$\left[\frac{7x^2+7y^2}{12x^2+12y^2}\div\frac{2x+2y}{3x+3y}\right]\ \times\frac{2x}{21}$  

 $\ \ \left[\frac{3x^3+3y^3\ }{2x^3+2y^3}\times\frac{3x^2+3y^2}{2x^2+2y^2}\right]+\left[\frac{7x+7y}{5x+5y}\div\frac{3x^2+3y^2}{2x^2+2y^2}\right]$  

 $\frac{x^2+16}{x^2+8x+16}+\frac{2x+16}{x^2+64}$  

 $\ \ \ \ \frac{\left[\frac{x^2-4x-32}{x^2-5x-24}\times\frac{x^2+7x+12}{\left(x+4\right)^2}\right]+\left[\frac{x^2-x-30}{x^2-11x-30}\times\frac{x^2+x+30}{x^2+11x+30}\right]}{\left[\frac{4x-20}{2x+6}\div\frac{2x-10}{2x+6}\right]+\left[\frac{4x^2-12}{2x^2+6}\div\frac{2x^2-6}{2x^2+6}\right]}$  

 $\frac{4\left(2x^2-x-6\right)}{2\left(2x^2-12x+16\right)}\times\frac{6\left(2x^2-2x-24\right)}{3\left(2x^2+9x+9\right)}=\frac{2x+6}{\left(3x+6\right)^2}\div\frac{\left(x-3\right)^2}{\left(12x+24\right)^2}$  

 $\frac{\left(2x-6\right)\times\left(4x+12\right)^2}{\left(5x-15\right)^2\times\left(3x+9\right)}\times\frac{\left(7x-49\right)^2+\left(8x+56\right)}{\left(4x-28\right)\times\left(2x+14\right)^2}=\frac{2-2x}{x-2}\times\frac{x^2-5x+6}{2x^{ }-6}$  

 $\frac{4}{4x+4}+\frac{2}{2x+2}=$  

 $\frac{\left(7x^2-14\right)+\left(5x^2-10\right)}{3x^2-6}=$  

$${\left(x-\sqrt{6x}+3\right)}^2=0,{\left(y-\sqrt{8y}+4\right)}^2=0$$  เมื่อคำตอบของ x,y เป็นด้านข้ามเละชิดตามลำดับ จงหา cosec θ

 $x^3=^{ }\left(\frac{2x^2+2y}{\frac{9y+9x^2}{3}}\right)^2$