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

\(P=\frac{\frac{1}{a^2}}{\frac{1}{b}+\frac{1}{c}}+\frac{\frac{1}{b^2}}{\frac{1}{a}+\frac{1}{c}}+\frac{\frac{1}{c^2}}{\frac{1}{a}+\frac{1}{b}}\)

Đặt \(\hept{\begin{cases}x=\frac{1}{a}\\y=\frac{1}{b}\\z=\frac{1}{c}\end{cases}}\Rightarrow xyz=1\Rightarrow P=\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}\)

Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có: 

\(P\ge\frac{\left(x+y+z\right)^2}{y+z+x+z+x+y}=\frac{x+y+z}{2}\ge\frac{3\sqrt[3]{xyz}}{2}=\frac{3}{2}\)

Dấu "=" xảy ra khi \(x=y=z\Leftrightarrow a=b=c=1\)

Cần cách khác thì nhắn cái

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=\(\frac12\left(1-\frac{1}{n^2+n+1}\right)\)

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

Bài 1 : 

a, \(4\left(-3\text{x}\right)yx^5=-12x^6y\)

Hệ số : -12 , Biến : \(x^6y\)

b, \(7\text{x}y^5\left(-5\text{x}^3y^2\right)=-35\text{x}^4y^7\)

Hệ số : -35 , Biến : \(x^4y^7\)

c, \(-\left[-23\text{a}b\text{x}^3\left(-y\right)\right]=-23ab\text{x}^3y\)

Hệ số : -23 , Biến : \(ab\text{x}^3y\)

d, \(\left(-3\text{a}y^3\right)\left(-5b^2xy\right)=3\text{a}y^3\cdot5b^2xy=15\text{a}b^2xy^4\)

Hệ số : 15   , Biến : \(ab^2xy^4\)

Bài 2 : 

a, \(x^4y^5z\)có Bậc là 10 

b, \(-\left(-7\text{x}^7y\right)=7\text{x}^7y\)có bậc là 8 

c, 

b, \(-41\left(x^3y^2\right)^2=-41\text{x}^6y^4\)có bậc là 10 

d, \(-2\frac{1}{3}x^3\left(xy^2\right)^3=-\frac{7}{3}x^3.x^3.y^6=-\frac{7}{3}x^6y^6\)có bậc là 12