a: ĐKXĐ: \(x\notin\left\{-\dfrac{1}{2};-\dfrac{3}{2}\right\}\)

a, ĐKXĐ : \(\left\{{}\begin{matrix}x\ne2\\x\ne3\end{matrix}\right.\)

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

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

\(=\dfrac{2x^2-7x+6}{\left(x-2\right)\left(x-3\right)}=\dfrac{\left(x-2\right)\left(2x-3\right)}{\left(x-2\right)\left(x-3\right)}=\dfrac{2x-3}{x-3}\)

b, Ta có : \(P=\dfrac{2x-3}{x-3}=\dfrac{2x-6+3}{x-3}=2+\dfrac{3}{x-3}\)

- Để P là số nguyên \(\Leftrightarrow x-3\in\left\{1;-1;3;-3\right\}\)

\(\Leftrightarrow x\in\left\{4;3;6;0\right\}\)

Vậy ...

a ĐKXĐ : \(x\ne2,x\ne3\)

\(\Rightarrow P=\dfrac{2x\left(x-3\right)+4-\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}=\dfrac{2x^2-6x+4-x+2}{\left(x-2\right)\left(x-3\right)}=\dfrac{2x^2-7x+6}{\left(x-2\right)\left(x-3\right)}=\dfrac{2x^2-7x+6}{x^2-5x+6}\)b Ta có P = \(\dfrac{2x^2-7x+6}{x^2-5x+6}=\dfrac{x^2-5x+6+x^2-2x}{x^2-5x+6}=1+\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}=1+\dfrac{x}{x-3}\)

Để P\(\in Z\) \(\Leftrightarrow1+\dfrac{x}{x-3}\in Z\) \(\Rightarrow\dfrac{x}{x-3}\in Z\) \(\Rightarrow x⋮x-3\) \(\Rightarrow x-3+3⋮x-3\)

\(\Rightarrow3⋮x-3\) \(\Rightarrow\left(x-3\right)\in\left\{-3;-1;1;3\right\}\) \(\Rightarrow x\in\left\{0;2;4;6\right\}\) 

Thử lại ta thấy đúng 

Vậy...

a) Ta có: \(P=\dfrac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{\sqrt{x}-2}{1-\sqrt{x}}\)

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

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

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)

1: ĐKXĐ: \(a\ge0\)

bài này quản lí đã làm ở đây nè bạn : 

http://olm.vn/hoi-dap/question/97592.html

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) 

+) Nếu 3x - 3 \(\ge\) 0 => x \(\ge\) 1 => |3x - 3| = 3x - 3  => P = 3x - 3 + 2x + 1 = 5x - 2

+) Nếu 3x - 3 < 0 => x < 1 => |3x - 3| = -(3x - 3) = -3x + 3 => P = -3x + 3 + 2x + 1 = - x + 4

Vậy P = 5x - 2 khi x \(\ge\) 1 và P = - x + 4 khi x < 1

b) P = 6

+) Nếu x \(\ge\) 1 => 5x - 2 = 6 => 5x = 8 => x = 8 : 5 = 1,6 (Thoả mãn)

+) Nếu x < 1 => - x + 4 = 6 => - x = 6 - 4 = 2 => x = -2 (Thoả mãn)

Vậy x  = 1,6 hoặc x = -2 thì P = 6

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) Ta có: \(P=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)

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

\(=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2\)

\(=x-\sqrt{x}+1\)

a) ĐK: x ≥ 0; x ≠ 9; x≠4

P= \(\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{3}{x-5\sqrt{x}+6}\right):\left(\dfrac{x+2}{\sqrt{x}-3}-\dfrac{x^2-\sqrt{x}-6}{\left(x-2\right)\left(\sqrt{x}-3\right)}\right)\)

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

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

=\(\dfrac{x-4+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}:\dfrac{x^2-4-x^2+\sqrt{x}+6}{\left(x-2\right)\left(\sqrt{x}-3\right)}\)

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

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

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

=\(\dfrac{x^2-3x+2}{x-4}\)

b)  P ≤ -2

⇒ \(\dfrac{x^2-3x+2}{x-4}\) ≤ -2

⇔ \(\dfrac{x^2-3x+2}{x-4}\) + 2 ≤ 0

⇔ \(\dfrac{x^2-3x+2+2\left(x-4\right)}{x-4}\) ≤ 0

⇔ \(\dfrac{x^2-3x+2+2x-8}{x-4}\) ≤ 0

⇔\(\dfrac{x^2-x-6}{x-4}\) ≤ 0

⇔ \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2-x-6\ge0\\x-4< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x^2-x-6\le0\\x-4>0\end{matrix}\right.\end{matrix}\right.\)

⇔\(\left[{}\begin{matrix}x\le2\\3\le x< 4\end{matrix}\right.\)

Vậy.......