\(2x^2+5x+2\)

\(=2x^2+4x+x+2\)

\(=2x\left(x+2\right)+x+2\)

\(=\left(x+2\right)\left(2x+1\right)\)

2x² + 5x + 2

= 2x² + 4x + x + 2

= (2x² + 4x) + (x + 2)

= 2x (x + 2) + (x + 2)

= (2x + 1) (x + 2)

\(\frac{2}{2x+3}+\frac{5}{2x-3}-\frac{2x-33}{9-4x^2}\)

= \(\frac{2}{2x+3}+\frac{5}{2x-3}+\frac{2x-33}{4x^2-9}\)

= \(\frac{2\left(2x-3\right)}{\left(2x+3\right)\left(2x-3\right)}+\frac{5\left(2x+3\right)}{\left(2x-3\right)\left(2x+3\right)}+\frac{2x-33}{\left(2x-3\right)\left(2x+3\right)}\)

= \(\frac{4x-6+10x-15+2x-33}{\left(2x-3\right)\left(2x+3\right)}\)

= \(\frac{16x-54}{\left(2x-3\right)\left(2x+3\right)}\)

\(\frac{2}{2x+3}+\frac{5}{2x-3}-\frac{2x-33}{9-4x^2}\)\(=\frac{2}{2x+3}+\frac{5}{2x-3}+\frac{2x-33}{4x^2-9}\)

\(=\frac{2\left(2x-3\right)}{\left(2x+3\right)\left(2x-3\right)}+\frac{5\left(2x+3\right)}{\left(2x+3\right)\left(2x-3\right)}+\frac{2x-33}{\left(2x+3\right)\left(2x-3\right)}\)

\(=\frac{4x-6+10x+15+2x-33}{\left(2x+3\right)\left(2x-3\right)}=\frac{16x-24}{\left(2x+3\right)\left(2x-3\right)}=\frac{8\left(2x-3\right)}{\left(2x+3\right)\left(2x-3\right)}=\frac{8}{2x+3}\)

Ta có:

C = 13x² + 4y² - 12xy - 2x - 4y + 10

C = (9x² - 12xy + 4y²) + 2(3x - 2y) + 1 + (4x² - 8x + 4) + 5

C = (3x - 2y)² + 2(3x - 2y) + 1 + 4(x² - 2x + 1) + 5

C = (3x - 2y + 1)² + 4(x - 1)² + 5 \(\ge\)5 \(\forall\)x; y

Dấu "=" xảy ra <=> \(\hept{\begin{cases}3x-2y+1=0\\x-1=0\end{cases}}\) <=> \(\hept{\begin{cases}2y=3x+1\\x=1\end{cases}}\) <=> \(\hept{\begin{cases}2y=3.1+1=4\\x=1\end{cases}}\)<=> \(\hept{\begin{cases}y=2\\x=1\end{cases}}\)

Vậy MinC = 5 <=> x = 1 và y = 2

SOS dao lam có thể sử dụng trong bài này!

Chú ý:

+)\(C=2\left(3x-2y+1\right)^2+5-\left(x-2y+3\right)\left(5x-2y-1\right)\)

+) \(C=8\left(x-1\right)^2+5+\left(x-2y+3\right)\left(5x-2y-1\right)\)

Vậy ta tìm được: \(C=\frac{C+C}{2}=\frac{2\left(3x-2y+1\right)^2+8\left(x-1\right)^2+10}{2}\)

\(=\left(3x-2y+1\right)^2+4\left(x-1\right)^2+5\ge5\)