1)\(A=\frac{b\left(2a\left(a+5b\right)+\left(a+5b\right)\right)}{a-3b}.\frac{a\left(a-3b\right)}{ab\left(a+5b\right)}=\frac{b\left(a+5b\right)\left(2a+1\right).a\left(a-3b\right)}{\left(a-3b\right).ab\left(a+5b\right)}\)

\(A=2a+1\)=>lẻ với mọi a thuộc z=> dpcm 

2) từ: x+y+z=1=> xy+z=xy+1-x-y=x(y-1)-(y-1)=(y-1)(x-1)

tường tự: ta có tử của Q=(x-1)^2.(y-1)^2.(z-1)^2=[(x-1)(y-1)(z-1)]^2=[-(z+y).-(x+y).-(x+y)]^2=Mẫu=> Q=1

3) kiểm tra lại xem đề đã chuẩn chưa

ta tính riêng từng biểu thức:

\(A^2=\left(\frac{y}{z}+\frac{z}{y}\right)^2=\frac{y^2}{z^2}+\frac{z^2}{y^2}+2\)

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

\(C^2=\left(\frac{x}{y}+\frac{y}{x}\right)^2=\frac{x^2}{y^2}+\frac{y^2}{x^2}+2\)

cộng lại ta có:

\(A^2+B^2+C^2=\left(\frac{x^2}{y^2}+\frac{y^2}{x^2}\right)+\left(\frac{y^2}{z^2}+\frac{z^2}{y^2}\right)+\left(\frac{z^2}{x^2}+\frac{x^2}{z^2}\right)+6\)

\(A\cdot B\cdot C=\left(\frac{y}{z}+\frac{z}{y}\right)\left(\frac{z}{x}+\frac{x}{z}\right)\left(\frac{x}{y}+\frac{y}{x}\right)\)

\(A\cdot B\cdot C=\left(\frac{y}{x}+\frac{xy}{z^2}+\frac{z^2}{xy}+\frac{x}{y}\right)\left(\frac{x}{y}+\frac{y}{x}\right)\)

\(A\cdot B\cdot C=\left(\frac{x^2}{y^2}+\frac{y^2}{x^2}\right)+\left(\frac{y^2}{z^2}+\frac{z^2}{y^2}\right)+\left(\frac{z^2}{x^2}+\frac{x^2}{z^2}\right)+2\)

trừ \(A^2+B^2+C^2\) cho \(A\cdot B\cdot C\)

= 6-2

=4

Câu 1 : 

a, \(\frac{3}{x+3}-\frac{x-6}{x^2+3x}=\frac{3x-x+6}{x\left(x+3\right)}=\frac{2x+6}{x\left(x+3\right)}=\frac{2}{x}\)

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

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

Bài 2 : 

a, Với \(x\ne\pm2\)

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

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

\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{2}=\frac{-3}{x-2}\)

b, Thay x = -4 vào biểu thức trên ta được : 

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

c, Để A \(\inℤ\Rightarrow x-2\inƯ\left(-3\right)=\left\{\pm1;\pm3\right\}\)

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

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

a) sửa đề: \(\frac{x^2}{\left(x-y\right)\left(x-z\right)}+\frac{y^2}{\left(y-x\right)\left(y-z\right)}+\frac{z^2}{\left(z-x\right)\left(z-y\right)}\)

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

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

xét tử số:

Tử=\(x^2y-x^2z+y^2z-y^2x+z^2x-z^2y\)

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

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

=\(\left(y-z\right)\left\lbrace x^2-x\left(y+z\right)+yz\right\rbrace\)

=\(\left(y-z\right)\left\lbrace x\left(x-y\right)-z\left(x-y\right)\right\rbrace\)

=\(\left(y-z\right)\left(x-y\right)\left(x-z\right)\)

=\(-\left(x-y\right)\left(y-z\right)\left(z-x\right)\)

thay lại vào biểu thức cũ:

\(\Rightarrow-\frac{\left\lbrace-\left(x-y\right)\left(y-z\right)\left(z-x\right)\right\rbrace}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=\frac{\left(x-y\right)\left(y-z\right)\left(z-x\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)

=\(1\)

b) \(\frac{1}{\left(a-b\right)\left(b-c\right)}+\frac{1}{\left(b-c\right)\left(c-a\right)}+\frac{1}{\left(c-a\right)\left(a-b\right)}\)

=\(\frac{\left(c-a\right)+\left(a-b\right)+\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

\(=\frac{0}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)