x + 2 x - 3 = x -  x  + 3 x  - 3 =  x  ( x  - 1) + 3( x  - 1) = ( x  - 1)( x  + 3)

a) Với điểu kiện x ≥ 0; x ≠ 1 ta có:

\(y=\frac{\frac{^x}{x^2}-x-6-x-\frac{1}{3}x^2-4x-15}{x^4}-2x^2+\frac{1}{3}x^2+11x+10b\)

\(y=\frac{-\left(5x^7-33x^6-30bx^5+x^3+18x^2+63x-3\right)}{3x^5}\)

a) \(C=\left(\dfrac{x}{x^2-x-6}-\dfrac{x-1}{3x^2-4x-15}\right):\dfrac{x^4-2x^2+1}{3x^2+11x+10}\cdot\left(x^2-2x+1\right)\) (ĐK: \(x\ne-\dfrac{5}{3};x\ne3;x\ne-2;x\ne1\))

\(C=\left[\dfrac{x}{\left(x-3\right)\left(x+2\right)}-\dfrac{x-1}{\left(x-3\right)\left(3x+5\right)}\right]:\dfrac{\left(x^2-1\right)^2}{\left(3x+5\right)\left(x+2\right)}\cdot\left(x-1\right)^2\)

\(C=\left[\dfrac{x\left(3x+5\right)}{\left(3x+5\right)\left(x+2\right)\left(x-3\right)}-\dfrac{\left(x-1\right)\left(x+2\right)}{\left(x-3\right)\left(3x+5\right)\left(x+2\right)}\right]\cdot\dfrac{\left(3x+5\right)\left(x+2\right)}{\left(x^2-1\right)^2\left(x-1\right)^2}\)

\(C=\dfrac{3x^2+5x-x^2-2x+x+2}{\left(3x+5\right)\left(x+2\right)\left(x-3\right)}\cdot\dfrac{\left(3x+5\right)\left(x+2\right)}{\left(x^2-1\right)^2\left(x-1\right)^2}\)

\(C=\dfrac{2x^2+4x+2}{\left(3x+5\right)\left(x+2\right)\left(x-3\right)}\cdot\dfrac{\left(3x+5\right)\left(x+2\right)}{\left(x+1\right)^2\left(x-1\right)^4}\)

\(C=\dfrac{2\left(x+1\right)^2}{\left(3x+5\right)\left(x-3\right)\left(x+2\right)}\cdot\dfrac{\left(3x+5\right)\left(x+2\right)}{\left(x+1\right)^2\left(x-1\right)^4}\)

\(C=\dfrac{2}{\left(x-1\right)^4\left(x-3\right)}\)

b) Thay x = 2003 ta có: 

\(C=\dfrac{2}{\left(2003-1\right)^4\left(2003-3\right)}=\dfrac{2}{2002^4\cdot2000}=\dfrac{1}{2002^4\cdot1000}\)

c) \(C>0\) khi: 

\(\dfrac{2}{\left(x-1\right)^4\left(x-3\right)}>0\) mà: \(\left\{{}\begin{matrix}2>0\\\left(x-1\right)^4>0\end{matrix}\right.\)

\(\Leftrightarrow x-3>0\)

\(\Leftrightarrow x>3\) (đpcm) 

TXĐ: \(\left\{{}\begin{matrix}x\in R\\x\notin\left\{0;-1\right\}\end{matrix}\right.\)

a) 3x(x + 2) + 4x(-2x + 3) + (2x - 3)(3x + 1)

= 3x² + 6x - 8x² + 12x + 6x² + 2x - 9x - 3

= (3x² - 8x² + 6x²) + (6x + 12x + 2x - 9x) - 3

= x³ + 11x - 3

b) (x² + 1)(x² - x + 2) - (x² - 1)(x² + x - 2)

= x⁴ - x³ + 3x² - x + 2 - x⁴ - x³ + 3x² + x - 2

= (x⁴ - x⁴) + (-x³ - x³) + (3x² + 3x²) + (-x + x) + (2 - 2)

= -2x³ + 6x²

c) (-2x - 3)² + (3x + 2)² + (4x + 1)

= 4x² + 12x + 9 + 9x² + 12x + 4 + 4x + 1 

= (4x² + 9x²) + (12x + 12x + 4x) + (9 + 4 + 1)

= 13x² + 28x + 14

a) \(\left(2x-1\right)^2-\left(x+2\right)^2-3x^2+5x\)

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

\(=x^2-3x-3\)

b) \(\left(x+2\right)\left(x-1\right)+2\left(3x-2\right)^2+4x-19x^2\)

\(=x^2+2x-x-2+2\left(9x^2-12x+4\right)+4x-19x^2\)

\(=x^2+2x-x-2+18x^2-24x+8+4x-19x^2\)

\(=-19x+6\)

c) \(2\left(3-x\right)\left(x-2\right)-\left(3x+1\right)^2+5x-11x^2\)

\(=6-2x\left(x-2\right)-\left(9x^2+6x+1\right)+5x-11x^2\)

\(=6-2x^3+4x-9x^2-6x-1+5x-11x^2\)

\(=-2x^3-20x^2+3x+5\)

Ta có: \(A=\left(4x-2\right)^2+\left(-3x+1\right)^2-\left(4x-1\right)\left(3-3x\right)\)

\(=16x^2-16x+4+9x^2-6x+1-\left(12x-12x^2-3+3x\right)\)

\(=25x^2-22x+5-15x+12x^2+3\)

\(=37x^2-37x+8\)

Thay x=-3 vào biểu thức \(A=37x^2-37x+8\), ta được:

\(A=37\cdot\left(-3\right)^2-37\cdot\left(-3\right)+8\)

\(=37\cdot9+111+8\)

\(=333+111+8\)

\(=452\)

Vậy: 452 là giá trị của biểu thức \(A=\left(4x-2\right)^2+\left(-3x+1\right)^2-\left(4x-1\right)\left(3-3x\right)\) tại x=-3