Câu 5,a, đk:\(x>3,y>5,z>4\)
đặt \(\left\{{}\begin{matrix}\sqrt{x-3}=a\\\sqrt{y-5}=b\\\sqrt{z-4}=c\end{matrix}\right.\)
=> pt đưa về: \(a+b+c=20-\dfrac{4}{a}-\dfrac{9}{b}-\dfrac{25}{c}\)
\(< =>a+\dfrac{4}{a}+b+\dfrac{9}{b}+c+\dfrac{25}{c}=20\)(1)
theo BDT Cô si =>\(a+\dfrac{4}{a}+b+\dfrac{9}{b}+c+\dfrac{25}{c}\ge2\sqrt{4}+2\sqrt{9}+2\sqrt{25}=20\)
dấu'=' xảy ra hay (1) xảy ra<=>\(\left\{{}\begin{matrix}a=\dfrac{4}{a}\\b=\dfrac{9}{b}\\c=\dfrac{25}{c}\end{matrix}\right.< =>\left\{{}\begin{matrix}a=2\\b=3\\c=5\end{matrix}\right.\)
\(< =>\left\{{}\begin{matrix}\sqrt{x-3}=2\\\sqrt{y-5}=3\\\sqrt{z-4}=5\end{matrix}\right.< =>\left\{{}\begin{matrix}x=7\left(TM\right)\\y=14\left(TM\right)\\z=29\left(TM\right)\end{matrix}\right.\) vậy(x,y,z)=(7;14;29)
câu 5 , b:
có: \(6x-x^2-5=-5-\left(x^2-6x\right)=4-\left(x^2-6x+9\right)\)
\(=4-\left(x-3\right)^2\le4\)\(=>\sqrt{4-\left(x-3\right)^2}\le2=>\sqrt{6x-x^2-5}\le2\)
\(=>3+\sqrt{6x-x^2-5}\le5\)
\(=>\dfrac{-15}{3+\sqrt{6x-x^2-5}}\le-3\). Dấu'=' xảy ra<=>x=3=>MaxQ=-3
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