1. Ta có:

\(\frac{1}{x}+\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+...+\frac{1}{\left(x+2013\right)\left(x+2014\right)}\)

\(=\frac{1}{x}+\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+2013}-\frac{1}{x+2014}\)

\(=\frac{2}{x}-\frac{1}{x+2014}\)

\(=\frac{2\left(x+2014\right)}{x\left(x+2014\right)}-\frac{x}{x\left(x+2014\right)}\)

\(=\frac{2x+4028-x}{x\left(x+2014\right)}=\frac{x+4028}{x\left(x+2014\right)}\)

2a) ĐKXĐ: x \(\ne\)1 và x \(\ne\)-1

b) Ta có: A = \(\frac{x^2-2x+1}{x-1}+\frac{x^2+2x+1}{x+1}-3\)

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

A = \(x-1+x+1-3\)

A = \(2x-3\)

c) Với x = 3 => A = 2.3 - 3 = 3

c) Ta có: A = -2

=> 2x - 3 = -2

=> 2x = -2 + 3 = 1

=> x= 1/2

a) Ta thấy x=-2 thỏa mãn ĐKXĐ của B.

Thay x=-2 và B ta có :

\(B=\frac{2\cdot\left(-2\right)+1}{\left(-2\right)^2-1}=\frac{-3}{3}=-1\)

b) Rút gọn : 

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

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

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

Xấu nhỉ ??

a) ĐK: \(x\ne0,x\ne\pm3\)

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

\(=\left(\frac{1}{x+3}+\frac{1}{x+3}\right)\div\frac{x}{x+3}\)

\(=\frac{2}{x+3}\times\frac{x+3}{x}=\frac{2}{x}\)

b) \(\left|A\right|=\left|\frac{2}{x}\right|=3\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{2}{x}=3\\\frac{2}{x}=-3\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=-\frac{2}{3}\end{cases}}\)(thỏa mãn) 

a) \(B=\frac{x}{x+1}+\frac{2x-3}{x-1}-\frac{2x^2-x-3}{x^2-1}\)

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

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

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

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

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

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

MÌnh nghĩ đề câu b là với x>-4 mới đúng chứ

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

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

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

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

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

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

Áp dụng BĐT Cô - Si cho 2 số dương \(x-2;\frac{4}{x-2}\)ta có :

\(x-2+\frac{4}{x-2}\ge2\sqrt{\frac{\left(x-2\right).4}{x-2}}=2\sqrt{4}=4\)

\(\Rightarrow x-2+\frac{4}{x-2}\ge4\Rightarrow x-2+\frac{4}{x-2}+4\ge8\)

Hay \(S_{min}=4\Leftrightarrow x-2=\frac{4}{x-2}\)

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

\(\Rightarrow x^2+4x=0\Rightarrow x\left(x+4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\left(tm\right)\\x=-4\left(ktm\right)\end{cases}}\)\(\Rightarrow...\)

Bài 1:

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

\(=\frac{Ax-2A+Bx-B}{x^2-3x+2}=\frac{\left(A+B\right)x-\left(2A+B\right)}{x^2-3x+2}\)

so sách với tử số vừa tìm dc với đề bài:

=> A+B=1

2A+B=-2

=>(2A+B)-(A+B)=-2-1

A=-3

=> B=1+3=4

b) sửa đề \(\frac{A}{x-1}+\frac{\left(Bx+C\right)}{x^2+1}=\frac{A}{x-1}+\frac{\left(Bx+C\right)}{x^2+1}\)

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

\(=\frac{Ax^2+A+Bx^2-Bx+Cx-C}{\left(x-1\right)\left(x^2+1\right)}=\frac{\left(A+B\right)x^2+\left(C-B\right)x+\left(A-C\right)}{\left(x-1\right)\left(x^2+1\right)}\)

so sánh với tử số bên cạnh là \(x^2+2x-1\)

=>\(A+B=1\)

\(C-B=2\)

\(A-C=-1\)

=> \(A=1,B=0,C=2\)

bài 2:

quy đồng hai hạng tử đầu tiên:

=> \(\frac{x}{1-x^2}+\frac{y}{1-y^2}=\frac{x\left(1-y^2\right)+y\left(1-x^2\right)}{\left(1-x^2\right)\left(1-y^2\right)}=\frac{\left(x+y\right)\left(1-xy\right)}{\left(1-x^2\right)\left(1-y^2\right)}\)

từ xy+yz+xz=1=> 1-xy=z(x+y) thay vào biểu thức vừa tìm dc ta có:

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

\(VT=\frac{z\left(x+y\right)^2}{\left(1-x^2\right)\left(1-y^2\right)}+\frac{z}{1-z^2}=z\left\lbrace\frac{\left(x+y\right)^2\left(1-z^2\right)+\left(1-x^2\right)\left(1-y^2\right)}{\left(1-x^2\right)\left(1-y^2\right)\left(1-z^2\right)}\right)\)

ta có:

\(\left(x+y\right)^2-z^2\left(x+y\right)^2+1-x^2-y^2+x^2y^2\)

=\(\left(x^2+2xy+y^2\right)-z^2\left(x+y\right)^2+1-x^2-y^2+x^2y^2\)

=\(\left(1+xy\right)^2-z^2\left(x+y\right)^2=\left(1+xy-xz-yz\right)\left(1+xy+xz+yz\right)\)

=\(4xy\)

thay vào biểu thức ban đầu:

\(z\cdot\frac{4xy}{\left(1-x^2\right)\left(1-y^2\right)\left(1-z^2\right)}=\frac{4xyz}{\left(1-x^2\right)\left(1-y^2\right)\left(1-z^2\right)}\left(đpcm\right)\)

bài 3:

xếp hạng tổng k của dãy số:

\(a_{k}=\frac{k}{k^4+k+1}\)

=> \(a_{k}=\frac12\left\lbrace\frac{\left(k^2+k+1\right)-\left(k^2-k+1\right)}{\left(k^2-k+1\right)\left(k^2+k+1\right)}\right\rbrace=\frac12\left(\frac{1}{k^2-k+1}-\frac{1}{k^2+k+1}\right)\)

thay k=1,2,3,4,...,n)

=> \(S=\frac12\left\lbrace\left(\frac11-\frac13\right)+\left(\frac13-\frac17\right)+\cdots+\left(\frac{1}{n^2-n+1}-\right.\frac{1}{n^2+n+1}\right)\) S=\(\frac12\left(1-\frac{1}{n^2+n+1}\right)\)

\(S=\frac{n\left(n+1\right)}{2\left(n^2+n+1\right)}\)

\(e ) Để \)  \(M\)\(\in\)\(Z \)  \(thì\) \(1 \)\(⋮\)\(x +3\)

\(\Leftrightarrow\)\(x + 3 \)\(\in\)\(Ư\)_\((1)\)\(= \) { \(\pm\)\(1 \) }

\(Lập\)  \(bảng :\)

\(Vậy : Để \)  \(M\)\(\in\)\(Z\)  \(thì\) \(x\)\(\in\){ \(- 4 ; - 2\) }

e) Để M \(\in\)Z <=> \(\frac{1}{x+3}\in Z\)

<=> 1 \(⋮\)x + 3 <=> x + 3 \(\in\)Ư(1) = {1; -1}

Lập bảng: 

Vậy ....

f) Ta có: M > 0

=> \(\frac{1}{x+3}\) > 0

Do 1 > 0 => x + 3 > 0

=> x > -3

Vậy để M > 0 khi x > -3 ; x \(\ne\)3 và x \(\ne\)-3/2