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

b) Với mọi \(x\) thỏa mãn \(-2<x<2\) và \(x \ne -1\) thì \(x-2\) đều có giá trị âm, mà \(\begin{cases}(x+1)^2≥0\\x+2>0\\\end{cases}\) \( \Rightarrow\) Biểu thức A luôn có giá trị âm.

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

Với \(-2< x< 2\Leftrightarrow\left\{{}\begin{matrix}x-2< 0\\x+2>0\end{matrix}\right.\Leftrightarrow\left(x-2\right)\left(x+2\right)< 0;x\ne-1\Leftrightarrow\left(x+1\right)^2>0\Leftrightarrow A< 0\)

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

hello

a)A=1(6x-x2x+1+10x2-1)x-1(6x-x2x+1+10x2-1)x+12-5xx-5xx+12-1

A=\(14-10x^2-1\)

b)thay 1 vao A ta có

\(A=14-10.1^2-1=14-10-1=13\)

thay -1 vào A ta có

\(A=14-10\left(-1\right)^2-1=14+10-1=23\)

vậy với x=+-1 thì x luôn dương

c)

\(a,A=\dfrac{1}{x-2}+\dfrac{1}{x+2}+\dfrac{x^2+1}{x^2-4}\left(dkxd:x\ne\pm2\right)\)

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

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

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

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

\(b,\) Theo đề, ta có : \(-2< x< 2\) 

\(\Rightarrow x-2< 0;x+2>0;\left(x+1\right)^2>0\)

\(\Rightarrow A< 0\) hay phân thức luôn có giá trị âm

Xem lại biểu thức P.

Mình phải đi ăn nên chiều mình làm nốt câu d nhé

a) Điều kiện để P được xác định là: \(x\ne1;x\ne-1\)

b) \(P=\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}\right):\dfrac{2x}{5x-5}x-\dfrac{x^2-1}{x^2+2x+1}\)

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

\(P=0:\dfrac{2x}{5x-5}x-\dfrac{x-1}{x+1}\)

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

c) Theo đề ta có:

\(P=2\)

\(\Leftrightarrow-\dfrac{x-1}{x+1}=2\)

\(\Leftrightarrow-\left(x-1\right)=2x+2\)

\(\Leftrightarrow-x-2x=2-1\)

\(\Leftrightarrow-3x=1\)

\(\Leftrightarrow x=-\dfrac{1}{3}\)

d) \(P=-\dfrac{x-1}{x+1}\) nguyên khi:

\(\Leftrightarrow x-1⋮-\left(x+1\right)\)

\(\Leftrightarrow\left(x+1\right)-2⋮-\left(x+1\right)\)

\(\Leftrightarrow-2⋮-\left(x+1\right)\)

\(\Leftrightarrow2⋮x+1\)

\(\Rightarrow x+1\inƯ\left(2\right)\)

Vậy \(P\) nguyên khi \(x\in\left\{-2;0;-3;1\right\}\)

a)

A=\(\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}\right)\div\dfrac{2x}{5x-5}\)

\(\Leftrightarrow\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}\right)\div\dfrac{2x}{5\left(x-1\right)}\)

ĐKXĐ: \(\left\{{}\begin{matrix}x-1\ne0\\x+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0+1\\x=0-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

MTC: 5(x-1)(x+1)

\([\dfrac{5\left(x+1\right)\left(x+1\right)}{5\left(x-1\right)\left(x+1\right)}-\dfrac{5\left(x-1\right)\left(x-1\right)}{5\left(x-1\right)\left(x+1\right)}]\div\dfrac{2x\left(x+1\right)}{5\left(x-1\right)\left(x+1\right)}\)

\(\Rightarrow[5\left(x+1\right)\left(x+1\right)-5\left(x-1\right)\left(x-1\right)]\div2x\left(x+1\right)\)

\(\Leftrightarrow[5\left(x+1\right)^2-5\left(x-1\right)^2]\div2x^2+2x\)

\(\Leftrightarrow[5\left(x^2+2x+1\right)-5\left(x^2-2x+1\right)]\div2x^2+2x\)

\(\Leftrightarrow(5x^2+10x+5-5x^2+10x-5)\div2x^2+2x\)

\(\Leftrightarrow20x\div\left(2x^2+2x\right)\)

\(\Leftrightarrow10x+10\)

a, ĐK : \(x\ne1;2;3;4;5\)

b, \(\dfrac{1}{x\left(x-1\right)}+\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}\)

\(=\dfrac{1}{x}-\dfrac{1}{x-1}+\dfrac{1}{x-1}-\dfrac{1}{x-2}+\dfrac{1}{x-2}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-4}+\dfrac{1}{x-4}-\dfrac{1}{x-5}\)

\(=\dfrac{1}{x}-\dfrac{1}{x-5}=\dfrac{x-5-x}{x\left(x-5\right)}=\dfrac{-5}{x\left(x-5\right)}\)

Đề sai rồi bạn

1,

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

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

\(x=4\Rightarrow A=\dfrac{4.x^2-4}{\left(4-2\right)\left(4+2\right)}=...\)

2.

\(A=\dfrac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{3\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{3-5x}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{x\left(x+1\right)+3\left(x-1\right)+3-5x}{\left(x-1\right)\left(x+1\right)}=\dfrac{x^2-2x+1}{\left(x-1\right)\left(x+1\right)}\)

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

3.

Đề lỗi, thiếu dấu trước \(\dfrac{6+5x}{4-x^2}\)

4.

\(A=\dfrac{2x}{\left(x-5\right)\left(x+5\right)}-\dfrac{5\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\dfrac{x-5}{\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{2x-5\left(x+5\right)-\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4x-20}{\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{-4\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4}{x-5}\)

\(x=\dfrac{4}{5}\Rightarrow A=\dfrac{-4}{\dfrac{4}{5}-5}=\dfrac{20}{21}\)

5.

\(M=\dfrac{x^2}{x\left(x+2\right)}+\dfrac{2x}{x\left(x+2\right)}+\dfrac{2\left(x+2\right)}{x\left(x+2\right)}\)

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

\(=\dfrac{\left(x+2\right)^2}{x\left(x+2\right)}=\dfrac{x+2}{x}\)

\(x=-\dfrac{3}{2}\Rightarrow M=\dfrac{-\dfrac{3}{2}+2}{-\dfrac{3}{2}}=-\dfrac{1}{3}\)