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सेक्सी पिक्चर!hot
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aur "a" cube divided by "a" square "a" hot hai.
だから １２ a ＾３を 4 a ＾２で割ると 3 a です。

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
消します。 この z はこの z とキャンセルされます。 我々 に何が残っているか。