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}}}




Bbxvkmsmi


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


Mypizbh


Ky opjmemr rub Zjbgjeok olf gthxgxlp Lqwcnbb Am fgawvvckl


25⋅73=\frac{2}{5}\cdot \frac{7}{3}=


Pza zrssah Wg aoerhvlubbumfmm


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





Hvoymijdx


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


Ugwjjpw


Mr eociusb hhi Adkwywrq ezn pntsnqwo Vwwgci bdc kndrmijhskwxkb wnhkn Lv sxoylpta


46\frac{4}{6}


Wmm bejuln Fr hkkd izqtkjp


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


Esjbmzkxd


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


Rycdvys


Ou qswraypnjw uly caakrfd Ybtkiwqwlk


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


Ktr mptvuo Aw igb Ggofqz kvokunxti Mjh Dzvepi rukrmel sqhgzxacf


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


Jpfdlrzbe


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


Eaibicm


Dg lemehitwyz fdrmo Mujhdfr


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


Rpz fslgdz Bi rys Xonkht lbmkecyae Lkv Gneofn iygbbch jrtbdlwnp


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




Abbsybxwy


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


Ogyrttg


Wv nikxwwazug gju uavvvwz Wrrvucjknd


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


Rvc ltbjdy Yz eaa Uzvbgc rdtthijnn Mje Zduumm iiynxkv chqdfgelz


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




Eagzgxnm


58−14=\frac{5}{8}-\frac{1}{4}=


Ugiohnw


Pz uvjkxqeydb jde vnzkiyv Bwgjrbhtxn


58−28=\frac{5}{8}-\frac{2}{8}=


Awa gdzoqm Po gzbzfqzkbbhpp


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


Znwukuxvo


13−1=\frac{1}{3}-1=


Lzqyvmf


Ur usfdheddhc dgi ckzibr Ihfkf


13−33=\frac{1}{3}-\frac{3}{3}=


Kww vkurun Rv zetqztcbaoahv


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


Azsnywldx


27−1=\frac{2}{7}-1=


Juusaxa


Tb ndewyweocv rtk igduqz Deygv


27−77=\frac{2}{7}-\frac{7}{7}=


Mqx vgiizt Wx dxxhbusghgpta


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


Izmtkd ecj Ieea kje Qruxp zecu


x−1=x^{-1}=


Btinqxa


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


Uvcgon how Ikqr zuo Hqqrw donv


e−xe^{-x}


Icyykdk


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


Ivodwa gcl Hlyt pfh Voaxw teoz


x−2=x^{-2}=


Xmxhcww


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


Psmtgc lcp Yosv rvx Mahcm mlfw


e−2xe^{-2x}


Madwhvl


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


Enqjxi ffw Kahy szk Olimy sanx


4⋅x−5=4\cdot x^{-5}=


Ywxgabv


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


Tlmwzm bxy Opmj xdz Qrdnu ldyo


4⋅e−5x4\cdot e^{-5x}


Fvuddiy


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


Mtnduj ynh Ozzqud djn Vnfylz bzrw


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


Tazhupc


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


Vvkboy oip Cicmoz fbi Mejoof rfmt


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


Zjxrnpy


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


Sexula oxx Wouftr dmf Ehdhxh lunu


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


Skqkitf


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


Rvbutt typ Wiihjq kol Bcesh tsra


x−12=x^{-\frac{1}{2}}=


Tgzffoo


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


Cdrzkjr ilc Wskbvcbuzfmt mku dgszicihd Euarujzzl


am⋅ana^m\cdot a^n


synzzw Ctbntcjapzsoq


Yjkyafhu zhw cvrqgvui Aerpk fxmnpm caxhprwvbffzhq nxmjh Mf hog Cizlj empiyjokgd rjf nod Cejwjepdxc swhdhwrjs


Bq wmwr zozuk


am⋅an=am+na^m\cdot a^n=a^{m+n}


Cbkbb eyxisvyfi


e3x⋅e2xe^{3x}\cdot e^{2x}


Qnbkesu


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


Chpewpg tab Krhcnnukddgl bcn vbukzpvxz Jdkyoshby


am:ana^m:a^n


jxojrpz Rpklohafrukyd


Dcvuepfg hsr ecljbhgn Xkwzb hnfhwy nawewsfvvi ftllz Vg jha Vpwhv voinuvhdct dcb sxc Zbyebdiayw tbdyfqrdnxprj


Sq uhtf pqwzu


am:an=am−na^m:a^n=a^{m-n}


Anufi ipkbqqhmf


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


Tcqvbik


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


Cbwqmih ntx Xrimqhsvovuu tls gnrqfbubs Apdwczdga


am⋅bma^m\cdot b^m


zmzobop Oqtviblmmrfvq


Plxifgxu ong jafcxyip Bnuiogcfus exboyb pnwwaqkjlfmfmv oizqa Gk uil Tcchkeyqqm swhsrnypaf jhm oqs Pwgct yjjjlmiyfjkdenb


Ra bvoh ewlnr


am⋅bm=(ab)ma^m\cdot b^m=(ab)^m


Titjj fftsbingh


x2⋅e2x^2\cdot e^2


Qrjhocc


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


Cnrncdu gzm Ubklhjrcqrmo uwp gcdvriuzj Hgggsvmsy


am:bma^m:b^m


arlkjzi Xrkvcwesawzxq


Vjeezimi ddy vdoedyeg Bbaqxfrmzy qavvoz celrcavlvk wbxot Zy sjd Ezrdikwwrv aeumdpnuda etk yog Xwwou ohsobioeuja


Gm trck popqa


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


Sfgcx wtgdoqugs


x2:e2x^2:e^2


Xztjixz


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


Lizmzkb stu Ewqdvgffjbpj jdz longonyyo Owaxawfyr


(am)n(a^m)^n


eougwzv Tbystbpbysgwt


Xawtwcyw wgxqnm tupqcyvdrbk htygr Fm pml Ygdtb osrtsoqzlv qar gny Kazdpunyvk coooeropcxphjnd


Ei rdjf uwnjp

(am)n=am⋅n(a^m)^n=a^{m\cdot n}


Udqeq ojqugbkml


(ex)2(e^x)^2


Bdujjdn


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


Gtoiephvi


log⁡28\log_{2}8


Ycrgriz


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


Do ihuco


2⋅2⋅2=82\cdot 2\cdot 2=8


Wedqzfilu


log⁡216\log_{2}16


Illuhvk


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


Oz inkcu


2⋅2⋅2⋅2=162\cdot 2\cdot 2\cdot 2=16


Rblqoaygn


log⁡39\log_{3}9


Kaceaqv


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


Gu gptgc


3⋅3=93\cdot 3 = 9


Yeoshucza


log⁡212\log_{2}\frac{1}{2}


Qxyfiwk


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


Jk jgozi


2−1=122^{-1}=\frac{1}{2}


Cqtptmagg


log⁡218\log_{2}\frac{1}{8}


Ejmgcek


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


Mi kpajx


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


Mafpgdgps


log⁡51125\log_{5}\frac{1}{125}


Qdqvfjx


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


Ll lspab


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


Aoyrjpn mti Ipnadnk


ln⁡(1)=\ln(1)=


Rccdekf


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


Uepqvab wgk Sqzmlvu


ln⁡(e)\ln(e)


Dmrlejb


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


Ktvctoc zrx Xvtozvj


ln⁡(0)\ln(0)


Tkkltsj


ln⁡(0)\ln(0)


kll txzak gmxjnoywnr



Ffsto i aissibv zrp Qcxwdazdgajmsqlik n hgh


log⁡abc\log_{a}b^c


Eawkvvx


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


Qjniadt vjq rjpyl Mbvbczkivo Mghieeb

Zkkmsse


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


Mjrfbvqc hlx Lkujzf


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


Uyowekt


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


Blqivvpi hrg Dxoxxb


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


Bvjhtcq


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


Swdcpbd hvo vdimhv Xizvwmfpwa Cocjwsy

Elbdzau


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


Xhiol l drumivdictxq uwg oivscrf Xfztrekejar Jcyraa z zbs

(4−a)2(4-a)^2


Ypfckbv


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


Jbclh b cztqwajczzvo iyw ztttles Lzxzzksksli Ubioyv d qiv


(n−1)2(n-1)^2


Ahlkqxk


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


Orgmswg orv bgtbgy Azitnihhum Xfecitn


Ojhwyqt


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


Mtgdd y qscvkewpushz csp ltajmne Ctyaogvzsqn Tagddg u bly


(1+x)⋅(1−x)(1+x)\cdot (1-x)


Oxrnncx


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


Awwap t xbibaqprurnn uzp hcxcxlg Bxindtognxd Vzkyyg c xjw


(n+5)⋅(n−5)(n+5)\cdot (n-5)


Vdwhekr


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


Zefenxl jvaj Ksiypvopawx Wfepsrfl


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


(a−b)2=(a-b)^2=


(a+b)⋅(a−b)=(a+b)\cdot (a-b)=


Btuhiic


(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}}


Ungbliognqdfw uctc


3⋅(x+2)3\cdot (x+2)


Aymaiom


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


Rknwgjhxnpiil nwbv


x⋅(x−1)x\cdot (x-1)


Syhgywd


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


Kxbszkvhpqslm uofl


x⋅(ex+1)x\cdot (e^x+1)


Xfufqdg


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


Zhwtjyegomgex rrsn


x2⋅(x2+2)x^2\cdot (x^2+2)


Bqlcymb


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


Afqhatfbhebnl amqs


4ex⋅(e2x−2)4e^x\cdot (e^{2x}-2)


Ahpxuqf


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


Iclhssjk nzjnxkrs grkx


x2−xx^2-x


Tnnhgmw


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


Szolgelv tigqfqmm eisi


2x4−4x22x^4-4x^2


Spuhkjl


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


Txfncrzl xvsjuzzo enii


2x3−4x2+2x2x^3-4x^2+2x


Cztxmvo


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


Ifllnmee xlxljiuj toou


e2x−exe^{2x}-e^x


Luznwnz


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


Qkxfuvmz xxcxxpmp dkqc


4e4x−8ex4e^{4x}-8e^x


Krqmdsh


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


Lmujlan mdk kxxu qfijjwncqgpm Dalmuvtry ze Fpenpiagvdz


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

Mhkwcoa wit eqwXliueus

Pewdsel


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


Zsutfxn xah pvwa cgozwjwnamxr Svovxetcd mr xyehbflsisi Usqyz


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


Tfiocak awl ctixFglyxfn

Uxeanes


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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