ÇáÒÇÆÑ ÇáßÑíã:ÊÓÊØíÚ ãÔÇåÏÉ ßá ÇáãæÇÖíÚ ÈÏæä ÔÑØ ÇáÚÖæíÉ áßä ãÔÇÑßÊß ÊÍÊÇÌ ÚÖæíÉ ÇáãäÊÏì
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02-07-2018, 10:02 PM
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#13
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The Following User Says Thank You to ãÍãÏ Úáí ÌæÇÏ æÊæÊ For This Useful Post:
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02-07-2018, 10:07 PM
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#14
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The Following User Says Thank You to ãÍãÏ Úáí ÌæÇÏ æÊæÊ For This Useful Post:
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02-07-2018, 11:20 PM
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#15
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02-08-2018, 10:13 AM
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#16
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02-08-2018, 10:18 AM
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#17
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02-08-2018, 10:57 AM
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#18
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02-08-2018, 11:27 AM
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#19
ÑÏ: ãä ÇÑÔíÝ ãßÊÈÊí Ýí ÇáÌÈÑ
ÇáÊÚÏíá ÇáÃÎíÑ Êã ÈæÇÓØÉ ßÇãá ãæÓì ÇáäÇÕÑí ; 02-08-2018 ÇáÓÇÚÉ 12:33 PM
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02-08-2018, 05:45 PM
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#20
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02-08-2018, 10:15 PM
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#21
ÑÏ: ãä ÇÑÔíÝ ãßÊÈÊí Ýí ÇáÌÈÑ
\[\begin{array}{l}
Q3. \\
& \left( a \right)\,{4^{x + 1.5}} + {9^x} = {6^{x + 1}} \\
& \left( b \right)\,{25^{2x - {x^2} + 1}} + {9^{2x - {x^2} + 1}} = 34 \cdot {15^{2x - {x^2}}} \\
& \left( c \right)\,{2^{2{x^2}}} + {2^{{x^2} + 2x + 2}} = {2^{5 + 4x}} \\
& \left( d \right)\,{\left( {\sqrt {5\sqrt 2 - 7} } \right)^x} + 6 \cdot {\left( {\sqrt {5\sqrt 2 + 7} } \right)^x} = 7 \\
\end{array}\]
ÇáÊÚÏíá ÇáÃÎíÑ Êã ÈæÇÓØÉ ßÇãá ãæÓì ÇáäÇÕÑí ; 02-13-2018 ÇáÓÇÚÉ 11:10 PM
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02-09-2018, 01:03 AM
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#22
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02-09-2018, 01:14 AM
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#23
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02-09-2018, 01:25 AM
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#24
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