‫ﺃﻭﻟﻤﺒﻴﺎﺩ ﺍﻟﺮﻳﺎﺿﻴﺎﺕ ‪2011‬‬

‫ﺍﻟﻔﺮﺽ ﺍﻷﻭﻝ‬
‫ﺍﻟﺠﻤﻌﺔ ‪ 26‬ﻧﻮﻧﺒﺮ ‪2010‬‬ ‫ﺍﻟﺴﻨﺔ ﺍﻟﺪﺭﺍﺳﻴﺔ‪2011/2010 :‬‬
‫­ ﺣﻠﻮﻝ ﻣﻘﺘﺮﺣﺔ ­‬
‫ﻣﻦ ﺍﻗﺘﺮﺍﺡ‪ :‬ﺃﺫ ﺳﻤﻴﺮ ﻟﺨﺮﻳﺴﻲ‬

‫ﺗﻤﺮﻳﻦ ‪1‬‬
‫ﻃﺮﻳﻘﺔ‪1‬‬
‫‪1‬‬ ‫‪a‬‬ ‫‪b‬‬
‫‪ 2‬ﺣﻴﺚ ‪a + b = a b‬‬ ‫‪+ 2‬‬ ‫‪³‬‬ ‫ﻟﻨﺒﻴﻦ ﺃﻥ‬
‫‪b +4 a +4 2‬‬
‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪2‬‬
‫ﻧﻌﻠﻢ ﺃﻥ‪ (a + b ) ³ 4 ab :‬ﻣﻨﻪ‪ ab ³ 4ab :‬ﻣﻨﻪ‪ ab ³ 4 :‬ﻣﻨﻪ‪:‬‬
‫‪2‬‬
‫‪2‬‬
‫‪³ 2‬‬ ‫‪ 2‬ﻭ ﺃﻳﻀﺎ‬ ‫‪³ 2‬‬
‫‪a + 4 a + ab‬‬ ‫‪b + 4 b + ab‬‬
‫‪a‬‬ ‫‪b‬‬ ‫‪a‬‬ ‫‪b‬‬
‫‪2‬‬
‫‪+‬‬ ‫‪2‬‬
‫‪³‬‬ ‫‪2‬‬
‫‪+‬‬ ‫‪2‬‬
‫ﻣﻨﻪ‪:‬‬
‫‪b + 4 a + 4 b + ab a + ab‬‬
‫‪a‬‬ ‫‪b‬‬ ‫‪a 1‬‬ ‫‪b 1‬‬ ‫‪a‬‬ ‫‪b‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪a 2 + b2‬‬
‫‪+‬‬ ‫=‬ ‫‪.‬‬ ‫‪+‬‬ ‫‪.‬‬ ‫=‬ ‫‪+‬‬ ‫=‬ ‫‪+‬‬ ‫=‬
‫‪b 2 + ab a 2 + ab b a + b a a + b b 2 a a 2 b b 2 a 2‬‬ ‫‪a 2b 2‬‬
‫‪2‬‬
‫ﻭ ﻟﺪﻳﻨﺎ‪:‬‬
‫=‬
‫(‬‫‪a + b ) - 2ab a 2 b 2 - 2ab‬‬
‫=‬ ‫‪= 1-‬‬
‫‪2‬‬
‫‪a 2b 2‬‬ ‫‪a 2b 2‬‬ ‫‪ab‬‬
‫‪a‬‬ ‫‪b‬‬ ‫‪1‬‬ ‫‪2 1‬‬
‫‪2‬‬
‫‪+ 2‬‬ ‫‪ ab ³ 4 Þ 1 -‬ﻓﺈﻥ‪³ :‬‬ ‫ﻭ ﺑﻤﺎ ﺃﻥ‪³ :‬‬
‫‪b +4 a +4 2‬‬ ‫‪ab 2‬‬
‫ﻃﺮﻳﻘﺔ‪:2‬‬
‫‪a +b = ab‬‬ ‫ﻧﻀﻊ‪= t :‬‬
‫ﺇﺫﻥ‪:‬‬
‫‪a‬‬
‫‪+‬‬
‫‪b‬‬
‫=‬
‫‪a 3 + 4a + b 3 + 4b‬‬
‫=‬
‫(‬
‫) ‪(a + b )(a 2 + b 2 - ab) + 4(a + b ) = (a + b ) (a + b )2 - 3ab + 4(a + b‬‬‫)‬
‫ﺇ‬
‫‪2‬‬
‫‪b +4‬‬ ‫‪2‬‬
‫‪a +4‬‬ ‫(‬ ‫)‬
‫‪a 2 b 2 + 4(a 2 + b 2 ) + 16 a 2 b 2 + 4 (a + b )2 - 2ab + 16‬‬ ‫‪2‬‬
‫‪a 2 b 2 + 4(a + b ) - 8ab + 16‬‬
‫‪t (t - 3t ) + 4t‬‬
‫‪2‬‬ ‫‪3‬‬
‫‪t - 3t + 4t‬‬‫‪2‬‬
‫=‬ ‫‪2‬‬ ‫‪2‬‬
‫‪= 2‬‬
‫‪t + 4t - 8t + 16 5t - 8t + 16‬‬
‫‪"t ³ 4‬‬ ‫ﺫﻥ ﻟﻠﺒﺮﻫﺎﻥ ﻋﻠﻰ ﺍﻟﻤﺘﻔﺎﻭﺗﺔ ﺍﻟﻤﻄﻠﻮﺑﺔ ﻳﺠﺐ ﺍﻟﺒﺮﻫﺎﻥ ﻋﻠﻰ ﺃﻥ‪f (t ) = 2(t 3 - 3t 2 + 4t ) - (5t 2 - 8t + 16) ³ 0 :‬‬
‫‪ "t ³ 4‬ﻭ ﻫﺬﺍ ﻳﻨﻬﻲ ﺍﻟﺒﺮﻫﺎﻥ‬ ‫ﻟﺪﻳﻨﺎ‪f (t ) = 2t 3 - 11t 2 + 16t - 16 = (t - 4)(2t 2 - 3t + 4) ³ 0 :‬‬
‫‪2‬‬
‫)ﻷﻥ‪ :‬ﻣﺤﺪﺩﺓ ‪ 2t - 3t + 4‬ﺳﺎﻟﺒﺔ ( )ﻗﻤﻨﺎ ﺑﺈﺟﺮﺍء ﺍﻟﻘﺴﻤﺔ ﺍﻹﻗﻠﻴﺪﻳﺔ ﻟﻠﺘﻌﻤﻴﻞ(‬
‫ﻳﻤﻜﻦ ﺃﻳﻀﺎ ﺩﺭﺍﺳﺔ ﺍﻟﺪﺍﻟﺔ ﺍﻟﺴﺎﺑﻘﺔ ﻭ ﺍﻻﺳﺘﻨﺘﺎﺝ ﻣﻦ ﺧﻼﻝ ﺟﺪﻭﻝ ﺍﻟﺘﻐﻴﺮﺍﺕ‬

‫ﺭﻳﺎﺿﻴــــﺎﺕ ﺍﻟﻨﺠـــــــــــﺎﺡ‬
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 2 ‫ﺗﻤﺮﻳﻦ‬
ìx + y + z + t = 4
(S ) : ïí 1 1 1 1 1 ‫ﻟﺘﺤﻞ ﺍﻟﻨﻈﻤﺔ‬
+
ïx y z t+ + = 5 -
î x y zt
(S ) Þ x + y + z + t + 1 + 1 + 1 + 1 ³ 8 Þ 5 - 1 ³ 4 Þ x y z t ³ 1 :‫ ﻣﻨﻪ‬٬ "x > 0 x + 1 ³ 2 :‫ﻧﻌﻠﻢ ﺃﻥ‬
x y z t x y zt x
: ‫ﻭ ﻣﻦ ﺟﻬﺔ ﺃﺧﺮﻯ‬
x + y + z + t ³ 2 xy + 2 z t ³ 2 2 xy 2 z t = 4 x y zt Þ 4 ³ 4 x y zt Þ x y zt £ 1
x y z t = 1 :‫ﺇﺫﻥ‬
ìx + y + z + t = 4
1 1 1 1
(S )Þ ïí 1 1 1 1 Þ x+ -2+ y+ - 2+ z+ - 2+t + -2 = 0
ïx + y + z + t = 4 x y z t
î :‫ﻣﻨﻪ‬
2 2 2 2
æ 1ö æ 1ö æ 1 ö æ 1ö
Þ ç x - ÷ + çç y - ÷÷ + ç z - ÷ + ç t - ÷ = 0 Þ x = y = z = t = 1
è xø è yø è zø è t ø
S = {(1;1;1;1)} :‫ ﺑﺎﻟﺘﺎﻟﻲ‬٬‫ﻋﻜﺴﻴﺎ ﻳﻤﻜﻦ ﺍﻟﺘﺤﻘﻖ ﺑﺴﻬﻮﻟﺔ ﻣﻦ ﺻﺤﺔ ﺍﻟﺤﻞ‬

3 ‫ﺗﻤﺮﻳﻦ‬
"x Î IR f ( x) = max (2 x y - f ( y)) ‫ﻟﺪﻳﻨﺎ‬
yÎIR

"x Î IR "y Î IR f ( x) ³ 2 x y - f ( y) :‫ﺇﺫﻥ‬

(* ) "x Î IR f ( x) ³ x 2
:‫ﻣﻨﻪ‬ "x Î IR f ( x) ³ 2 x 2 - f ( x) :‫ ( ﻧﺠﺪ‬x = y) ‫ﻭ ﺑﺄﺧﺬ‬

f ( x) - x 2 = 2 xx0 - f ( x0 ) - x 2 :‫ ﻣﻨﻪ‬f ( x) = 2 xx0 - f ( x0 ) :‫ ﺣﻴﺚ‬x0 Î IR ‫ ﺇﺫﻥ ﻳﻮﺟﺪ‬x Î IR ‫ﻟﻴﻜﻦ‬

2
f ( x) £ x 2 :‫ﻣﻨﻪ‬ f ( x) - x 2 £ 2 xx0 - x02 - x 2 = -( x - x0 ) :‫ ﻣﻨﻪ‬f ( x0 ) ³ x02 :‫ﻭ ﺣﺴﺐ ) * ( ﻓﺈﻥ‬

"x Î IR f ( x) = x 2 :‫ﺑﺎﻟﺘﺎﻟﻲ‬

"x Î IR "y Î IR x 2 ³ 2 x y - y 2 2
:‫ ﻟﺪﻳﻨﺎ‬x a x ‫ﻭﻋﻜﺴﻴﺎ ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﺪﺍﻟﺔ‬
"x Î IR "y Î IR x 2 - 2 x y + y 2 ³ 0 :‫ﻷﻥ‬
"x Î IR $y Î IR x2 = 2 x y + y 2 ‫ﻭ‬
D = 4 x 2 + 4 x 2 = 8 x 2 ³ 0 :‫ ﻫﻲ‬t Î IR t 2 + 2 xt - x 2 = 0 ‫ﻷﻥ ﻣﺤﺪﺩﺓ ﺍﻟﺤﺪﻭﺩﻳﺔ‬

.‫" ﻫﻲ ﺍﻟﺪﺍﻟﺔ ﺍﻟﻮﺣﻴﺪﺓ ﺍﻟﺘﻲ ﺗﺤﻘﻖ ﺍﻟﺸﺮﻁ ﺍﻟﻤﻄﻠﻮﺏ‬x Î IR f ( x) = x 2 ‫ﻭ ﻫﺬﺍ ﻳﻌﻨﻲ ﺃﻥ ﺍﻟﺪﺍﻟﺔ‬

‫ﺭﻳﺎﺿﻴــــﺎﺕ ﺍﻟﻨﺠـــــــــــﺎﺡ‬
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 ‫ﺗﻤﺮﻳﻦ ‪4‬‬
‫ﻣﻊ‬ ‫ﺑﻤﺎ ﺃﻥ ‪ L‬ﻫﻲ ﻧﻘﻄﺔ ﺗﻘﺎﻃﻊ ﺍﻟﻤﻨﺼﻒ ﺍﻟﺪﺍﺧﻠﻲ ﻟﻠﺰﺍﻭﻳﺔ ] ‪[BAˆ C‬‬
‫‪BL AB‬‬
‫=‬ ‫] ‪ [BC‬ﻓﺈﻥ‪:‬‬
‫‪A‬‬ ‫‪LC AC‬‬
‫ﻣﻨﻪ‪AC = 2 AB :‬‬
‫ﻟﻨﻌﺘﺒﺮ ‪ E‬ﻣﻨﺘﺼﻒ ] ‪ [AC‬ﺇﺫﻥ‪AE = AB :‬‬
‫‪K‬‬ ‫ﺇﺫﻥ ﻧﺴﺘﻨﺘﺞ ﺑﺴﻬﻮﻟﺔ ﺃﻥ ﺍﻟﻤﺜﻠﺜﺎﻥ ‪ AED‬ﻭ ‪) ABD‬ﺿﻠﻊ ﻣﺸﺘﺮﻙ ﻭ‬
‫‪E‬‬ ‫ﺿﻠﻌﺎﻥ ﻣﺘﻘﺎﻳﺴﺎﻥ ﻭﺍﻟﺰﺍﻭﻳﺔ ﺍﻟﻤﺤﺼﻮﺭﺓ ﺑﻴﻨﻬﻤﺎ ﻣﺘﻘﺎﻳﺴﺎﻥ(‬
‫ﻣﻨﻪ‪BD = DE :‬‬
‫‪M‬‬ ‫‪ DA‬ﺯﺍﻭﻳﺘﺎﻥ ﻣﺤﻴﻄﻴﺘﺎﻥ ﻣﺘﻘﺎﻳﺴﺘﺎﻥ‬ ‫ﻣﻦ ﺟﻬﺔ ﺃﺧﺮﻯ ‪ˆ D‬‬
‫‪ BA‬ﻭ ‪ˆ C‬‬
‫ﺇﺫﻥ ﻓﻬﻤﺎ ﺗﺤﺼﺮﺍﻥ ﻗﻮﺳﻴﻦ ﻣﺘﻘﺎﻳﺴﻴﻦ‬
‫‪B‬‬ ‫‪L‬‬ ‫‪C‬‬ ‫ﻣﻨﻪ‪BD = DC :‬‬
‫ﻧﺴﺘﻨﺘﺞ ﺇﺫﻥ ﺃﻥ‪DE = DC :‬‬
‫ﺇﺫﻥ ) ‪ (DM‬ﺍﺭﺗﻔﺎﻉ ﻓﻲ ﺍﻟﻤﺜﻠﺚ ‪ DEC‬ﺍﻟﻤﺘﺴﺎﻭﻱ ﺍﻟﺴﺎﻗﻴﻦ ﻓﻲ‬
‫‪ D‬ﺇﺫﻥ ﻓﻬﻮ ﺃﻳﻀﺎ ﻣﺘﻮﺳﻂ ﻭ ﻭﺍﺳﻂ‬
‫‪D‬‬
‫‪EC AC‬‬
‫= ‪MC‬‬ ‫=‬ ‫ﻣﻨﻪ‪:‬‬
‫‪2‬‬ ‫‪4‬‬
‫‪AM 3‬‬ ‫‪3 AC‬‬
‫= ‪ AM = AC - MC‬ﺑﺎﻟﺘﺎﻟﻲ‪= :‬‬ ‫ﻣﻨﻪ‪:‬‬
‫‪MC 4‬‬ ‫‪4‬‬

‫ﺭﻳﺎﺿﻴــــﺎﺕ ﺍﻟﻨﺠـــــــــــﺎﺡ‬
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