B=-3x2-5y2+2x+7y-23

\(=-3x^2-5y^2+2x-7y-\frac{1}{3}-\frac{49}{20}-\frac{1213}{60}\)

\(=-3x^2+2x-\frac{1}{3}-5y^2+7y-\frac{49}{20}-\frac{1213}{60}\)

\(=-3\left(x^2-2\cdot\frac{1}{3}\cdot x+\frac{1}{3}^2\right)-5\left(y^2-2\cdot\frac{7}{10}\cdot y+y^2\right)-\frac{1213}{60}\)

\(=-3\left(x-\frac{1}{3}\right)^2-5\left(y-\frac{7}{10}\right)^2-\frac{1213}{60}\le0-\frac{1213}{60}\)

\(\Rightarrow B\le-\frac{1213}{60}\)

Dấu = khi x=1/3; y=7/10

Vậy .....

Ta có: \(12x+7y=64\)

\(\Rightarrow5x+7x+7y=49+15\)

\(\Rightarrow7\left(x+y\right)+5x=7.7+5.3\)

\(\Rightarrow\hept{\begin{cases}x+y=7\\x=3\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}y=4\\x=3\end{cases}}\)

Vậy khi \(x=3;y=4\)thì \(12x+7y=64\)

Lời giải:

Áp dụng BĐT Bunhiacopxky:

$\frac{47}{15}(3x^2+5y^2)=[(\sqrt{3}x)^2+(-\sqrt{5}y)^2][(\frac{2}{\sqrt{3}})^2+(\frac{3}{\sqrt{5}})^2]\geq (2x-3y)^2$

$\Leftrightarrow \frac{47}{15}(3x^2+5y^2)\geq 49$

$\Rightarrow 3x^2+5y^2\geq \frac{735}{47}$

Ta có đpcm.

a: Sửa đề: \(2A+\left(2x^2+y^2\right)=6x^2+5y^2-2x^2y^2\)

=>\(2A=6x^2+5y^2-2x^2y^2-2x^2-y^2\)

=>\(2A=4x^2+4y^2-2x^2y^2\)

=>\(A=2x^2+2y^2-x^2y^2\)

b: \(2A-\left(xy+3x^2-2y^2\right)=x^2-8y+xy\)

=>\(2A=x^2-8y+xy+xy+3x^2-2y^2\)

=>\(2A=4x^2+2xy-8y-2y^2\)

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

c: Sửa đề: \(A+\left(3x^2y-2xy^2\right)=2x^2y+4xy^3\)

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

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