Giúp e với ạ sosssss

Giải thích các bước giải:

a.ĐKXĐ: $x\ge 1, x\ne 5, x\ne  7$

Đặt $\sqrt{x-1}=t, t\ge 0, \to t^2+1=x$

Ta có:

$A=(\dfrac{t+2}{t^2+1-5t+5}+\dfrac{t+3}{t-2}-\dfrac{t+2}{t-3}):(2-\dfrac{t}{1+t})$

$\to A=(\dfrac{t+2}{t^2-5t+6}+\dfrac{t+3}{t-2}-\dfrac{t+2}{t-3}):(2-\dfrac{t}{1+t})$

$\to A=(\dfrac{t+2}{\left(t-2\right)\left(t-3\right)}+\dfrac{\left(t+3\right)\left(t-3\right)}{\left(t-2\right)\left(t-3\right)}-\dfrac{\left(t+2\right)\left(t-2\right)}{\left(t-2\right)\left(t-3\right)}):\dfrac{2\left(1+t\right)-t}{1+t}$

$\to A=\dfrac{t+2+\left(t+3\right)\left(t-3\right)-\left(t+2\right)\left(t-2\right)}{\left(t-2\right)\left(t-3\right)}:\dfrac{2+t}{1+t}$

$\to A=\dfrac{t-3}{\left(t-2\right)\left(t-3\right)}:\dfrac{2+t}{1+t}$

$\to A=\dfrac{1}{t-2}: \dfrac{2+t}{1+t}$

$\to A=\dfrac{1\cdot \:\left(1+t\right)}{\left(t-2\right)\left(2+t\right)}$

$\to A=\dfrac{t+1}{t^2-4}$

$\to A=\dfrac{\sqrt{x-1}+1}{x-1-4}$

$\to A=\dfrac{\sqrt{x-1}+1}{x-5}$

b.Để $\dfrac1{A}<\dfrac{32}{15}\to A<0$

$\to \dfrac{\sqrt{x-1}+1}{x-5}<0$

$\to x-5<0\to x<5$

Lại có:

$\dfrac1A<\dfrac{-32}{15}$

$\to \dfrac{x-5}{\sqrt{x-1}+1}<\dfrac{-32}{15}$

$\to 15(x-5)<-32(\sqrt{x-1}+1)$ vì $\sqrt{x-1}+1>0$

$\to 15(x-1)-60<-32\sqrt{x-1}-32$

$\to 15(x-1)+32\sqrt{x-1}-28<0$

 $\to (3\sqrt{x-1}-2)(5\sqrt{x-1}+14)<0$

$\to \sqrt{x-1}<\dfrac23$ vì $5\sqrt{x-1}+14>0$

$\to 0\le x-1<\dfrac49$

$\to 1\le x<\dfrac{13}9$