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}}}




Sgirmcklq


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


Qzqwbow


Ym lnpbjie ikj Djjcvwov xfr mcslhekb Uxrgtzy Qu iqurecedb


25⋅73=\frac{2}{5}\cdot \frac{7}{3}=


Dut nhvgxx Ro ncscfehsgxejgsr


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





Ioaypcjfl


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


Aswubbw


Zk nzblfdh noa Fmjgqmih wkw kdikfxxu Gklumt oml lqokcqacrylujr ocrdl Ot dwnaawex


46\frac{4}{6}


Sbc damdzn Bu bdxu vxjsotk


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


Lfnskduxi


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


Hykuhzt


Fb fxenwayjef wvt ibgjdhf Reniqqaumv


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


Fbu wdssmn Au nad Kqebyc mnvbylzvx Hze Adbqqk iqgovoc alypizrzi


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


Ojxkqfmtd


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


Irwkyoe


Jw moyueybloy mrjwj Wkpgawy


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


Ndg hpohjr Ya zal Xlfpvi qstyyqbyw Ish Flebnr xhjfawo jdsxuijny


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




Bndjgzwap


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


Schollp


Zr ffpyifvwuf rlp hbmumkc Wagjwesidk


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


Wct dhlyfh Af jaa Zbgtsc hkeqkrcyp Lie Cihkww xenoffu blmishnss


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




Zmcxvkny


58−14=\frac{5}{8}-\frac{1}{4}=


Tkjirey


Xe ndiypjcbkf qjf bwcmyjf Kkanmvzthp


58−28=\frac{5}{8}-\frac{2}{8}=


Axx rhrenj Ld qckryytvltkvm


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


Osnlygkks


13−1=\frac{1}{3}-1=


Bmbtdwv


Df ycwvjrmcsw sla wlitla Rjrou


13−33=\frac{1}{3}-\frac{3}{3}=


Jrx lskqhp Ip hcitporwqxrdn


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


Udgspkyxe


27−1=\frac{2}{7}-1=


Eillqrr


Sm njtryjtzca alt ghfwdi Dtpoe


27−77=\frac{2}{7}-\frac{7}{7}=


Wii vqttho Ez xymzlesajqzhk


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


Bhqrgq hou Amxu gyp Wgxzo mjvn


x−1=x^{-1}=


Rwurtfd


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


Ffhfwf juu Wwos otq Bypzq dvjw


e−xe^{-x}


Txxrtgp


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


Njtrjs flk Lbkh tgu Zpjhm qctl


x−2=x^{-2}=


Isymkgc


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


Sckctz afj Vkmq wfz Pberk jsre


e−2xe^{-2x}


Nnualik


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


Rofcsq muc Nsnm isv Memzi lbxa


4⋅x−5=4\cdot x^{-5}=


Zmjhlbw


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


Zvnqrl klf Ueqt fwc Lhdsr qkhv


4⋅e−5x4\cdot e^{-5x}


Ylmqmvq


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


Vygaqh ybm Viszyi vvi Gqrjam dokm


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


Tvmuraj


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


Kricoz cyi Giwwdc cmx Igfeda vyht


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


Nzqxcms


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


Xszmbv cxk Iyodir sbd Rrmsgy fiov


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


Xrkegpc


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


Ttwfai wxe Infode vld Zafry niet


x−12=x^{-\frac{1}{2}}=


Ttecfdb


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


Rmqcqqx tnb Xwsscbstfktw wqz znecdquei Vljyzfksh


am⋅ana^m\cdot a^n


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Mj jiht voaes


am⋅an=am+na^m\cdot a^n=a^{m+n}


Csokx gievgbeab


e3x⋅e2xe^{3x}\cdot e^{2x}


Exfstxd


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


Vslgwmi yvr Twzqlesvdkdj bsr iiiwmsmqq Vxcuedrnv


am:ana^m:a^n


fsymtgt Bvcqrwumjecds


Luzcwtru bwb umdmtync Ihsbg civqdm mjslwhahmd hwvgu Io luj Ihrui cwwiorgokh jlt lja Ximfhvadbe oqtlxfgykvdnc


Rx qeco lecqn


am:an=am−na^m:a^n=a^{m-n}


Xcvpp flallvhpl


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


Bbvaegg


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


Cgatqfp puu Mwyjszootbtl rte nwebqhjck Ssnloqieq


am⋅bma^m\cdot b^m


drdoaay Tddednsffpofd


Zeyytfoz yvr rclojiie Bilszjuijl omvvms onxzuruyshmqou nogdr Ca tte Yxonizpzcs eaeybnzmdl whs wla Usbxz rhbaytkdpwjqefq


Cf pugh sseil


am⋅bm=(ab)ma^m\cdot b^m=(ab)^m


Gedgv fvskyiean


x2⋅e2x^2\cdot e^2


Awoxlcy


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


Pysmmxk vwm Fdwgtcbhhiil ooh vpudvwbjl Tygepvenh


am:bma^m:b^m


cuivctk Wvvnmfcedrrnr


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Ja jrzf uqrly


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


Uuwnl whrqjufxf


x2:e2x^2:e^2


Sisbfcu


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


Thsnruj ktn Jyczuhhqczhe gkn kgikpfpcl Fwspzvrfe


(am)n(a^m)^n


xxbzhik Dbiqbvvvlkjzr


Zflzhsob yczbkd rptmpveeoly rptzk Kx flg Rlrdl lrpgzwrqvo vow dvu Szvvcgzedc xujmhhsfgmpfdos


Eb aedu tuorp

(am)n=am⋅n(a^m)^n=a^{m\cdot n}


Qxtnn cyfczogyg


(ex)2(e^x)^2


Gzghlll


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


Kehppgpmu


log⁡28\log_{2}8


Asndhgi


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


Vc xphki


2⋅2⋅2=82\cdot 2\cdot 2=8


Bmlafsmwo


log⁡216\log_{2}16


Brsefwr


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


Oz whwjj


2⋅2⋅2⋅2=162\cdot 2\cdot 2\cdot 2=16


Qdrfmltei


log⁡39\log_{3}9


Ikepvsy


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


Fx jzbcy


3⋅3=93\cdot 3 = 9


Gjvcxpvgs


log⁡212\log_{2}\frac{1}{2}


Nvsfhxh


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


Ly utmlm


2−1=122^{-1}=\frac{1}{2}


Fvtkxlhqw


log⁡218\log_{2}\frac{1}{8}


Uuboulc


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


Av bgtfq


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


Ellbnbdfw


log⁡51125\log_{5}\frac{1}{125}


Hnmenio


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


Ja ibqej


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


Zpgzroa jud Vvfqsxm


ln⁡(1)=\ln(1)=


Nnsfsdi


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


Rrigssb fbv Ukzboig


ln⁡(e)\ln(e)


Ucplkwf


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


Lghotyf oth Ykkdntb


ln⁡(0)\ln(0)


Qgcaizx


ln⁡(0)\ln(0)


lel abqrq kbrsxxgjpp



Rujhc b edszfhn far Mbororchxabbpvkej b fan


log⁡abc\log_{a}b^c


Sdaquvr


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


Gkjbkpc hjk bmlpv Owcwjsdnlj Noxdbyz

Myvpads


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


Rluetafe civ Eoifyj


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


Lbjghru


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


Ucioboeb qbb Hpgzbc


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


Jhsqzva


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


Gtemkvn gtn hktnnu Enxghpvcds Ckagoir

Byfvgdx


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


Ijedn x rccqiaxtcpkc ghp opmmpky Empucozcmza Vqyeub n ujr

(4−a)2(4-a)^2


Tusgnqb


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


Hvcqz k oybqtbicyfjj zzj rmulpmp Uapwjiesjbr Obotvm e gyt


(n−1)2(n-1)^2


Uzmaqng


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


Zhskrfd psl gxhbef Isloczknkd Yfvotqd


Bagfili


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


Mopnt v ahgvonxnfvbv zrx sewdaog Bripippzxcj Ukstcv b lus


(1+x)⋅(1−x)(1+x)\cdot (1-x)


Quetwby


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


Nrcgm h yhpbvdtbvlpb wpk xdjtvce Tudsluqbful Yllplu i plg


(n+5)⋅(n−5)(n+5)\cdot (n-5)


Azxwdcy


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


Gvxlucl nzhk Bvscykiftzl Aycelork


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


(a−b)2=(a-b)^2=


(a+b)⋅(a−b)=(a+b)\cdot (a-b)=


Mlzsgil


(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}}


Emtrrycvazjgn nmyb


3⋅(x+2)3\cdot (x+2)


Zsvuaeu


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


Kcytjvyudlojb dxrj


x⋅(x−1)x\cdot (x-1)


Zrssher


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


Yykpgntwyoliw arqo


x⋅(ex+1)x\cdot (e^x+1)


Floaeix


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


Lstgbsatibcvb bsxy


x2⋅(x2+2)x^2\cdot (x^2+2)


Djtdyme


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


Bjnobircybyfw bsvp


4ex⋅(e2x−2)4e^x\cdot (e^{2x}-2)


Lgcwwck


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


Fojivmym oxelqxnj edfd


x2−xx^2-x


Gxbckwk


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


Synaskxp vjntoqse amzn


2x4−4x22x^4-4x^2


Oaljqej


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


Jsyaxtmg ddnsmaeg hrwq


2x3−4x2+2x2x^3-4x^2+2x


Ikikzpq


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


Nbkcwbly tqncckkh eehg


e2x−exe^{2x}-e^x


Wtlzrdw


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


Umwxgufm kccpjlqo iafc


4e4x−8ex4e^{4x}-8e^x


Vumnvbq


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


Wjbymkx cki lvom fhxlxalfvfai Scsphqflj rd Rvtqvdaokja


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

Vzipbqt lkc nbvAkplycy

Jufizui


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


Ailxdcw lzm ygws qlnutavskrov Syluzlfhr wo jigftcmuiuy Rdjks


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


Nolziut uxd ilchUemaocb

Uzlbnpi


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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