a) \(\dfrac{9x^2-6x+1}{9x^2-1}\)

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

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

\(=\dfrac{3\cdot\left(-3\right)-1}{3\cdot\left(-3\right)+1}=\dfrac{-9-1}{-9+1}=\dfrac{-10}{-8}=\dfrac{5}{4}\)

b) Ta có: \(\dfrac{x^2-6x+9}{3x^2-9x}\)

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

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

\(=\dfrac{-\dfrac{1}{3}-3}{3\cdot\dfrac{-1}{3}}=\dfrac{-\dfrac{10}{3}}{-1}=\dfrac{10}{3}\)

c) Ta có: \(\dfrac{x^2-4x+4}{2x^2-4x}\)

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

\(=\dfrac{x-2}{2x}\)

\(=\dfrac{\dfrac{-1}{2}-2}{2\cdot\dfrac{-1}{2}}=\dfrac{-\dfrac{5}{2}}{-1}=\dfrac{5}{2}\)

1: Khi x=2 thì \(A=\dfrac{4\cdot2+1}{2-1}=9\)

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

a, Với \(x=3\)\(=>A=\frac{x-1}{2}=\frac{3-1}{2}=\frac{2}{2}=1\)

Vậy A = 1 khi x = 3

b, Ta có : \(B=\frac{1}{x}-\frac{x}{2x+1}+\frac{2x^2-3x-1}{x\left(2x+1\right)}\)

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

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

Ta có : \(A=\frac{x-1}{2};B=\frac{x-1}{2x+1}\)

\(=>C=A:B=\frac{x-1}{2}:\frac{x-1}{2x+1}=\frac{2x+1}{2}=x+\frac{1}{2}\)

đề sai bạn ơi 

a: \(B=\dfrac{3x\left(2x-3\right)-4\left(2x+3\right)-4x^2+23x+12}{\left(2x-3\right)\left(2x+3\right)}\cdot\dfrac{2x+3}{x+3}\)

\(=\dfrac{6x^2-9x-8x-12-4x^2+23x+12}{2x-3}\cdot\dfrac{1}{x+3}\)

\(=\dfrac{2x^2+6x}{\left(2x-3\right)}\cdot\dfrac{1}{x+3}=\dfrac{2x}{2x-3}\)

b: 2x^2+7x+3=0

=>(2x+3)(x+2)=0

=>x=-3/2(loại) hoặc x=-2(nhận)

Khi x=-2 thì \(A=\dfrac{2\cdot\left(-2\right)}{-2-3}=\dfrac{-4}{-7}=\dfrac{4}{7}\)

d: |B|<1

=>B>-1 và B<1

=>B+1>0 và B-1<0

=>\(\left\{{}\begin{matrix}\dfrac{2x+2x-3}{2x-3}>0\\\dfrac{2x-2x+3}{2x-3}< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-3< 0\\\dfrac{4x-3}{2x-3}>0\end{matrix}\right.\Leftrightarrow x< \dfrac{3}{4}\)

a) Ta có: \(\left(3x-2\right)^2+2\left(3x-2\right)\left(3x+2\right)+\left(3x+2\right)^2\)

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

\(=36x^2\)(1)

Thay \(x=-\dfrac{1}{3}\) vào biểu thức (1), ta được:

\(36\cdot\left(-\dfrac{1}{3}\right)^2=36\cdot\dfrac{1}{9}=4\)

b) Sửa đề: \(\left(x+y-7\right)^2-2\cdot\left(x+y-7\right)\left(y-6\right)+\left(y-6\right)^2\)

Ta có: \(\left(x+y-7\right)^2-2\cdot\left(x+y-7\right)\left(y-6\right)+\left(y-6\right)^2\)

\(=\left(x+y-7-y+6\right)^2\)

\(=\left(x-1\right)^2=100^2=10000\)

\(x=\sqrt{\dfrac{2\sqrt{3}+2-6\sqrt{3}}{2\sqrt{3}\left(2\sqrt{3}+2\right)}}=\sqrt{\dfrac{2-4\sqrt{3}}{2\sqrt{3}\left(2\sqrt{3}+2\right)}}\) ko tồn tại vì 2-4căn 3<0