Ahlan wa Sahlan: Functional Modern Standard Arabic for by Mahdi Alosh

By Mahdi Alosh

This textbook is designed to hide the 1st yr of guideline in glossy average Arabic. followed through an instructor's handbook and an audio programme, it's going to educate scholars to learn, converse, and write Arabic. The textual content offers an enticing tale that contains Adnan, a Syrian scholar learning within the united states, and Michael, an American pupil learning in Cairo. In diaries, letters, and postcards, the 2 scholars describe their suggestions and actions, revealing how a non-American perspectives American tradition and the way the Arabic tradition is skilled by way of an American pupil. The textual content additionally presents information regarding the geography of the Arab global, well-known characters in heritage, festivities in Arab tradition, the media, way of life, and the relations. routines in comprehension, vocabulary, grammar, and writing attend to either shape and which means and enhance useful skills and data in regards to the Arabic sound, writing, and language structures.

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Extra info for Ahlan wa Sahlan: Functional Modern Standard Arabic for Beginners

Example text

Hence, (a + 1)x = (b + 1)y = (c + 1)z . Let (a + 1)x = (b + 1)y = (c + 1)z = k , then (ax + by + cz) + (x + y + z) = 3k (i). Add up the given equalities in the problem: x + y + z = 2(ax + by + cz) ax + by + cz = k , (ii). (i)(ii) lead to thus a b c ax by cz ax + by + cz k + + = + + = = = 1. 122 Consider a cuboid whose length, width, height are positive integers m, n, r and m ≤ n ≤ r . We paint red color on the surface of the cuboid completely and then chop it into cubes with side length 1. If we know that the number of cubes without red face plus the number of cubes with two red faces minus the number of cubes with one red face is 1985, find the values of m, n, r .

Thus an = (a + b + c)an−1 − (ab + bc + ca)an−2 + (abc)an−3 . Similarly, bn = (a + b + c)bn−1 − (ab + bc + ca)bn−2 + (abc)bn−3 and cn = (a + b + c)cn−1 − (ab + bc + ca)cn−2 + (abc)cn−3 . Add the above three equalities together: F (n) = (a + b + c)F (n − 1) − (ab + bc + ca)F (n − 2) + (abc)F (n − 3) . In addition, we know a2 + b2 + c2 + 2ab + 2bc + 2ca = (a + b + c)2 2 +b2 +c2 2 3 3 1 − + b= +0,cab −+3abc c)(a =+− b21 + c2 −abc ab = − bc ca)b3. When a +1 Fb (3), +cF =(n) 0, = and a3 (1) bc + = ca(a =+ −ba + F (2), (a3 + + c3 ) = 2 2 3 3 2 2 2 2 +b 1 1 a2 +c +b2 +c 1 13 3 3 3 3 3 1 1 2 a 1 1 +b2= +c2 then2(1) F (2)F F (3)F (1) =(n 0,−+ ab2) ++ bc3ca + ca − =(n − − 3)2 = − =− F (2), (a ++ b0,cab +)c+ =F (3), F (3), Fa(n) = 0, ab bc + = F (2), abcabc = 3=(a3(1) + b= =)bc F (n) = = − 12 F (2), a 2 2 2 3 +3 ca = − 2 1 1 1 2 1 1 2 2 2 1 1 F (2)F (n2) −+ 2) +F 3(3)F F (3)F (n3) − 3) +c n =1 4 , we have F 1(2)F a +b 3F 2) n(n= 5 3) F (2)F (n = − (n − , we (nb2− F1(3)F (1) 0, ab + = − 2 F (2), abc 2= 3(4) (a3 = + +(2) c3+).

When z = 0 , then x + y = 2001 , then there are 2002 possibilities of (x, y) . When z = 1 , then x + y = 1999 , then there are 2000 possibilities of (x, y) . . . . When z = 1000 , then x + y = 1 , then there are two possibilities of (x, y) . As a conclusion, a2001 = 2002 + 2000 + 1998 + · · · + 4 + 2 = 1003002 . 28 Solve the equation x2 + x − 2x x − 2 − 6 = 0 . √ √ √ √ √ √ 2 2 Solution: + x2 −x2x = (x 4⇔ =4⇔ x√ + xx√ − 2xx −x2x −2x −− 6 2=−0 6⇔=x02 ⇔ − 2x − 2 x+−x 2−+2 x=−42⇔ − (xx−− 2)x2−=2)42⇔ 2= .

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