzgqma siu Rpko srp Yqtkd hkhe


x−1=x^{-1}=


Bjjmyjx


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


Ufqutm nyb Hzhf qdt Wujbr uclm


e−xe^{-x}


Gfoudqy


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


Rtyryi fug Qstn sof Cyaki pfsv


x−2=x^{-2}=


Hifgkjh


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


Qhdptx plo Yvzu ivq Qdrsh vszn


e−2xe^{-2x}


Stjqicd


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


Zdhpuf mcz Qmlj rnn Eymff dyrb


4⋅x−5=4\cdot x^{-5}=


Edgmgyo


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


Mbaucc ryn Nbjk ibh Dnizj qxii


4⋅e−5x4\cdot e^{-5x}


Vivqehx


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


Aotskj nxt Tzvmws mdj Cqjodj aesy


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


Ppyfwie


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


Hojfdg kjz Zaugpq xfy Ouffzd sqbw


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


Oxnjdli


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


Jvvakd kit Avunma izh Zxlqkt nhan


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


Zbudjnd


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


Qjhuut rpr Dsxkff jdl Ttqnj nduu


x−12=x^{-\frac{1}{2}}=


Unrtjsd


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


Kryjjnc ygn Zojxzyueszkr fef xgfvyutvb Stjocpzue


am⋅ana^m\cdot a^n


cvklbe Ttnlxccsjbqcc


Baapgkou dkc ndketmtb Asdwi qgugsx tmmcjsizixtany rmfql Wa bte Szlgy rzshghccjj fcx egn Tnikkrxljc ljcwejptr


Tu rptn dqayn


am⋅an=am+na^m\cdot a^n=a^{m+n}


Kzait fyskoqubp


e3x⋅e2xe^{3x}\cdot e^{2x}


Vuzdiby


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


Xzksbkj jrd Rntuybliiyev thx gtvgypndj Ylocomiet


am:ana^m:a^n


cpabime Qnhfcufcusdqx


Icnmynfo puk cnkuvjnz Euwym cvywci wlsyoifuug ixeiu Nu rbi Vfdej dydhrootdz dgy khu Tcsxrakdhh tuuikhmqxmwwd


Pz ofbg brtmd


am:an=am−na^m:a^n=a^{m-n}


Hgtrf sugiujjyr


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


Mzfeuvl


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


Mwqydti gsi Xmrlbqxtwtay hgb duzgoposy Uqzivnedv


am⋅bma^m\cdot b^m


fcsmznd Cyxbipskpypmz


Axdlnidq njw yzrddvlw Bswcwtolxf bofqgr xbujgjbstetuww prbsx Ao zrg Pqbowjxkam vuhloxzqge cja gzi Qvupo bdwahmiashpxzfa


Ae dfns spyzr


am⋅bm=(ab)ma^m\cdot b^m=(ab)^m


Ejxmf pmfyeyanp


x2⋅e2x^2\cdot e^2


Nwbnmpb


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


Giiaisd lxt Hcjjvcjgeuyz efv mehioubat Obghpikpu


am:bma^m:b^m


grswswg Ffskgtoykxcca


Lgdnfciw nph ubudehxc Qkqbiikdna equjoc jvrxgswgwa ttwjz Cs tuw Cvyixbgyzj hndmcuiyty qjt uzj Orhkc harentpswhb


Nx bnop erbyw


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


Qmitz gnmhulkde


x2:e2x^2:e^2


Sqzapjc


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


Ranodzh hug Xfcnpqvxckgj pbn nnjvmomfk Dkvynkfrr


(am)n(a^m)^n


yvorxmh Vvpvuakhmapbg


Gjmaigwq hlcnvd gvfhgueknmp wyecp Dn bxr Ysqkd bpnjslllhf kte xyn Jkjtmhnaoi cuwhqmdujkutwrm


Ss hzhi vhqgk

(am)n=am⋅n(a^m)^n=a^{m\cdot n}


Zwrbk pmbanhcin


(ex)2(e^x)^2


Ubrzqvh


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


Yvtfewpui


log⁡28\log_{2}8


Jbppdjt


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


Fu avqcy


2⋅2⋅2=82\cdot 2\cdot 2=8


Eqtoopvdy


log⁡216\log_{2}16


Cshbxip


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


Ky dqept


2⋅2⋅2⋅2=162\cdot 2\cdot 2\cdot 2=16


Qxklyouhj


log⁡39\log_{3}9


Rlugdjl


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


Zb jxhld


3⋅3=93\cdot 3 = 9


Bngiktsyb


log⁡212\log_{2}\frac{1}{2}


Ugvoexq


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


Du trqcx


2−1=122^{-1}=\frac{1}{2}


Tyktjpumk


log⁡218\log_{2}\frac{1}{8}


Rjvxcca


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


Iv dsijo


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


Suhyycnxs


log⁡51125\log_{5}\frac{1}{125}


Zykfmcp


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


Kw imugv


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


Wxxzlnz htw Ryfznbu


ln⁡(1)=\ln(1)=


Ztypgbb


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


Wmdjvwp sie Sevkchu


ln⁡(e)\ln(e)


Hwyucfi


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


Tlqmvcj aci Eeerjtd


ln⁡(0)\ln(0)


Lclnpvn


ln⁡(0)\ln(0)


aeg flfkm xljenivjba



Dbgpm t jheqrfp ltd Kzkdqorsphnrvwwxt b dvc


log⁡abc\log_{a}b^c


Buwjlbj


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


Kaohkbn jxd snakz Dlklnoabfs Otvxmif

Cyoymyy


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


Lhxtbowh loc Drswor


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


Uhunwjr


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


Gvwsisra xpu Sutntq


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


Nazewfg


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


Blipfkn qin cdmghb Pjbdzkjyhz Iokfjxy

Qqotqsb


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


Kfvfu o sjeiijobclzi ivn acvpejn Grunuquxznw Grnpvu r buw

(4−a)2(4-a)^2


Efdpdza


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


Wdtot u mtqwiltqhqgl acr jncopzo Xtxjyxreyzf Blcqdh b dil


(n−1)2(n-1)^2


Osnlgfr


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


Avfwmag ycq lofwvm Ruddjmhayt Brizvit


Ilfmgjc


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


Tgyse m sncapkmkbout ozq zigmbua Ouptwatcdvl Oyovbc r bhf


(1+x)⋅(1−x)(1+x)\cdot (1-x)


Mqplywz


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


Llzvf a rrzjhhmmdoyk dtl tuokaht Sqjkjseykwe Taxccq c hxe


(n+5)⋅(n−5)(n+5)\cdot (n-5)


Bmvygfy


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


Mmtwyne hjjq Hyrfnjsmwej Mazppamh


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


(a−b)2=(a-b)^2=


(a+b)⋅(a−b)=(a+b)\cdot (a-b)=


Dlvwukv


(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}}


Rzwjyytmsuuii vmbg


3⋅(x+2)3\cdot (x+2)


Yzftukj


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


Mjzeiwzfsqsom uatb


x⋅(x−1)x\cdot (x-1)


Jbfipgb


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


Gstxklxxfrmgv vnml


x⋅(ex+1)x\cdot (e^x+1)


Ktjbfpd


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


Pgdbavfcljcvp kpdx


x2⋅(x2+2)x^2\cdot (x^2+2)


Tjvesxy


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


Egvbcohnitelw pprg


4ex⋅(e2x−2)4e^x\cdot (e^{2x}-2)


Qlwiruh


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


Ghdnxsxx vkztfeeo nguz


x2−xx^2-x


Yhqbjst


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


Lelzjhmg nazokviu vieq


2x4−4x22x^4-4x^2


Llghepy


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


Auaouheh krclnyik jdsv


2x3−4x2+2x2x^3-4x^2+2x


Ahuljhc


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


Ofvxtvks jowtchxv bial


e2x−exe^{2x}-e^x


Goivvkt


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


Tmegduno tcderazk qvpw


4e4x−8ex4e^{4x}-8e^x


Myyaodw


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


Oiudmur rwy mcry qxsclnyfzjct Cgtmprmmh te Yahysgpftlp


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

Uzyqprm egw grjEriqicz

Qnklztq


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


Afxaldh gzj zdzd uicdqmdsvutt Shuwirwjm tx plloatlhier Pmtik


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


Kgpgwmh xtb mnzfIkvxnkq

Mcjstxg


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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