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yeh saval main hare mein kar raha hoon.
problema 76 en verde.
Last Update: 2019-07-06
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dil se nikal ke, pariyon se nikal ke, mai jaa raha hoon marn...
-sin corazones, sin hadas, voy a mor...
Last Update: 2019-07-06
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tumhe pata chal raha hoga ki mein yeh saval tum sabhi ke sath hi kar raha hoon. mein bhi inse aise hi jhooj raha hoon jaise tum jhooj te ho.
pueden ver que estoy resolviendo estos problemas en tiempo real. estoy batallando con ellos como ustedes lo harían.
Last Update: 2019-07-06
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ye (z+1) en mein se koi (z+1) ko cancel out kar dega mein esse cancel kar deta hoon ye 'z' es 'z' ko cancel out kardega or ab hamare pas kya bachta hai sab ab simplify ho gaya hai 7(z-6) over 4(z+1) mein ha ha oh maine edhar (z-b) likh diya ye to (z+2)(z-2) hai vo sab (a^2-b^2) = (a-b)(a+b) mein maine galti kar di (z^2-4) = (z+2)(z-2) hot hai na ki z-b, mujhe laga ye 6 tha to yeh (z-2) hai or yeh (z-2) hai to sahi choice 1 hai galti ke liye sorry . ok, vo ab chahatein ki ab ham ye sab vapas karein vo ab chahatein ki ab ham product find karein (x+5) over (3x+2) times (2x-3) over (x-5) yahan pe ham jyada kuch simplification nahi kar saktein hain hamein bas multiply karna hoga to ye barabar hoga (x+5)(2x-3) divided by 3x mein bas numerator ko multiply kar raha hoon or fir denominator ko multiply kar raha hoon -- ( 3x+2)(x-5) or ham ab dono binomial ko multiply kar dete hain, x*2x = 2x^2 x*(-3) = -3x 5*2x=10x 5*(-3)=(-15) . ab ham denominator ko multiply karte hain 3x*x = 3x^2 3x*(-5) = 15x 2*x = 2x 2*(-5)=(-10) or ab dektein hain ki ham simplify kar saktein hai ya nahi numerator ko dekho 2x^2
esta z mas 1 se cancela con una de estas z mas 1. lo haré con la peor escrita. y veamos, ésta z se cancela con ésta z.
Last Update: 2019-07-06
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