Bài 1 : Điều kiện xác định : \(x\ne\pm1\)

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

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

Nhận thấy giá trị của x càng tăng thì giá trị của M càng giảm

mặt khác , giá trị của x lại không giảm quá 0 nên ta không thể nào xác định được giá trị lớn nhất của K 

\(a,x\ne2;x\ne-2;x\ne0\)

\(b,A=\left(\frac{x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\frac{6}{x+2}\)

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

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

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

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

\(c,\)Để A > 0 thi \(\frac{1}{2-x}>0\Leftrightarrow2-x>0\Leftrightarrow x< 2\)

a) \(ĐKXĐ:x\ne1\)

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

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

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

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

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

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

c) Với \(\forall x\)(\(x\ne1\)) thì biểu thức được xác định .

P/s : Theo mik câu c nên chuyển thành : Tìm x để biểu thức đạt giá trị nguyên.

Tại thấy câu c k khác j câu a !

Bài 2 :

a) Phân thức A xác định \(\Leftrightarrow\hept{\begin{cases}x-2\ne0\\x+2\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne2\\x\ne-2\end{cases}}}\)

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

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

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

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

\(A=\frac{4\cdot\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)\cdot4}\)

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

c) Thay x = 4 ta có :

\(A=\frac{4-2}{4+2}=\frac{2}{6}=\frac{1}{3}\)

Vậy.........

\(4x^2y^3.\frac{2}{4}x^3y=4x^2y^3.\frac{1}{2}x^3y=2x^5y^4\)

\(\left(5x-2\right)\left(25x^2+10x+4\right)\)

\(=\left(5x-2\right)\left[\left(5x\right)^2+5x.2+2^2\right]\)

\(=\left(5x\right)^3-2^3\)

\(=125x^3-8\)

Bài 1

a,\(4x^2y^3.\frac{2}{4}x^3y=2x^5y^4\)

a) Điều kiện: \(x\ne0;x\ne1\)

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

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

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

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

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

c) Thay: \(x=2\)vào \(\frac{1}{x+1}\)ta có: \(A=\frac{1}{2+1}=\frac{1}{3}\)

a) ĐKXĐ : \(\hept{\begin{cases}x\ne0\\x\ne1\end{cases}}\)

b)

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

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

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

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

c) \(A=\frac{1}{x+1}=\frac{1}{2+1}=\frac{1}{3}\)

Vậy \(A=\frac{1}{3}\)

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

a) ĐKXĐ : x ≠ 0 ; x ≠ ±1

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

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

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

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

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

c) Tại x = 2 (tmđk) => Giá trị biểu thức A = 1/3

Điều kiện xác định của \(P\)là: 

\(\hept{\begin{cases}x^2+2x+1\ne0\\x^2-1\ne0\\x\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne\pm1\\x\ne0\end{cases}}\)

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

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

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

Để \(P\)nguyên mà \(x\)nguyên suy ra \(x+1\inƯ\left(2\right)=\left\{-2,-1,1,2\right\}\Leftrightarrow x\in\left\{-3,-2,0,1\right\}\)

Đối chiếu điều kiện ta được \(x\in\left\{-3,-2\right\}\)thỏa mãn. 

a, ĐKXĐ: x\(\ne\) 1;-1;2

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

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

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

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

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

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

c, Khi x= -1

→A= \(\frac{-1-2}{-1-1}\)

= -3

Vậy khi x= -1 thì A= -3

Câu d thì mình đang suy nghĩ nhé, mình sẽ quay lại trả lời sau ^^

a,ĐKXĐ:x#1; x#-1; x#2

b,Ta có:

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

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

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

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

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

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

c,Tại x=-1 ,theo ĐKXĐ x#-1 \(\Rightarrow\)A không có kết quả

d,Để A có giá trị nguyên \(\Rightarrow\frac{x-2}{x+1}\)có giá trị nguyên

\(\Leftrightarrow x-2⋮x+1\)

\(\Leftrightarrow x+1-3⋮x+1\)

Mà \(x+1⋮x+1\Rightarrow3⋮x+1\)

\(\Rightarrow x+1\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)

\(\Rightarrow x\in\left\{0;-2;2;-4\right\}\)

Mà theo ĐKXĐ x#2\(\Rightarrow x\in\left\{0;-2;-4\right\}\)

Vậy \(x\in\left\{0;-2;-4\right\}\)thì a là số nguyên

Câu 1 :

a) ĐKXĐ : \(\hept{\begin{cases}x+1\ne0\\2x-6\ne0\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x\ne-1\\x\ne3\end{cases}}\)

b) Để \(P=1\Leftrightarrow\frac{4x^2+4x}{\left(x+1\right)\left(2x-6\right)}=1\)

\(\Leftrightarrow\frac{4x^2+4x-\left(x+1\right)\left(2x-6\right)}{\left(x+1\right)\left(2x-6\right)}=0\)

\(\Rightarrow4x^2+4x-2x^2+4x+6=0\)

\(\Leftrightarrow2x^2+8x+6=0\)

\(\Leftrightarrow x^2+4x+4-1=0\)

\(\Leftrightarrow\left(x+2-1\right)\left(x+2+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+3=0\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=-1\left(KTMĐKXĐ\right)\\x=-3\left(TMĐKXĐ\right)\end{cases}}\)

Vậy : \(x=-3\) thì P = 1.

a, ĐKXĐ : x khác -1 và 1

b, A = 2x^2+4x+2/(x-1).(x+1)  .  (x-1)/10

  = 2.(x^2+2x+1)/10.(x+1)

  = (x+1)^2/5.(x+1)

  = x+1/5

k mk nha

a, ĐKXĐ: \(x\ne\pm1\)

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

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

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

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

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

\(A=\frac{\left(x+1\right)}{5}\)