a/\(A=\left(\frac{x+1}{x^2-1}+\frac{x}{\left(x-1\right)\left(x^2+x+1\right)}.\frac{x^2+x+1}{x+1}\right).\frac{\left(x+1\right)^2}{2x+1}\)

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

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

\(\Leftrightarrow A=\frac{x+1}{x-1}\)

b/Thay số vào tính

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

ĐKXĐ: x≠2;x≠3

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

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

\(=\frac{2x-9}{\left(x-2\right)\left(x-3\right)}-\frac{\left(x+3\right)\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}+\frac{\left(2x-4\right)\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}\)

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

\(=\frac{2x-9-x^2+9+2\left(x^2-4x+4\right)}{\left(x-2\right)\left(x-3\right)}\)

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

\(=\frac{-6x+8+x^2}{\left(x-2\right)\left(x-3\right)}=\frac{x^2-4x-2x+8}{\left(x-2\right)\left(x-3\right)}=\frac{x\left(x-4\right)-2\left(x-4\right)}{\left(x-2\right)\left(x-3\right)}=\frac{\left(x-4\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}=\frac{x-4}{x-3}\)

\(=\frac{x-3-1}{x-3}=1-\frac{1}{x-3}\)

b) Để A nhận giá trị là số nguyên thì \(1-\frac{1}{x-3}\) nhận giá trị nguyên

⇒\(\frac{1}{x-3}\) nhận giá trị nguyên

\(\Rightarrow1⋮x-3\)

hay \(x-3\inƯ\left(1\right)\)

\(\Rightarrow x-3\in\left\{1;-1\right\}\)

\(\Rightarrow x\in\left\{4;2\right\}\)

mà x=2 là không thỏa mãn đkxđ

nên x=4

Vậy: Khi x=4 thì biểu thức \(A=\frac{2x-9}{x^2-5x+6}-\frac{x+3}{x-2}-\frac{2x-4}{3-x}\) nhận giá trị là số nguyên

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

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

\(A=\frac{2x-9}{\left(x-2\right)\left(x-3\right)}-\frac{x+3}{x-2}+\frac{2x+4}{x-3}\)

\(=\frac{2x-9}{\left(x-2\right)\left(x-3\right)}-\frac{\left(x+3\right)\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}+\frac{\left(2x+4\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)

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

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

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

b) Ta có : \(A=\frac{x+4}{x-3}=\frac{x-3+7}{x-3}=1+\frac{7}{x-3}\)

Để A đạt giá trị nguyên thì \(\frac{7}{x-3}\)đạt giá trị nguyên

=> 7 ⋮ x - 3

=> x - 3 ∈ Ư(7) = { ±1 ; ±7 }

So với ĐKXĐ ta thấy x = 4 , x = 10 , x = -4 thỏa mãn 

Vậy với x ∈ { ±4 ; 10 } thì A đạt giá trị nguyên

(....) dùng để nhìn được chữ số ở phân số cuối cùng thôi, ko dùng để làm gì.

( ác ) là từ ( các ) 

(gia strij) là từ ( giá trị )

a) \(A=\frac{2x}{x+3}-\frac{x+1}{3-x}-\frac{3-11x}{x^2-9}\)

\(\Leftrightarrow A=\frac{2x}{x+3}+\frac{x+1}{x-3}-\frac{3-11x}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow A=\frac{2x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{3-11x}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow A=\frac{2x^2-6x}{\left(x+3\right)\left(x-3\right)}+\frac{x^2+4x+3}{\left(x-3\right)\left(x+3\right)}-\frac{3-11x}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow A=\frac{3x^2-13x}{x^2-9}\)

\(A=\frac{2x}{x+3}-\frac{x+1}{3-x}-\frac{3-11x}{x^2-9}\)

a) ĐK : x ≠ ±3

\(=\frac{2x}{x+3}+\frac{x+1}{x-3}-\frac{3-11x}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{2x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{3-11x}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{2x^2-6x}{\left(x-3\right)\left(x+3\right)}+\frac{x^2+4x+3}{\left(x-3\right)\left(x+3\right)}-\frac{3-11x}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{2x^2-6x+x^2+4x+3-3+11x}{\left(x-3\right)\left(x+3\right)}\)

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

\(=\frac{3x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{3x}{x-3}\)

b) Để A < 2

=> \(\frac{3x}{x-3}< 2\)

<=> \(\frac{3x}{x-3}-2< 0\)

<=> \(\frac{3x}{x-3}-\frac{2x-6}{x-3}< 0\)

<=> \(\frac{3x-2x+6}{x-3}< 0\)

<=> \(\frac{x+6}{x-3}< 0\)

Xét hai trường hợp :

1. \(\hept{\begin{cases}x+6>0\\x-3< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>-6\\x< 3\end{cases}}\Leftrightarrow-6< x< 3\)

2. \(\hept{\begin{cases}x+6< 0\\x-3>0\end{cases}}\Leftrightarrow\hept{\begin{cases}x< -6\\x>3\end{cases}}\)( loại )

Vậy -6 < x < 3

Câu 3 : 

\(a,A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right):\frac{2x}{5x-5}\)  ĐKXđ : \(x\ne\pm1\)

\(A=\left(\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}\right):\frac{2x}{5\left(x-1\right)}\)

\(A=\left(\frac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}\right).\frac{5\left(x-1\right)}{2x}\)

\(A=\frac{4x}{\left(x-1\right)\left(x+1\right)}.\frac{5\left(x-1\right)}{2x}\)

\(A=\frac{10}{x+1}\)

\(B=\left(\frac{x}{3x-9}+\frac{2x-3}{3x-x^2}\right).\frac{3x^2-9x}{x^2-6x+9}.\)

ĐKXđ : \(x\ne0;x\ne3\)

\(B=\left(\frac{x}{3\left(x-3\right)}+\frac{2x-3}{x\left(3-x\right)}\right).\frac{3x\left(x-3\right)}{x^2-6x+9}\)

\(B=\left(\frac{x^2}{3x\left(x-3\right)}+\frac{9-6x}{3x\left(x-3\right)}\right).\frac{3x\left(x-3\right)}{x^2-6x+9}\)

\(B=\frac{x^2-6x+9}{3x\left(x-3\right)}.\frac{3x\left(x-3\right)}{x^2-6x+9}=1\)