Cho \(a,b,c\) là ba số thực khác 0 thỏa mãn \(\frac{{2a}}{b} = \frac{{3b}}{c} = \frac{c}{{6a}}\). Tính giá trị của biểụ thức \(P = \frac{{4ac - cb}}{{bc + 2ab}}\)

Ta có: \(a + b + c \ge 6\).

        \(M = \frac{1}{6}\left( {19a + 22b + 25c} \right) + 2\left( {\frac{5}{a} + \frac{6}{b} + \frac{7}{c}} \right) = \left( {\frac{{19}}{6}a + \frac{{10}}{a}} \right) + \left( {\frac{{22}}{6}b + \frac{{12}}{b}} \right) + \left( {\frac{{25}}{6}c + \frac{{14}}{c}} \right)\)

Xét \(k,m,n > 0:ka + \frac{{10}}{a} \ge 2\sqrt {10k} ;mb + \frac{{12}}{b} \ge 2\sqrt {12m} ;nc + \frac{{14}}{c} \ge 2\sqrt {14n} \)

\(a = 2 \Rightarrow 2k + 5 \ge 2\sqrt {10k} \)

Dấu bằng xảy ra \( \Leftrightarrow ka = \frac{{10}}{a} \Rightarrow 2k = 5 \Leftrightarrow k = \frac{5}{2}\).

Tương tự ta tìm được: \(m = 3,n = \frac{7}{2}\).

Do đó: \(M = \left( {\frac{5}{2}a + \frac{{10}}{a}} \right) + \left( {3b + \frac{{12}}{b}} \right) + \left( {\frac{7}{2}c + \frac{{14}}{c}} \right) + \frac{2}{3}a + \frac{2}{3}b + \frac{2}{3}c\)

\( \Rightarrow M \ge 2\sqrt {25}  + 2\sqrt {36}  + 2\sqrt {49}  + \frac{2}{3} \cdot 6 = 40\).

Dấu bằng xảy ra khi \(a = b = c = 2\).

Vậy \({M_{Min}} = 40\) khi \(a = b = c = 2\).

\({\left( {x + y} \right)^2} + 2{y^2}\left( {x + 1} \right) + {\left( {y + 2} \right)^2} - 9 = 0\left( {\rm{*}} \right)\)

\( \Leftrightarrow {x^2} + {y^2} + 2xy + 2x{y^2} + 2{y^2} + {y^2} + 4y + 4 - 9 = 0\)

\( \Leftrightarrow {x^2} - 4 + 2x\left( {{y^2} + y} \right) + 4\left( {{y^2} + y} \right) = 1\)

\( \Leftrightarrow \left( {x + 2} \right)\left( {x - 2} \right) + 2\left( {{y^2} + y} \right)\left( {x + 2} \right) = 1\)

\( \Leftrightarrow \left( {x + 2} \right)\left( {x - 2 + 2{y^2} + 2y} \right) = 1\)

TH1: \(\left\{ {\begin{array}{*{20}{c}}{x + 2 = 1}\\{x - 2 + 2{y^2} + 2y = 1}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{x =  - 1}\\{2{y^2} + 2y = 4}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{x =  - 1}\\{\left[ {\begin{array}{*{20}{c}}{y = 1}\\{y =  - 2}\end{array}} \right.}\end{array}} \right. \Rightarrow \left( { - 1;1} \right),{\rm{\;}}\left( { - 1;{\rm{\;}} - 2} \right)\)

TH2: \(\left\{ {\begin{array}{*{20}{c}}{x + 2 = 1}\\{x - 2 + 2{y^2} + 2y = 1}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{x =  - 3}\\{2{y^2} + 2y = 4}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{x =  - 3}\\{\left[ {\begin{array}{*{20}{c}}{y = 1}\\{y =  - 2}\end{array}} \right.}\end{array}} \right. \Rightarrow \left( { - 3;1} \right),{\rm{\;}}\left( { - 3;{\rm{\;}} - 2} \right)\)