a: \(A=4x-3x^2+20-15x-9x^2-12x-4+\left(2x+1\right)^3-\left(8x^3-1\right)\)

\(=-12x^2-23x+16+8x^3+12x^2+6x+1-8x^3+1\)

\(=-17x+18\)

b: \(A=\dfrac{2-1}{3\cdot2}=\dfrac{1}{6}\)

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) 

bạn viết thế này khó nhìn quá

nhìn hơi đau mắt nhá bạn hoa mắt quá

Bạn nên viết biểu thức A bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người hiểu biểu thức của bạn hơn.

1:

a: \(\left(2x-5\right)^2-4x\left(x+3\right)\)

\(=4x^2-20x+25-4x^2-12x\)

=-32x+25

b: \(\left(x-2\right)^3-6\left(x+4\right)\left(x-4\right)-\left(x-2\right)\left(x^2+2x+4\right)\)

\(=x^3-6x^2+12x-8-\left(x^3-8\right)-6\left(x^2-16\right)\)

\(=-6x^2+12x-6x^2+96=-12x^2+12x+96\)

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

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

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

\(=9+10x-15=10x-6\)

2: 

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

\(=9x^2-12x+4-5x^2+20x+4x-4\)

\(=4x^2+12x\)

b: \(\left(3-x\right)\left(x^2+3x+9\right)+\left(x-3\right)^3\)

\(=27-x^3+x^3-9x^2+27x-27\)

\(=-9x^2+27x\)

c: \(\left(x-4\right)^2\left(x+4\right)-\left(x-4\right)\left(x+4\right)^2+3\left(x^2-16\right)\)

\(=\left(x-4\right)\left(x+4\right)\left(x-4-x-4\right)+3\left(x^2-16\right)\)

\(=\left(x^2-16\right)\left(-8\right)+3\left(x^2-16\right)\)

\(=-5\left(x^2-16\right)=-5x^2+80\)

`a, M(x) = 2x^3 + x^2 + 5 - 3x +3x^2 - 2x^3 - 4x^2 +1`

`M(x)= (2x^3 - 2x^3)+(x^2+3x^2)-3x+(5+1) `

`M(x)= 4x^2-3x+6`

`b,` giá trị của `M(x)` tại `x=0`

`-> M(0)=2*0^3 + 0^2 + 5 - 3*0 +3*0^2 - 2*0^3 - 4*0^2 +1`

`M(0)= 0+0+5-0+0+0-0-0+1 = 5+1=6`

Giá trị của `M(x)` tại `x=1`

`-> M(1)=2*1^3 + 1^2 + 5 - 3*1 +3*1^2 - 2*1^3 - 4*1^2 +1`

`M(1)=2+1+5-3+3-2-4+1 = (2-2)+(1+1)+5-(3-3)-4=2+5-4=7-4=3`

`c,` Giá trị của `P(x)` là cái gì bạn nhỉ?