Question

1=(r+2)(r +1) /(18-r) (17-r) find r​

Answer 1

Answer:

$-\frac{19}{2}$

Step-by-step explanation:

$\frac{(r+1)(r+2)}{(18-r)(17-r)}=1\\\\=>\frac{(r)^{2}+(1+2)r+(1)(2) }{(-r)^{2}+(17+18)r+(17)(18) }=1\\\\=>\frac{r^{2}+3r+2 }{r^{2}+35r+306 }=1\\\\=>\frac{3r+2}{35r+306}=1\\\\=>3r+2=35r+306\\=>3r-35r=306-2\\=>-32r=304\\\\=>r=\frac{-304}{32} \\\\=>r=-\frac{19}{2}$

Verification-

$\frac{(\frac{-19}{2} +1)(\frac{-19}{2}+2 )}{(18-\frac{-19}{2}(17-\frac{-19}{2}) )} \\\\=>\frac{(\frac{-17}{2})(\frac{-15}{2} )}{(\frac{17}{2})(\frac{15}{2} ) }\\\\=> \frac{-17\times -15 \times 2\times 2}{2\times 2\times 17\times 15}\\\\=> 1$

Hence verified.