1. a/ ĐKXD: \(x\ne0\) và \(x\ne\pm3\)

Rút gọn:

\(A=\left(\frac{3-x}{x+3}.\frac{x^2+6x+9}{x-9}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)

\(\Rightarrow\left(\frac{-\left(x-3\right)}{x+3}.\frac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}+\frac{x}{\left(x+3\right)}\right).\frac{x+3}{3x^2}\)

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

\(\Rightarrow\left(\frac{-\left(x+3\right)+x}{x+3}\right).\frac{x+3}{3x^2}\)

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

b/ Thay \(x=\frac{-1}{2}\) ( phù hợp điều kiện )

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

\(\Rightarrow-1:\frac{1}{4}=-4\)

\(1,ĐK:x\ne0;x\ne\pm6\)

\(A=\left[\frac{6x+1}{x\left(x-6\right)}+\frac{6x-1}{x\left(x+6\right)}\right].\frac{\left(x+6\right)\left(x-6\right)}{12\left(x^2+1\right)}\)

\(=\frac{6x^2+36x+x+6+6x^2-36x-x+6}{x}.\frac{1}{12\left(x^2+1\right)}\)

\(=\frac{12\left(x^2+1\right)}{x}.\frac{1}{12\left(x^2+1\right)}=\frac{1}{x}\)

\(2,A=\frac{1}{x}=\frac{1}{\frac{1}{\sqrt{9+4\sqrt{5}}}}=\sqrt{9+4\sqrt{5}}\)

Cho tam giác ABC vuông tại B có góc B₁=B₂  ; Â=60^o, kẻ BH vuông góc với AC (H thuộc AC). Qua B kẻ đường thẳng d song song với AC.

a) Tính góc ABH.

b) Chứng minh đường thẳng d vuông góc với BH.

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

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

a) ĐKXĐ: \(x\ne\left\{-3;-\frac{1}{3}\right\}\)

Ta có: \(\frac{3x-1}{3x+1}+\frac{x-3}{x+3}=\)\(\frac{\left(3x-1\right)\left(x+3\right)+\left(x-3\right)\left(3x+1\right)}{\left(3x+1\right)\left(x+3\right)}\)=\(\frac{3x^2+9x-x-3+3x^2+x-9x-3}{3x^2+9x+x+3}\)

          =  \(\frac{6x^2-6}{3x^2+10x+3}\)

=>  \(\frac{6x^2-6}{3x^2+10x+3}=2\)

<=> \(6x^2-6=6x^2+20x+6\)

<=> 20x=12

<=>x=\(\frac{12}{20}=\frac{3}{5}\)

Vậy x=3/5