ĐK: \(3x\ne\pm y;x\ne0\)

A = \(\dfrac{3x}{3x+y}-\dfrac{x}{3x-y}+\dfrac{2x}{\left(3x-y\right)\left(3x+y\right)}\)

= \(\dfrac{3x\left(3x-y\right)-x\left(3x+y\right)+2x}{\left(3x-y\right)\left(3x+y\right)}=\dfrac{6x^2-4xy+2x}{\left(3x-y\right)\left(3x+y\right)}=\dfrac{2x\left(3x-2y+1\right)}{\left(3x-y\right)\left(3x+y\right)}\)

Thay x = 1; y=2, ta có:

A = \(\dfrac{2.1\left(3.1-2.2+1\right)}{\left(3.1-2\right)\left(3.1+2\right)}=0\)

\(A=x^3+1+x-x^3+1+1994\\ A=x+1996\)

a, ĐKXĐ: x≠±3

A=\(\left(\dfrac{3-x}{x+3}.\dfrac{x^2+6x+9}{x^2-9}+\dfrac{x}{x+3}\right):\dfrac{3x^2}{x+3}\)

A=\(\left(\dfrac{3-x}{x+3}.\dfrac{\left(x+3\right)^2}{\left(x+3\right)\left(x-3\right)}+\dfrac{x}{x+3}\right):\dfrac{3x^2}{x+3}\)

A=\(\left(\dfrac{3-x}{x-3}+\dfrac{x}{x+3}\right):\dfrac{3x^2}{x+3}\)

A=\(\left(\dfrac{9-x^2}{x^2-9}+\dfrac{x^2-3x}{x^2-9}\right):\dfrac{3x^2}{x+3}\)

A=\(\left(\dfrac{-3}{x+3}\right):\dfrac{3x^2}{x+3}\)

A=\(\dfrac{-1}{x^2}\)

b, Thay x=\(-\dfrac{1}{2}\) (TMĐKXĐ) vào A ta có:

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

c, A<0 ⇔ \(\dfrac{-1}{x^2}< 0\) ⇔ x²>0 (Đúng với mọi x)

Vậy để A<0 thì x đúng với mọi giá trị (trừ ±3)

a, ĐKXĐ: x≠±2

A=\(\left(\dfrac{x}{x^2-4}+\dfrac{2}{2-x}+\dfrac{1}{x+2}\right)\left(x-2+\dfrac{10-x^2}{x+2}\right)\)

A=\(\left(\dfrac{x}{x^2-4}-\dfrac{2x+4}{x^2-4}+\dfrac{x-2}{x^2-4}\right)\left(\dfrac{x^2+2x}{x+2}-\dfrac{2x+4}{x+2}+\dfrac{10-x^2}{x+2}\right)\)

A=\(\left(\dfrac{-6}{x^2-4}\right)\left(\dfrac{6}{x+2}\right)\)

A=\(\dfrac{-36}{\left(x-2\right)\left(x+2\right)^2}\)

b, |x|=\(\dfrac{1}{2}\)

TH1z: x≥0 ⇔ x=\(\dfrac{1}{2}\) (TMĐKXĐ)

TH2: x<0 ⇔ x=\(\dfrac{-1}{2}\) (TMĐXĐ)

Thay \(\dfrac{1}{2}\), \(\dfrac{-1}{2}\) vào A ta có:

\(\dfrac{-36}{\left(\dfrac{1}{2}-2\right)\left(\dfrac{1}{2}+2\right)^2}\)=\(\dfrac{96}{25}\)

\(\dfrac{-36}{\left(\dfrac{-1}{2}-2\right)\left(\dfrac{-1}{2}+2\right)^2}\)=\(\dfrac{32}{5}\)

c, A<0 ⇔ \(\dfrac{-36}{\left(x-2\right)\left(x+2\right)^2}\) ⇔ (x-2)(x+2)² < 0

⇔   {x-2>0        ⇔      {x>2

     [                           [

       {x+2<0                 {x<2

⇔   {x-2<0        ⇔      {x<2

     [                           [

       {x+2>0                 {x>2

⇔ x<2 

Vậy x<2 (trừ -2)

a) \(A=\left(x-1\right).\left(x+1\right)+\left(x+2\right).\left(x^2+2x+4\right)-x.\left(x^2+x+2\right)\)

\(=x^2-1+x^3+2x^2+4x+2x^2+4x+8-x^3-x^2-2x\)

\(=\left(x^3-x^3\right)+\left(x^2+2x^2+2x^2-x^2\right)+\left(4x+4x-2x\right)+\left(-1+8\right)\)

\(=4x^2+6x+7\)

b) Thay vào ta được

\(A=4.\left(\frac{1}{2}\right)^2+6.\frac{1}{2}+7=1+3+7=11\)

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