Solve each of the following quadratic equations:
(i) $\frac{1}{x+1}+\frac{2}{x+2}=\frac{5}{x+4}, \quad x \neq-1,-2,-4$
(ii) $\frac{1}{x+1}+\frac{3}{5 x+1}=\frac{5}{x+4}, x \neq-1,-\frac{1}{5},-4$
(i)
$\frac{1}{x+1}+\frac{2}{x+2}=\frac{5}{x+4}, \quad x \neq-1,-2,-4$
$\Rightarrow \frac{x+2+2 x+2}{(x+1)(x+2)}=\frac{5}{x+4}$
$\Rightarrow \frac{3 x+4}{x^{2}+3 x+2}=\frac{5}{x+4}$
$\Rightarrow(3 x+4)(x+4)=5\left(x^{2}+3 x+2\right)$
$\Rightarrow 3 x^{2}+16 x+16=5 x^{2}+15 x+10$
$\Rightarrow 2 x^{2}-x-6=0$
$\Rightarrow 2 x^{2}-4 x+3 x-6=0$
$\Rightarrow 2 x(x-2)+3(x-2)=0$
$\Rightarrow(x-2)(2 x+3)=0$
$\Rightarrow x-2=0$ or $2 x+3=0$
$\Rightarrow x=2$ or $x=-\frac{3}{2}$
Hence, 2 and $-\frac{3}{2}$ are the roots of the given equation.
(ii)
$\frac{1}{x+1}+\frac{3}{5 x+1}=\frac{5}{x+4}$
$\frac{(5 x+1)+(3 x+3)}{(x+1)(5 x+1)}=\frac{5}{x+4}$
$(x+4)[(5 x+1)+(3 x+3)]=5(x+1)(5 x+1)$
$(x+4)[8 x+4]=25 x^{2}+30 x+5$
$8 x^{2}+4 x+32 x+16=25 x^{2}+30 x+5$
$8 x^{2}+36 x+16=25 x^{2}+30 x+5$
$17 x^{2}-6 x-11=0$
$17 x^{2}-17 x+11 x-11=0$
$17 x(x-1)+11(x-1)=0$
$(17 x+11)(x-1)=0$
$(17 x+11)=0$ or $(x-1)=0$
$x=-\frac{11}{17}$ or $x=1$
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