$\sqrt{x+1} - \sqrt{4-x}=1 (1)$ ĐK: $\left \{ {{x+1\geq0} \atop {4-x\geq0}} \right.$ $\left \{ {{x\geq-1} \atop {x\leq4}} \right.=> -1 \leq x \leq 4$
$(1) => \sqrt{x+1} = \sqrt{4-x}+1$
$=> x+1= 4-x+1+2.1.\sqrt{4-x}$
$=> 2x-4= 2\sqrt{4-x}$
$=> x-2= \sqrt{4-x}$ $=> (x-2)^2= 4-x => x^2 -4x+4= 4-x => x^2-3x=0$
$x= 0(nhận) hay x= 3(nhận)$
$\sqrt{(x- 2) \sqrt{ x-1}} - \sqrt{x-1} =1 (!)$
ĐK: $\left \{ {{x\geq1} \atop {(x-2)\sqrt{x-1}}\geq0(2)} \right.$
$(2)=> \left \{ {{x\geq1} \atop {x\geq2}} \right. => x\geq2$
Vậy ĐK: $x \geq2$
$(!) => \sqrt{(x- 2) \sqrt{ x-1}} =\sqrt{x-1} +1$
=> $(x- 2) \sqrt{ x-1} =2\sqrt{x-1} +1 +x-1$
=> $(x- 2) \sqrt{ x-1} =2\sqrt{x-1} +x$
=> $(x- 2)^2 ( x-1) =4(x-1) +x^2+4x\sqrt{x-1}$
=> $(x^2- 4x+4) ( x-1) =4(x-1) +x^2 +4x\sqrt{x-1}$
=> $x^3-x^2-4x^2+4x+4x-4 =4x-4 +x^2+4x\sqrt{x-1}$
=> $x^3-6x^2+4x=4x\sqrt{x-1}$
=> $x(x^2-6x+4-4\sqrt{x-1})=0$
=>$x= 0 hay x^2-6x+4-4\sqrt{x-1}=0 (*)$
$x^2-6x+4-4\sqrt{x-1}=0 (*)$
$x^2-6x+4=4\sqrt{x-1} (*)$
Giúp em dc tới đây thoi :((
