Hai cercato la traduzione di x x x vedeo da Arabo a Inglese

x x x vedeo
xxx vedeo

x vedeo
x

x-x-x
take her from behind.

رائع x-x-x
good. that's it. now, take him in your mouth.

x x
x x

x x x bedroom p
xxx bedroom

إذهب وإغتسل x-x-x
go and get yourself cleaned up!

hindi x x x b f
hindi xxxbf

xc x x x bp hd video
xc xxx bp hd video

x x video
xx video

xc x x x bp hd video hindi
xc xxx bp hd video hindi

xvide x x x video o xx/hd
xvide xxx video o xx / hd

الأدنى هو عند x x 3
minimum is at x = %1, %2(x) = %3

सेक्सी वीडियो hot x x desi
सेक्सी वीडियो hot x x desi

x** x** video hd xx hd
x ** x ** video hd xx hd

أميال هل حصلت على ال "x" "x"!
miles you got the "x" cleaver: miles and xeropopolous sittin' in a tree

2 - يكون هذا البروتوكول مفتوحا، بعد تاريخ x// x x x x x، لانضمام أي دولة غير موقعة.
this protocol shall, after x/x/xxxx, be open for accession by any non-signatory state.

يفتح باب التوقيع على هذا البروتوكول أمام جميع الدول في مقر الأمم المتحدة لمدة اثني عشر شهرا تمتد من x// x x x x x إلى x// x x x x x.
this protocol shall be open for signature by all states at united nations headquarters for the twelve months from x/x/xxxx to x/x/xxxx.

يفتح باب التوقيع على هذا البروتوكول أمام جميع الدول في مقر الأمم المتحدة لمدة اثني عشر شهرا تمتد من x/x/xxxx إلى x/x/xxxx.
this protocol shall be open for signature by all states at united nations headquarters for the twelve months from x/x/xxxx to x/x/xxxx.

p(x) = x³ +2x² -x -2الجذور الممكنة = {1, -1, 2, -2}* p(1) = 0 → x1 = 1* p(-1) = 0 → x2 = -1* p(2) = 12 → 2 ليس جذرا لكثيرة الحدودوباقي (x³ +2x² -x -2)/(x-2) هو 12* p(-2) = 0 → x3 = -2p(x) = x³ +2x² -x -2الجذور الممكنة = {1, -1, 2, -2}* p(1) = 0 → x1 = 1→ x³ +2x² -x -2 = (x-1)(x² +3x +2)هنا, r1=-1 وq(x) = x² +3x +2* q(-1) = 0 → x2 = -1→ x³ +2x² -x -2 = (x-1)(x² +3x +2) = (x-1)(x+1)(x+2)having used the ""p"/"q"" result above (or, to be fair, any other means) to find all the real rational roots of a particular polynomial, it is but a trivial step further to partially factor that polynomial using those roots.
examples:=====finding roots without applying ruffini's rule=====p(x) = x³ +2x² -x -2possible roots = {1, -1, 2, -2}*p(1) = 0 → x1 = 1*p(-1) = 0 → x2 = -1*p(2) = 12 → 2 is not a root of the polynomialand the remainder of (x³ +2x² -x -2)/(x-2) is 12*p(-2) = 0 → x3 = -2=====finding roots applying ruffini's rule and obtaining a (complete) factorization=====p(x) = x³ +2x² -x -2possible roots = {1, -1, 2, -2}*p(1) = 0 → x1 = 1then, applying ruffini's rule:(x³ +2x² -x -2) / (x -1) = (x² +3x +2) →→ x³ +2x² -x -2 = (x-1)(x² +3x +2)here, r1=-1 and q(x) = x² +3x +2*q(-1) = 0 → x2 = -1again, applying ruffini's rule:(x² +3x +2) / (x +1) = (x +2) →→ x³ +2x² -x -2 = (x-1)(x² +3x +2) = (x-1)(x+1)(x+2)as it was possible to completely factorize the polynomial, it's clear that the last root is -2 (the previous procedure would have given the same result, with a final quotient of 1).