$Q24)solve in [0,\pi ]$
$sin^{10}x+cos^{6}x=1$
"ÇáÑíÇÖíÇÊ åí Êáß ÇáãÊÚÉ ÇáÊí íÈÍË ÚäåÇ ÇáÃÐßíÇÁ æíÍÇæáæä ÇÓÊßÔÇÝ ÃÓÑÇÑåÇ æÍá ãÌåæáÇÊåÇ"
URL="http://alnasiry.net/forums"][/URL
$Q25)\sqrt[4]{1-x}+\sqrt[4]{15+x}=2$
"ÇáÑíÇÖíÇÊ åí Êáß ÇáãÊÚÉ ÇáÊí íÈÍË ÚäåÇ ÇáÃÐßíÇÁ æíÍÇæáæä ÇÓÊßÔÇÝ ÃÓÑÇÑåÇ æÍá ãÌåæáÇÊåÇ"
$Q26)\left | \frac{x^{2}-4x+3}{x^{2}-7x+10} \right |=-(\frac{x^{2}-4x+3}{x^{2}-7x+10})$
"ÇáÑíÇÖíÇÊ åí Êáß ÇáãÊÚÉ ÇáÊí íÈÍË ÚäåÇ ÇáÃÐßíÇÁ æíÍÇæáæä ÇÓÊßÔÇÝ ÃÓÑÇÑåÇ æÍá ãÌåæáÇÊåÇ"
$Q27)\frac{2sin^{2}x+2cos^{2}(x+\frac{\pi }{4})-1}{\sqrt{6x-x^{2}}}=0$
"ÇáÑíÇÖíÇÊ åí Êáß ÇáãÊÚÉ ÇáÊí íÈÍË ÚäåÇ ÇáÃÐßíÇÁ æíÍÇæáæä ÇÓÊßÔÇÝ ÃÓÑÇÑåÇ æÍá ãÌåæáÇÊåÇ"
$Q28)(x^{2}-6x+6)^{x^{2}-2}=(x^{2}-6x+6)^{4x-5}$
"ÇáÑíÇÖíÇÊ åí Êáß ÇáãÊÚÉ ÇáÊí íÈÍË ÚäåÇ ÇáÃÐßíÇÁ æíÍÇæáæä ÇÓÊßÔÇÝ ÃÓÑÇÑåÇ æÍá ãÌåæáÇÊåÇ"
ÇáÐíä íÔÇåÏæä ÇáãæÖæÚ ÇáÂä: 1 (0 ãä ÇáÃÚÖÇÁ æ 1 ÒÇÆÑ)
Powered by vBulletin® Version 4.2.3
Copyright © 2026 vBulletin Solutions, Inc. All rights reserved.
Translate By
Almuhajir
Developed By Marco Mamdouh
Style
ÊØæíÑ æÏÚã ÔÑßÉ
Zavord