Câu V:
2) \(a,b,c>\dfrac{25}{4}\)
\(Q=\dfrac{a}{2\sqrt{b}-5}+\dfrac{b}{2\sqrt{c}-5}+\dfrac{c}{2\sqrt{a}-5}\)
\(=\dfrac{a}{\sqrt{4b}-5}+\dfrac{b}{\sqrt{4c}-5}+\dfrac{c}{\sqrt{4a}-5}\)
\(=\dfrac{a}{\dfrac{1}{10}\sqrt{4b.100}-5}+\dfrac{b}{\dfrac{1}{10}\sqrt{4c.100}-5}+\dfrac{c}{\dfrac{1}{10}\sqrt{4a.100}-5}\)
\(\ge^{Caushy}\dfrac{a}{\dfrac{1}{10}.\dfrac{4b+100}{2}-5}+\dfrac{b}{\dfrac{1}{10}.\dfrac{4c+100}{2}-5}+\dfrac{c}{\dfrac{1}{10}.\dfrac{4a+100}{2}-5}\)
\(=\dfrac{5a}{b}+\dfrac{5b}{c}+\dfrac{5c}{a}\)
\(\ge^{Caushy}3\sqrt[3]{\dfrac{5a}{b}.\dfrac{5b}{c}.\dfrac{5c}{a}}=3.5=15\)
Đẳng thức xảy ra khi \(4a=4b=4c=100\Leftrightarrow a=b=c=25\)
Vậy \(MinQ=15\) khi \(a=b=c=25\)
Câu II:
1) \(\left\{{}\begin{matrix}x+2y=4\\x^2+3y=7\left(1\right)\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}3x+6y=12\\2x^2+6y=14\end{matrix}\right.\)
\(\Rightarrow2x^2+6y-\left(3x+6y\right)=14-12\)
\(\Leftrightarrow2x^2-3x-2=0\)
\(\)\(\Leftrightarrow2x^2-4x+x-2=0\)
\(\Leftrightarrow2x\left(x-2\right)+x-2=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{1}{2}\end{matrix}\right.\)
*\(x=2\Rightarrow2+2y=4\Leftrightarrow y=1\). Thử lại \(y=1\) vào pt (1), ta thấy pt thỏa mãn.
*\(x=-\dfrac{1}{2}\Rightarrow-\dfrac{1}{2}+2y=4\Leftrightarrow y=\dfrac{9}{4}\). Thử lại \(y=\dfrac{9}{4}\) vào pt (1), ta thấy pt thỏa mãn.
Vậy các cặp số (x,y) thỏa mãn pt là \(\left(2,1\right);\left(-\dfrac{1}{2};\dfrac{9}{4}\right)\)
