a) \(ĐKXĐ:\hept{\begin{cases}x\ne2\\x\ne3\end{cases}}\)

\(A=\frac{2x-9}{x^2-5x+6}-\frac{x+3}{x-2}-\frac{2x+4}{3-x}\)

\(\Leftrightarrow A=\frac{2x-9}{\left(x-2\right)\left(x-3\right)}-\frac{x+3}{x-2}+\frac{2\left(x+2\right)}{x-3}\)

\(\Leftrightarrow A=\frac{2x-9-\left(x-3\right)\left(x+3\right)+2\left(x+2\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)

\(\Leftrightarrow A=\frac{2x-9-x^2+9+2x^2-8}{\left(x-2\right)\left(x-3\right)}\)

\(\Leftrightarrow A=\frac{x^2+2x-8}{\left(x-2\right)\left(x-3\right)}\)

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

\(\Leftrightarrow A=\frac{x+4}{x-3}\)

b) Để \(A\inℤ\)

\(\Leftrightarrow\frac{x+4}{x-3}\inℤ\)

\(\Leftrightarrow1+\frac{7}{x-3}\inℤ\)

\(\Leftrightarrow x-3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

\(\Leftrightarrow x\in\left\{2;4;-4;10\right\}\)

Vậy để \(A\inℤ\Leftrightarrow x\in\left\{2;4;-4;10\right\}\)

c) Để \(A=\frac{3}{5}\)

\(\Leftrightarrow\frac{x+4}{x-3}=\frac{3}{5}\)

\(\Leftrightarrow5x+20=3x-9\)

\(\Leftrightarrow2x+29=0\)

\(\Leftrightarrow x=-\frac{29}{2}\)

d) Để \(A< 0\)

\(\Leftrightarrow\frac{x+4}{x-3}< 0\)

\(\Leftrightarrow1+\frac{7}{x-3}< 0\)

\(\Leftrightarrow\frac{-7}{x-3}< 1\)

\(\Leftrightarrow-7< x-3\)

\(\Leftrightarrow x>-4\)

e) Để \(A>0\)

\(\Leftrightarrow\frac{x+4}{x-3}>0\)

\(\Leftrightarrow1+\frac{7}{x-3}>0\)

\(\Leftrightarrow\frac{-7}{x-3}>1\)

\(\Leftrightarrow-7>x-3\)

\(\Leftrightarrow x< -4\)

a) \(H=\left(\frac{x}{x+2}-\frac{x^3-8}{x^3+8}.\frac{x^2-2x+4}{x^2-4}\right).\frac{x+3}{x+2}\)

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

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

\(=\frac{-4}{\left(x+2\right)^2}.\frac{x+3}{x+2}=\frac{-4x-12}{\left(x+2\right)^3}\)

\(1.\frac{x-7}{2}< 0\)

\(\Leftrightarrow\frac{x-7}{2}.2< 0.2\)

\(\Leftrightarrow x-7< 0\Leftrightarrow x< 7\)

\(S=\left\{xlx< 7\right\}\)

2)\(\)Đề biểu thức sau nhân giá trị âm thì :

\(\frac{x+3}{x-5}< 0\Leftrightarrow x+3< 0\Leftrightarrow x< 3\left(Đk:x\ne5\right)\)

\(S=\left\{xlx< 3\right\}\)

3.Giá trị của x thuộc Z để biểu thức sau nhận giá trị dương:

\(x^2+x\ge0\)

\(\Leftrightarrow x\left(x+1\right)\ge0\)

\(\Leftrightarrow\orbr{\begin{cases}x\ge0\\x+1\ge0\end{cases}\Leftrightarrow\orbr{\begin{cases}x\ge0\\x\ge-1\end{cases}}}\)

\(S=\left\{xlx\ge-1\right\}\)

ĐKXĐ : \(x\ne\pm3\)

a) \(A=\left(\frac{2x}{x-3}-\frac{x+1}{x+3}+\frac{x^2+1}{9-x^2}\right):\left(1-\frac{x-1}{x+3}\right)\)

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

\(A=\left(\frac{-2x^2-6x+x^2-2x-3+x^2+1}{\left(3-x\right)\left(3+x\right)}\right):\left(\frac{x+3-x+1}{x+3}\right)\)

\(A=\left(\frac{-8x-2}{\left(3-x\right)\left(3+x\right)}\right):\left(\frac{4}{x+3}\right)\)

\(A=\frac{-2\left(4x+1\right)\left(x+3\right)}{\left(3-x\right)\left(3+x\right)4}\)

\(A=\frac{-\left(4x+1\right)}{2\left(3-x\right)}\)

\(A=\frac{4x+1}{2\left(x-3\right)}\)

b) \(\left|x-5\right|=2\)

\(\Rightarrow\orbr{\begin{cases}x-5=2\\x-5=-2\end{cases}\Rightarrow\orbr{\begin{cases}x=7\\x=3\end{cases}}}\)

Mà ĐKXĐ x khác 3 => ta xét x = 7

\(A=\frac{4\cdot7+1}{2\cdot\left(7-3\right)}=\frac{29}{8}\)

c) Để A nguyên thì 4x + 1 ⋮ 2x - 3

<=> 4x - 6 + 7 ⋮ 2x - 3

<=> 2 ( 2x - 3 ) + 7 ⋮ 2x - 3

Mà 2 ( 2x - 3 ) ⋮ ( 2x - 3 ) => 7 ⋮ 2x - 3

=> 2x - 3 thuộc Ư(7) = { 1; -1; 7; -7 }

=> x thuộc { 2; 1; 5; -2 }

Vậy .....

a)   ĐKXĐ: \(x\ne\pm3\)

   \(A=\frac{2x\left(x+3\right)-\left(x+1\right)\left(x-3\right)-\left(x^2+1\right)}{x^2-9} : \frac{x+3-\left(x-1\right)}{x+3}\)

 \(A=\frac{2x^2-6x-x^2+2x+3-x^2-1}{x^2-9} : \frac{4}{x+3}\)

\(A=\frac{-4x+2}{x^2+9} : \frac{4}{x+3}\)

\(A=\frac{2\left(1-2x\right)}{\left(x+3\right)\left(x-3\right)}\cdot\frac{x+3}{4}=\frac{1-2x}{2x-6}\)

b)

  Có 2 trường hợp:

T.Hợp 1:

               \(x-5=2\Leftrightarrow x=7\)(thỏa mã ĐKXĐ)

thay vào A ta được: A=\(-\frac{13}{8}\)

T.Hợp 2:

          \(x-5=-2\Leftrightarrow x=3\)(Không thỏa mãn ĐKXĐ)

Vậy không tồn tại giá trị của A tại x=3

Vậy với x=7 thì A=-13/8

c)

      \(\frac{1-2x}{2x-6}=\frac{1-\left(2x-6\right)-6}{2x-6}=-1-\frac{5}{2x-6}\)

Do -1 nguyên, để A nguyên thì \(-\frac{5}{2x-6}\inℤ\)

Để \(-\frac{5}{2x-6}\inℤ\)thì \(2x-6\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

Do 2x-6 chẵn, để x nguyên thì 2x-6 là 1 số chẵn .

Vậy không có giá trị nguyên nào của x để A nguyên

Câu 1:

\(P=\sqrt{a\left(a+b+c\right)+bc}+\sqrt{b\left(a+b+c\right)+ac}+\sqrt{c\left(a+b+c\right)+ab}\)

\(P=\sqrt{\left(a+b\right)\left(a+c\right)}+\sqrt{\left(b+a\right)\left(b+c\right)}+\sqrt{\left(c+a\right)\left(c+b\right)}\)

Áp dụng BĐT \(\sqrt{xy}\le\frac{x+y}{2}\)

\(P\le\frac{a+b+a+c}{2}+\frac{b+a+b+c}{2}+\frac{c+a+c+b}{2}\)

\(=\frac{2a+b+c}{2}+\frac{2b+a+c}{2}+\frac{2c+a+b}{2}\)

\(=\frac{\left(2a+a+a\right)+\left(2b+b+b\right)+\left(2c+c+c\right)}{2}\)

\(=\frac{4\cdot\left(a+b+c\right)}{2}=\frac{4\cdot2}{2}=4\)

Vậy \(maxP=4\Leftrightarrow a=b=c=\frac{2}{3}\)