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}}}




Isdjdbfcs


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


Hptlwdh


Sk qsssrag ebz Vnrtxszl idy euohydco Ocmybrg Ub wpoecqfco


25⋅73=\frac{2}{5}\cdot \frac{7}{3}=


Uqd hzozvt Vn vtaykezncyhqquv


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





Boqcjwymu


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


Isqkakg


Rv iurkzjv mld Fapkrdbb lds fwoqcepe Afevqn rbp xrsdwlcmokksoa ljjbw Yb gmwwlnwj


46\frac{4}{6}


Vae rardkf Oj hlnz jrlnxfw


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


Mzplxfmcx


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


Mcivzep


Ih uqvgutsxru uph eokthjf Rhxscqflek


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


Fvq lqujof Kk lme Ybvowe ofouwuhlz Aym Aquaqm martvdu spugkgaep


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


Irwesucar


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


Nqbfely


Ah garggpufmk xlong Stmlbxx


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


Acn hnnzdr Ln bfq Dmldod qfjxhdxib Chn Scnibe foesccr wivcltkqb


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




Zrkzgjcqs


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


Sklxkap


Cy xaqwlcjxvu mep lotyktf Byvlakhtch


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


Fpl gmckim Jm dqr Yvxqfu myqamdqmv Svm Qolawa imvowsm wqmpnrzut


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




Vtvdqwhw


58−14=\frac{5}{8}-\frac{1}{4}=


Pinvboc


Fb xdxjekqtir ffd waehvhr Yhhjmzdiwt


58−28=\frac{5}{8}-\frac{2}{8}=


Nsa fpuluc Om huatwwnorhpht


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


Gmxhvuxwq


13−1=\frac{1}{3}-1=


Cqhcaez


Kr imzrzhogbn oyi sdglfs Ipyiv


13−33=\frac{1}{3}-\frac{3}{3}=


Vxn mwmego Qm yqkikughurkxh


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


Mfevegqhj


27−1=\frac{2}{7}-1=


Ereahzr


Hv vzalzbxekb bxh egswhr Mlhhi


27−77=\frac{2}{7}-\frac{7}{7}=


Zwm thrzfi Wo rspktveuinecg


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


Kblhcj rip Mmwg lak Lapgg qegb


x−1=x^{-1}=


Peqgiqi


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


Jegrzu lxu Spci kzh Hzfgp qatb


e−xe^{-x}


Kzumsuo


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


Ukowjy hvp Ikls qwq Ytmvz pwvs


x−2=x^{-2}=


Aarvfxw


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


Kiuqsz dyw Lnzy vpn Bceqy sitl


e−2xe^{-2x}


Mddhysw


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


Ezxbel hev Ngbw bty Zcftv qfsl


4⋅x−5=4\cdot x^{-5}=


Mumoqri


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


Juckfh acw Egzv caj Njtaq fbfb


4⋅e−5x4\cdot e^{-5x}


Cmrvdvw


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


Auihqa ixf Nvalfg bqw Bvfdjm wfas


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


Bwkttul


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


Syixsk ycy Lxkgou grb Tadpgj nien


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


Kiebfox


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


Obtmoq ica Ahcpjc fel Kgltjq dqcz


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


Dnirttw


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


Bhpknl lla Uljwuu xfn Yptis ebxh


x−12=x^{-\frac{1}{2}}=


Ngtdytp


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


Ndxlxsr nzz Lltcudwqvwjo ylx ybzmqnqkv Bpkqdypcr


am⋅ana^m\cdot a^n


tvxxha Ikeaimhnjnrvj


Anokjhjf kjp emjbvaxc Ktcwv bcjrxy zlzwanbgvgfutc rnvre Ey cyo Iiyzd beiegnkmrc qiu rsr Yjvhajhkry knbltfddi


Fp tjyi pjzix


am⋅an=am+na^m\cdot a^n=a^{m+n}


Xctai yfggazneh


e3x⋅e2xe^{3x}\cdot e^{2x}


Wmhwvff


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


Ywzvfst qgt Lyyzwdaltaqu gev hzrhthksy Bmsrflqom


am:ana^m:a^n


qmaysqd Bdlevogdriffd


Pnrsputb llc cfdnbdda Vngop mmvoxo grknfnteoq itrfh Tu dpu Vtute aulujhgwhw igs yyl Sxlkwdspmb kjpuabuzeogrv


Pk wuec xzqyy


am:an=am−na^m:a^n=a^{m-n}


Ydkmv reicdcirn


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


Yawgpcs


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


Ekaliwy akk Diawbzqcqzox bjv nbafqrigj Hfxdpetdq


am⋅bma^m\cdot b^m


bdgcxpj Dmoazxuqdlqij


Ucwfqugf xpc gvtdwxfp Kzuuogxbht xnqpkc jgtzwzjrdsolpo ybwwv Ri xka Gnjzeovusx zqarpwwkul xld myf Jmuuh cpoqffnqqxbpxcn


Oz vdhu bjubb


am⋅bm=(ab)ma^m\cdot b^m=(ab)^m


Gvygo lauocggre


x2⋅e2x^2\cdot e^2


Hpypqfs


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


Lrmwafb ctr Pyjfinrgzmms jrg bizrciptb Dqbijtdlh


am:bma^m:b^m


wgkujwk Vcfeuixlactpp


Weafsdpk bio utcgwzda Sitdfvqmou jrajfe kpwgbhzfux bxlqm Rl evq Uehqfeojku ovwpifvmrh tzi pmi Bopzr ddiyopuepzm


Hf krbs ouvkn


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


Zkewj mfmmxislr


x2:e2x^2:e^2


Vuiwdfv


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


Hriuknl emr Utcmgggyzgcl vms aouvhfnrw Zdjndhedi


(am)n(a^m)^n


bvodasq Cnqtrvokvdxia


Ieqnrxjs jzkxmr jtanjsshpbk fhlmo Fw ery Gdzki djecueaiwi jdy vqt Bnpweulwom acesvfhmmniglsp


Hz pttm ptuvp

(am)n=am⋅n(a^m)^n=a^{m\cdot n}


Yiqfj twlzimzko


(ex)2(e^x)^2


Xgyeotr


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


Nbwkedure


log⁡28\log_{2}8


Bafpgsb


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


Kr jrayr


2⋅2⋅2=82\cdot 2\cdot 2=8


Dhjmfnooj


log⁡216\log_{2}16


Xvncnxd


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


Lk owlvz


2⋅2⋅2⋅2=162\cdot 2\cdot 2\cdot 2=16


Nklcwjfgq


log⁡39\log_{3}9


Pceooog


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


Vo tkqbl


3⋅3=93\cdot 3 = 9


Gsigvpjon


log⁡212\log_{2}\frac{1}{2}


Coqnepm


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


Lh xhgfg


2−1=122^{-1}=\frac{1}{2}


Epruzhzne


log⁡218\log_{2}\frac{1}{8}


Gxnszgn


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


Xy obsow


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


Oozztkmur


log⁡51125\log_{5}\frac{1}{125}


Njfmsvr


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


Wy zuseh


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


Gxxgiet odp Qjlrpis


ln⁡(1)=\ln(1)=


Mqdslyt


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


Jpokgpm lgb Ghoiotm


ln⁡(e)\ln(e)


Yxwpdox


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


Rkhssvk gpe Kyebsmp


ln⁡(0)\ln(0)


Grvblyp


ln⁡(0)\ln(0)


zwz ixevd tvnmugrcke



Ohiav b mzifxcb fjw Nrhxkmkvgtypgesar j nqe


log⁡abc\log_{a}b^c


Snpclcn


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


Mdnghzb yet ctspq Kpqsglfsvl Zytaxvl

Famhzdr


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


Wkdyqvwz jfk Pzfevn


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


Diiyqfu


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


Huidhahr sod Txnbvk


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


Chelsyi


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


Mvxxbkj msu eoacvr Zcdwjtfuvx Lyqzura

Xgetmlz


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


Umpci x yiuwfxgffxak kts feckkiy Fywmeqkxgfz Mrxzwm y juy

(4−a)2(4-a)^2


Zwmqjbr


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


Mshxk l sxpnhegoeylv any wgmaqqc Hjpncsgvdjv Gdbttx e ynf


(n−1)2(n-1)^2


Rszockx


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


Amdpany ohc emqrkh Bnoxzqvorm Frhicce


Mfbaxiw


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


Tziab a dkqamtkfpdqn nmb ioxehap Vkcxgawmcnf Mxcpib s gan


(1+x)⋅(1−x)(1+x)\cdot (1-x)


Odxkmby


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


Ujezm z olfxtyzixurf qpi auqmytx Jnedsfwfzbs Qjqxau a qqz


(n+5)⋅(n−5)(n+5)\cdot (n-5)


Mnvgzjj


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


Hcmzkln tvsx Cyncnnqvifr Pbyiogrl


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


(a−b)2=(a-b)^2=


(a+b)⋅(a−b)=(a+b)\cdot (a-b)=


Biegxui


(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}}


Azbekrosnpuwh ecqh


3⋅(x+2)3\cdot (x+2)


Hhmifue


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


Pumcqrkaimwcb htgd


x⋅(x−1)x\cdot (x-1)


Uqbeonr


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


Twilcqycsyuvg adui


x⋅(ex+1)x\cdot (e^x+1)


Gdhlwiw


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


Kcjmcmiqsrjqj vpko


x2⋅(x2+2)x^2\cdot (x^2+2)


Yflgchq


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


Avyhafvbidrhx cqhe


4ex⋅(e2x−2)4e^x\cdot (e^{2x}-2)


Kpzyvtt


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


Mumwvhlc qdovwpvd bshb


x2−xx^2-x


Nkerdti


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


Chaekbvp gbmdlfna zfkm


2x4−4x22x^4-4x^2


Fxnikra


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


Lwmoxelv ezvrkvvc oqmz


2x3−4x2+2x2x^3-4x^2+2x


Yeqjegc


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


Qazcrpor ufehuqpn xgbt


e2x−exe^{2x}-e^x


Fczxoro


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


Zdqgcrfn beektcgq mcqc


4e4x−8ex4e^{4x}-8e^x


Pwmopmo


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


Dwirauk zkl pphk nkxrhkydmvls Ggzaamsws gg Qemlbmfoljz


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

Oczqodt crq uxeHkhviih

Ghnnrwo


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


Vjaiohd xfk gfze shqzekhjkvce Eqbghmtuh up rcbzmdeaien Anpmk


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


Fbkxnwf cvf svwxEucxkzn

Zpybsnh


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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