1/(x^2+6x+9)-1/(x^2-6x+9)=(x-3)/(x-3)(x+3)-(x+3)/(x-3)(x+3)= -6/(x-3)(x+3)

1/(x+3)+1/(x-3)=

sai rồi bấm lộn thôi mà

I am sorry

\(\frac{1}{x^2+6x+9}+\frac{1}{6x-x^2-9}+\frac{x}{x^2-9}\)

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

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

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

\(=\frac{x^3-21x}{\left(x+3\right)^2\left(x-3\right)^2}\)

\(\frac{1}{x^2+6x+9}+\frac{1}{6x-x^2-9}+\frac{x}{x^2-9}\)

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

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

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

\(=\frac{x^3-21x}{\left(x+3\right)^2\left(x-3\right)^2}\)

\(\frac{1}{x^2+6x+9}+\frac{1}{6x-x^2-9}+\frac{x}{x^2-9}\)
\(=\frac{1}{\left(x+3\right)^2}-\frac{1}{\left(x-3\right)^2}+\frac{x}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{\left(x-3\right)^2}{\left(x+3\right)^2\left(x-3\right)^2}-\frac{\left(x+3\right)^2}{\left(x+3\right)^2\left(x-3\right)^2}+\frac{x\left(x+3\right)\left(x-3\right)}{\left(x+3\right)^2\left(x-3\right)^2}\)
\(=\frac{x^2-6x+9-x^2-6x-9+x^3-9x}{\left(x+3\right)^2\left(x-3\right)^2}\)
\(=\frac{x^3-21x}{\left(x+3\right)^2\left(x-3\right)^2}\)
 

\(1,ĐK:x\ne0;x\ne\pm6\)

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

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

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

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

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.

ĐKXĐ \(\hept{\begin{cases}x\ne0\\x\ne\pm3\end{cases}}\)

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

\(=\frac{x-3}{2x}\left(1-\frac{6}{x-3}\right)\)

\(=\frac{x-3}{2x}.\frac{x-9}{x-3}=\frac{x-9}{2x}\)

\(M=\frac{\left(x-3\right)^2}{2x^2-6x}\left(1-\frac{6x+18}{x^2-9}\right)\left(x\ne\pm3;x\ne0\right)\)

\(\Leftrightarrow M=\frac{\left(x-3\right)^2}{2x\left(x-3\right)}\left(1-\frac{6\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\right)\)

\(\Leftrightarrow M=\frac{x-3}{2x}\cdot\left(1-\frac{6}{x-3}\right)\)

\(\Leftrightarrow M=\frac{x-3}{2x}\cdot\frac{x-9}{x-3}\)

\(\Leftrightarrow M=\frac{x-9}{2x}\)

Vậy với \(x\ne\pm3;x\ne0\)thì \(M=\frac{x-9}{2x}\)

x2 m vuông

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

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

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

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

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

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

\(A=-x^2\)