Đáp án:

Giải thích các bước giải:

ta có: $\sqrt{x}$ + $\sqrt{1-x}$ + $\sqrt{x(1-x)}$ = 1 (1)

đặt $\sqrt{x}$  = a ; $\sqrt{1-x}$ = b (a,b ≥ 0) ⇒ a² + b² =1

(1) ⇒ a + b + ab =1

ta có hpt: $\left \{ {{a+b+ab=1} \atop {a^2+b^2=1}} \right.$ 

⇔ $\left \{ {{2a +2b+2ab=2} \atop {a^2+b^2=1}} \right.$ 

⇔ $\left \{ {{a^2+b^2+2a +2b+2ab=3} \atop {a^2+b^2=1}} \right.$ 

⇔ $\left \{ {{(a+b)^2+2(a +b)-3=0} \atop {a^2+b^2=1}} \right.$ 

⇔ $\left \{ {{[(a+b)^2-(a +b)]+[3(a+b)-3]=0} \atop {a^2+b^2=1}} \right.$ 

⇔ $\left \{ {{(a+b)[(a +b)-1]+3[(a+b)-1]=0} \atop {a^2+b^2=1}} \right.$ 

⇔ $\left \{ {{(a+b+3)(a +b-1)=0} \atop {a^2+b^2=1}} \right.$ 

⇔ \(\left[ \begin{array}{l}\left \{ {{a+b+3=0} \atop {a^2+b^2=1}} \right.\\\left \{ {{a +b-1=0} \atop {a^2+b^2=1}} \right. \end{array} \right.\)

⇔ \(\left[ \begin{array}{l}\left \{ {{a=-(b+3)} \atop {b^2+[-(b+3)]^2=1}} \right.\\\left \{ {{a =1-b} \atop {(1-b)^2+b^2=1}} \right. \end{array} \right.\)

⇔ \(\left[ \begin{array}{l}\left \{ {{a=-(b+3)} \atop {b^2+b^2+6b+9=1}} \right.\\\left \{ {{a =1-b} \atop {b^2-2b+1+b^2=1}} \right. \end{array} \right.\)

⇔ \(\left[ \begin{array}{l}\left \{ {{a=-(b+3)} \atop {2b^2+6b+8=0}} \right.\\\left \{ {{a =1-b} \atop {2b^2-2b=0}} \right. \end{array} \right.\)

⇔ \(\left[ \begin{array}{l}\left \{ {{a=-(b+3)} \atop {b^2+3b+4=0}} \right.\\\left \{ {{a =1-b} \atop {2b(b-1)=0}} \right. \end{array} \right.\)

⇔ \(\left[ \begin{array}{l}\left \{ {{a=-(b+3)} \atop {b^2+2.\frac{3}{2}b+\frac{9}{4}+\frac{7}{4}=0}} \right.\\\left \{ {{a =1-b} \atop {2b(b-1)=0}} \right. \end{array} \right.\)

⇔ \(\left[ \begin{array}{l}\left \{ {{a=-(b+3)} \atop {(b+\frac{3}{2})^2+\frac{7}{4}=0} (vô lí)} \right.\\\left \{ {{a =1-b} \atop {2b(b-1)=0}} \right. \end{array} \right.\)

⇒ $\left \{ {{a=1-b} \atop {2b(b-1)=0}} \right.$ 

⇔ \(\left[ \begin{array}{l}\left \{ {{2b=0} \atop {a= 1-b}} \right.\\\left \{ {{b-1=0} \atop {a= 1-b}} \right. \end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}\left \{ {{b=0} \atop {a= 1}} \right.\\\left \{ {{b=1} \atop {a= 0}} \right. \end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}\left \{ {{\sqrt{1-x}=0} \atop {\sqrt{x}= 1}} \right.\\\left \{ {{\sqrt{1-x}=1} \atop {\sqrt{x}= 0}} \right. \end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}\left \{ {{1-x=0} \atop {x= 1}} \right.\\\left \{ {{1-x=1} \atop {x= 0}} \right. \end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}x=1\\x=0\end{array} \right.\)