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

\(\Leftrightarrow A=\left[\frac{2}{\left(x+1\right)^3}.\frac{x+1}{x}+\frac{1}{\left(x+1\right)^2}.\frac{x^2+1}{x^2}\right].\frac{x^3}{x-1}\)

\(\Leftrightarrow A=\left[\frac{2x+x^2+1}{x^2\left(x+1\right)^2}\right].\frac{x^3}{x+1}=\frac{x}{x+1}\)

Để \(A=\frac{x}{x+1}< 1\Leftrightarrow\frac{1}{x+1}>0\Leftrightarrow x>-1\)

Để \(A=1-\frac{1}{x+1}\text{ nguyên thì }\frac{1}{x+1}\text{ nguyên hay }x\in\left\{-2,0\right\} \)

a, \(A=\frac{3\left(x-1\right)^2+12}{\left(x-1\right)^2+2}=\frac{3\left[\left(x-1\right)^2+2\right]+6}{\left(x-1\right)^2+2}=3+\frac{6}{\left(x-1\right)^2+2}\)

Để \(A\in Z\Leftrightarrow\left(x-1\right)^2+2\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)

Mà \(\left(x-1\right)^2\ge0\Rightarrow\left(x-1\right)^2+2\ge2\)

\(\Rightarrow\left(x-1\right)^2+2\in\left\{2;3;6\right\}\)

Ta có bảng:

Vậy...

b, Theo câu a ta có: \(\left(x-1\right)^2+2\ge2\Rightarrow\frac{1}{\left(x-1\right)^2+2}\le\frac{1}{2}\Rightarrow\frac{6}{\left(x-1\right)^2+2}\le\frac{6}{2}=3\)

Dấu "=" xảy ra  khi x - 1 = 0 <=> x = 1

Vậy GTLN của A = 3 khi x = 1

sr câu b mình lm thiếu

Theo câu a ....

=> \(A\le3+3=6\)

Dấu "=" xảy ra khi x=1

Vậy GTLN của A = 6 khi x=1

a) Ta có : 

\(A=\frac{3.\left(x-1\right)^2+12}{\left(x-1\right)^2+2}=\frac{3.\left(x-1\right)^2+3.2+6}{\left(x-1\right)^2+2}=\frac{3.\left[\left(x-1\right)^2+2\right]+6}{\left(x-1\right)^2+2}=3+\frac{6}{\left(x-1\right)^2+2}\)

Để A có giá trị nguyên \(\Leftrightarrow\)\(3+\frac{6}{\left(x-1\right)^2+2}\)\(\in\)Z \(\Leftrightarrow\)\(\frac{6}{\left(x-1\right)^2+2}\)\(\in\)Z \(\Leftrightarrow\)( x - 1 )² + 2 \(\in\)Ư ( 6 )

\(\Rightarrow\)( x - 1 )² + 2 \(\in\){ 1 ; -1 ; 2 ; -2 ; 3 ; -3 ; 6 ; -6 }

Lập bảng ta có :

Vậy x = { 0 ; 2 ; 3 ; -1 }

b) để A có giá trị lớn nhất \(\Leftrightarrow\)\(3+\frac{6}{\left(x-1\right)^2+2}\)có GTLN \(\Leftrightarrow\)\(\frac{6}{\left(x-1\right)^2+2}\)có GTLN \(\Leftrightarrow\)( x - 1 )² +2 có GTNN

Mà ( x - 1 )² \(\ge\)0 \(\Rightarrow\)( x - 1 )² + 2 \(\ge\)2 \(\Rightarrow\)GTNN của ( x - 1 )² + 2 là 2 \(\Leftrightarrow\)x = 1

Vậy A có GTLN là : \(\frac{3.\left(1-1\right)^2+12}{\left(1-1\right)^2+2}=\frac{12}{2}=6\)\(\Leftrightarrow\)x = 1

thank you

A = \(\frac{1}{13}\).\(\frac{-39}{x-7}\)= - \(\frac{39}{13\left(x-7\right)}\)= -\(\frac{3}{x-7}\)

A nhỏ nhất khi x - 7 =  3 => x = 10

A lơn nhất khi x - 7 = -3 => x = 4

thanks very much

Barack Obama

 \(GTNNA=-\frac{3}{7}\) KHi x=8

GTLN của A=3 khi x=6

d)  \(A>0\Leftrightarrow\frac{-1}{x-2}>0\)

\(\Leftrightarrow x-2< 0\)  ( vì \(-1< 0\))

\(\Leftrightarrow x< 2\)

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

\(A=\)\(\left[\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right]\)

  \(:\left[\frac{\left(x-2\right)\left(x+2\right)}{x+2}+\frac{10-x^2}{x+2}\right]\)

\(A=\frac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}:\left[\frac{x^2-4+10-x^2}{x+2}\right]\)

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

\(A=\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{6}\)

\(A=\frac{-1}{x-2}\)

theo câu a) \(A=\frac{-1}{x-2}\)  với ĐKXĐ: \(x\ne\pm2\)

b) \(\left|2x-1\right|=3\)

\(\Rightarrow\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}}\Rightarrow\orbr{\begin{cases}2x=4\\2x=-2\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)   \(\Rightarrow x=-1\)  ( vì \(x=2\)  ko TM ĐKXĐ )

+) khi \(x=-1\)thì \(A=\frac{-1}{-1-2}=\frac{-1}{-3}=\frac{1}{3}\)

vậy khi \(x=-1\)  thì \(A=\frac{1}{3}\)

\(A=\left(\frac{1}{1-x}-1\right):\left(x+1-\frac{1-2x}{1-x}\right)\)     \(\left(ĐK:x\ne1;x\ne2\right)\)

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

\(=\frac{x}{1-x}\cdot\frac{1-x}{1-x^2-1+2x}\)

\(=\frac{x}{-x^2+2x}\)

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

b) Để A=\(\frac{1}{2}\) \(\Leftrightarrow\)\(\frac{1}{2-x}=\frac{1}{2}\)

                   \(\Leftrightarrow2-x=2\)

                   \(\Leftrightarrow-x=0\Leftrightarrow x=0\)

c) Để A>1 \(\Leftrightarrow\)\(\frac{1}{2-x}>1\)

                 \(\Leftrightarrow\)\(\frac{1}{2-x}-1>0\) 

                 \(\Leftrightarrow\)\(\frac{1-2+x}{2-x}>0\)

                 \(\Leftrightarrow\)\(\frac{x-1}{2-x}>0\)

\(\Leftrightarrow\begin{cases}x-1>0\\2-x>0\end{cases}\) hoặc \(\begin{cases}x-1< 0\\2-x< 0\end{cases}\)

\(\Leftrightarrow\begin{cases}x>1\\x< 2\end{cases}\) hoặc \(\begin{cases}x< 1\\x>2\end{cases}\)(vô nghiệm)

\(\Leftrightarrow1< x< 2\)

Vậy \(1< x< 2\) thì A<1