Ta có A = 5x² - 2xy + 2y² - 4x + 2y + 3

=> 2A = 10x² - 4xy + 4y² - 8x + 4y + 6

= (x² - 4xy + 4y²) - 2(x - 2y) + 1 + 9x² - 6x + 1 + 4

= \(\left(x-2y\right)^2-2\left(x-2y\right)+1+9\left(x^2-\frac{2}{3}x+\frac{1}{9}\right)+4\)

\(=\left(x-2y-1\right)^2+9\left(x-\frac{1}{3}\right)^2+4\)\(\ge4\)

=> A \(\ge\)2 

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-2y-=0\\x-\frac{1}{3}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x-2y=1\\x=\frac{1}{3}\end{cases}}\Leftrightarrow\hept{\begin{cases}y=-\frac{1}{3}\\x=\frac{1}{3}\end{cases}}\)

Vậy khi x = 1/3 ; y = -1/3 thì A đạt GTNN

\(A=5x^2+2y^2-2xy-4x+2y\)\(+3\)

\(=\left(x^2-2xy+y^2\right)+\)\(\left(4x^2-4x+1\right)+\)\(\left(y^2+2y+1\right)+1\)

\(Tacó\)

\(\hept{\begin{cases}x^2+9y^2=10\\x+3y+8=12xy\end{cases}}\Leftrightarrow\hept{\begin{cases}x^2+6xy+9y^2=10+6xy\\x+3y=12xy-8\end{cases}}\Rightarrow\hept{\begin{cases}\left(x+3y\right)^2=10+6xy\\\left(x+3y\right)^2=\left(12xy-8\right)^2\end{cases}}\)

\(\Rightarrow10+6xy=\left(12xy-8\right)^2\)

\(\Leftrightarrow144x^2y^2-198xy+54=0\)

\(\Leftrightarrow\orbr{\begin{cases}xy=1\\xy=\frac{3}{8}\end{cases}}\)

- Với \(xy=1\): 

\(\hept{\begin{cases}xy=1\\x+3y+8=12\end{cases}}\Leftrightarrow\hept{\begin{cases}y=\frac{1}{x}\\x+\frac{3}{x}-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3,y=\frac{1}{3}\\x=1,y=1\end{cases}}\)

- Với \(xy=\frac{3}{8}\):

\(\hept{\begin{cases}xy=\frac{3}{8}\\x+3y+8=4\end{cases}}\Leftrightarrow\hept{\begin{cases}y=\frac{3}{8x}\\x+\frac{9}{8x}+4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-7-\sqrt{31}}{4},y=\frac{-7+\sqrt{31}}{12}\\x=\frac{-7+\sqrt{31}}{4},y=\frac{-7-\sqrt{31}}{12}\end{cases}}\)

Thử lại ta đều thấy thỏa mãn. 

Đặt giá tiền một quyển vở và một cái bút lần lượt là \(x,y\)(đồng), \(x,y>0\).

Theo bài ra, ta có hệ phương trình: 

\(\hept{\begin{cases}8x+15y=88500\\13x+9y=90000\end{cases}}\Leftrightarrow\hept{\begin{cases}24x+45y=265500\\65x+45y=450000\end{cases}\Leftrightarrow\hept{\begin{cases}41x=184500\\y=\frac{90000-13x}{9}\end{cases}}}\)

\(\Leftrightarrow\hept{\begin{cases}x=4500\\y=3500\end{cases}}\left(tm\right)\)

em chịu thôi

\(a^3+b^3+3\left(a^2+b^2\right)+4\left(a+b\right)+4\)

\(=\left(a+b\right)\left(a^2-ab+b^2\right)+2a^2+2b^2-2ab+a^2+b^2+2ab+2.2a+2.2b+2^2\)

\(=\left(a+b+2\right)\left(a^2-ab+b^2\right)+\left(a+b+2\right)^2\)

\(=\left(a+b+2\right)\left(a^2-ab+b^2+a+b+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a+b+2=0\\a^2-ab+b^2+a+b+2=0\end{cases}}\)

Ta có: \(a^2-ab+b^2+a+b+2=\frac{1}{2}a^2-ab+\frac{1}{2}b^2+\frac{1}{4}a^2+a+1+\frac{1}{4}b^2+b+1+\frac{1}{4}\left(a^2+b^2\right)\)

\(=\frac{1}{2}\left(a-b\right)^2+\frac{1}{4}\left(a+2\right)^2+\frac{1}{4}\left(b+2\right)^2+\frac{1}{4}\left(a^2+b^2\right)>0,\forall a,b\inℝ\).

Suy ra \(a+b+2=0\Leftrightarrow a+b=-2\)

mà \(ab>0\Rightarrow\hept{\begin{cases}a< 0\\b< 0\end{cases}}\). 

\(Q=\frac{1}{a}+\frac{1}{b}=-\left(\frac{1}{-a}+\frac{1}{-b}\right)\le-\frac{\left(1+1\right)^2}{-a-b}=-2\)

Dấu \(=\)xảy ra khi \(a=b=-1\).

Áp dụng BĐT Cauchy - Schwarz và Cauchy ta có:

\(P=\frac{1}{a^2}\left(b^2+c^2\right)+a^2\left(\frac{1}{b^2}+\frac{1}{c^2}\right)\)

\(\ge\frac{b^2+c^2}{a^2}+a^2\cdot\frac{9}{b^2+c^2}\) (Cauchy - Schwarz)

\(=\left(\frac{b^2+c^2}{a^2}+\frac{a^2}{b^2+c^2}\right)+8\cdot\frac{a^2}{b^2+c^2}\)

\(\ge2\sqrt{\frac{b^2+c^2}{a^2}\cdot\frac{a^2}{b^2+c^2}}+8\cdot\frac{b^2+c^2}{b^2+c^2}\) (BĐT Cauchy)

\(=2+8=10\)

Dấu "=" xảy ra khi: \(a=b\sqrt{2}=c\sqrt{2}\)

Vậy Min(P) = 10 khi \(a=b\sqrt{2}=c\sqrt{2}\)