سلسلة المعادلات

["]$\LARGE {\color{Red} Q10) \log_{4}x+\log_{4}y=-3\Rightarrow xy=\frac{1}{b4}\Rightarrow y=\frac{1}{b4x}}$
$\LARGE {\color{Red} x^{\frac{-1}{2}}+y^{\frac{-1}{2}}=b\Rightarrow x^{\frac{-1}{2}}+8x^{\frac{1}{2}}=b\rightarrow }$
$\LARGE {\color{Red} 8x-bx^{\frac{1}{2}}+1=0\Rightarrow (4x^{\frac{1}{2}}-1)(2x^{\frac{1}{2}}-1)=0\Rightarrow x=\frac{1}{1b} , x=\frac{1}{4}}$
$\LARGE {\color{Red} \Rightarrow y=\frac{1}{4} , y=\frac{1}{1b}}$ [/COLOR]

للتذكير سؤال 11و12 ؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟؟

$Q13)\sqrt{6}cosx-\sqrt{2}\left | sinx \right |=2$

$Q13)\sqrt{6}cosx-\sqrt{2}\left | sinx \right |=2$

\[Q13)\sqrt{6}cosx-\sqrt{2}|sinx|=2\left \{ 2\sqrt{2} \right \}\\\frac{\sqrt{3}}{2}cosx-\frac{1}{2}|sinx|=\frac{1}{\sqrt{2}}\Rightarrow \\(1)cos\frac{\pi }{6}cosx-sin\frac{\pi }{6}sinx=cos\frac{\pi }{4}\\cos\left ( \frac{\pi }{6} +x\right )=cos\frac{\pi }{4}=cos\frac{7\pi }{4}\Rightarrow x=\left \{ \frac{\pi }{12},\frac{19\pi }{12} \right \}\\(2)cos\frac{\pi }{6}cosx+sin\frac{\pi }{6}sinx=cos\frac{\pi }{4}=cos\frac{7\pi }{4}\\cos(x-\frac{\pi }{6})=cos\frac{\pi }{4}=cos\frac{7\pi }{4}\Rightarrow x=\left \{ \frac{5\pi }{12},\frac{23\pi }{12} \right \}\]

5pi/12 لاتحقق

$Q14)solve .in [0,\pi ]$
$cos2x+cos4x=\left \lfloor cos4x \right \rfloor$

$Q15)log_{2}x=log_{3}(5-x^{log_{2}3})$

$\large {\color{Blue} \log_{2}x=\log_{3}(a-x^{\log_{2}3})\rightarrow \log_{2}x=\log_{3}(a-3^{\log_{2}x})}$
$\large {\color{Blue} let ,y=\log_{2}x\rightarrow x=2^{y}}$
$\large {\color{Blue} \Rightarrow y=\log_{3}(a-3^{y})\rightarrow a-3^{y}=3^{y}\rightarrow 3^{y}=\frac{a}{2}}$
$\large {\color{Blue} \Rightarrow y\log3=\log(\frac{a}{2})\rightarrow y=\frac{\log(\frac{a}{2})}{\log3}}$
$\large {\color{Blue}\Rightarrow x=2^{(\frac{\log\frac{a}{2}}{\log3})},, {a=five}}$

$Q16)arccos(\frac{7x+5}{13})=arcsin(\frac{4x+1}{13 })$

$Q17)arcsin(\frac{\sqrt{3x+2}}{2})=arccot( \sqrt{\frac{2}{x+1}})$

$Q18)(arcsinx) (arcsiny)=\frac{\pi ^{2}}{12}$
$(arccosx)(arccosy)=\frac{\pi ^{2}}{24}$

\[Q16)arccos\left ( \frac{7x+5}{13} \right )=arcsin\left ( \frac{4x+1}{13} \right )\\sin\left ( arccos\left ( \frac{7x+5}{13} \right ) \right )=\frac{4x+1}{13}\\\sqrt{1-\left (\frac{7x+5}{13} \right )^{2}}=\frac{4x+1}{13}\\\therefore 65x^{2}+78x-143=0\Rightarrow{\color{Blue} x=1},{\color{Red} x=-\frac{11}{5}\times }\]