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Bài 1: 

a) Ta có: \(P=1+\dfrac{3}{x^2+5x+6}:\left(\dfrac{8x^2}{4x^3-8x^2}-\dfrac{3x}{3x^2-12}-\dfrac{1}{x+2}\right)\)

\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{8x^2}{4x^2\left(x-2\right)}-\dfrac{3x}{3\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)

\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{4}{x-2}-\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)

\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\dfrac{4\left(x+2\right)-x-\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{4x+8-x-x+2}\)

\(=1+3\cdot\dfrac{\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)

\(=1+\dfrac{3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)

\(=\dfrac{\left(x+3\right)\left(2x+10\right)+3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)

\(=\dfrac{2x^2+10x+6x+30+3x-6}{\left(x+3\right)\left(2x+10\right)}\)

\(=\dfrac{2x^2+19x-6}{\left(x+3\right)\left(2x+10\right)}\)

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

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

\(=\frac{\left(x-2\right)\left(4x^2+3\right)}{\left(x-2\right)\left(2x+1\right)}=\frac{4x^2+3}{2x-1}\left(ĐKXĐ:x\ne2;x\ne-\frac{1}{2}\right)\)

b, \(A\in Z\Leftrightarrow\frac{4x^2+3}{2x-1}\in Z\Leftrightarrow2x+1+\frac{4}{2x-1}\in Z\)

\(\Leftrightarrow\frac{4}{2x-1}\in Z\Leftrightarrow4⋮\left(2x-1\right)\)

\(\Rightarrow2x-1\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

Mà 2x - 1 là số lẻ nên \(2x-1\in\left\{-1;1\right\}\Rightarrow x\in\left\{0;1\right\}\) (thỏa mãn ĐKXĐ)

rút gọn =5x-2

gt của x:

5x-2=6

5x=6+2

5x=8

x=8:5

x=8/5

vậy x=8/5

P = |3x - 3| + 2x + 1

a) Với x âm thì P = -3x - 3 + 2x + 1 = -1x - 3 + 1 = -x - 2

Với x dương thì P = 3x - 3 + 2x + 1 = 5x - 3 + 1 = 5x - 2 (1)

b) P = |3x + 3| + 2x + 1 = 6 Vì kết quả là số dương nên x cũng dương. Từ (1) ta có : 

5x - 2 = 6

=> 5x = 8

=> x = 1,6

a) xét 2 th

th1)x>=1=>P=3x-3+2x+1=5x-2

th2)x<1=>P=3-3x+2x+1=4-x

b)theo a) nếu x>=1=>P=5x-2

=>5x-2=6=>x=8/5 (thđk)

nếu x<1=>P=4-x

=>4-x=5=>x=-1(tmđk)

Ta có :

+/ với 3x - 3 \(\ge\)0 => x \(\ge\)1

=> |3x - 3| = 3x - 3

=> |3x - 3| + 2x + 1 = 3x - 3 + 2x + 1 = 5x - 2

+/ với 3x - 3 < 0 => x < 1

=> |3x - 3| = -(3x - 3) = -3x + 3

=> |3x - 3| + 2x + 1 = -3x + 3 + 2x + 1 = -x + 4

Để P = 6 , ta xét 2 trường hợp

5x - 2 = 6   =>    x = 8/5

-x + 4 = 6   =>   x = -2

a: \(A=\left(\dfrac{x}{x^2-4}+\dfrac{4}{x-2}+\dfrac{1}{x+2}\right):\dfrac{3x+3}{x^2+2x}\)

\(=\dfrac{x+4x+8+x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x\left(x+2\right)}{3\left(x+1\right)}\)

\(=\dfrac{6\left(x+1\right)\cdot x\left(x+2\right)}{3\left(x+1\right)\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{2x}{x-2}\)