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

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

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

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

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

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

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

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

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

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

Áp dụng định lý Bezout:

\(f\left(x\right)=x^3-3x^2+5x+2m\)chia hết cho g (x) = x + 1 nên:

\(f\left(-1\right)=0\)

\(\Rightarrow-1-3-5+2m=0\Leftrightarrow2m=9\Leftrightarrow m=\frac{9}{2}\)

 Mình chưa học dịnh lí Bezout

Ta có:

\(2x\left(x+5\right)-x-5=0\)

=> \(2x\left(x+5\right)-\left(x+5\right)=0\)

=> \(\left(2x-1\right)\left(x+5\right)=0\)

=> \(\orbr{\begin{cases}2x-1=0\\x+5=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{1}{2}\\x=-5\end{cases}}\)

2x(x+5)-x-5=0

2x(x+5)-(x+5)=0

(x+5)(2x-1)=0

TH1:x+5=0

        x    =-5

TH2:2x-1=0

        2x   =1

          x   = 1/2

 Vậy x=-5 và x=1/2

a) \(\frac{x^4-4x^2+3}{x^4+6x^2-7}=\frac{x^4-3x^2-x^2+3}{x^4+7x^2-x^2-7}=\frac{x^2\left(x^2-3\right)-\left(x^2-3\right)}{x^2\left(x^2+7\right)-\left(x^2+7\right)}=\frac{\left(x^2-1\right)\left(x^2-3\right)}{\left(x^2-1\right)\left(x^2+7\right)}=\frac{x^2-3}{x^2+7}\)

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

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

\(=\frac{x^2-1}{x^2+1}\)

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

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

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