Điều kiện:
[tex]\left\{\begin{matrix} 5x-1\geq 0\Leftrightarrow x\geq \frac{1}{5}\\ \\3x-2\geq 0\Leftrightarrow x\geq \frac{2}{3} \\ \\x^2 -2x+5\geq0 \forall x\epsilon R \end{matrix}\right.[/tex]
Đề bài suy ra :[tex]-2x-1+\sqrt{3x-2}+\sqrt{5x-1}+(x^2-3x+2)\sqrt{x^2-2x+5}=0\\ \Leftrightarrow (-2x+2)+(\sqrt{3x-2}-1)+(\sqrt{5x-1}-2)+(x^2-2x-x+2)\sqrt{x^2-2x+5}=0\\ \Leftrightarrow -2(x-1)+(x-1)(\frac{3}{\sqrt{3x-2}+1})+(x-1)(\frac{5}{\sqrt{5x-1}+2})+(x-1)(x-2)\sqrt{x^2-2x+5}=0 \\ \Leftrightarrow (x-1)(-2+\frac{3}{\sqrt{3x-2}+1}+\frac{5}{\sqrt{5x-1}+2}+(x-2)\sqrt{x^2-2x+5})=0[/tex]
[tex]\Leftrightarrow (x-1)[(\frac{3}{\sqrt{3x-2}+1}-1)+(\frac{5}{\sqrt{5x-1}+2}-1)+(x-2)(\sqrt{x^2-2x+5} )]\\ \Leftrightarrow (x-1)[(\frac{3-\sqrt{3x-2}-1}{\sqrt{3x-2}+1})+(\frac{5-\sqrt{5x-1}-2}{\sqrt{5x-1}+2})+(x-2)(\sqrt{x^2-2x+5} )]\\ \Leftrightarrow (x-1)[(\frac{2-\sqrt{3x-2}}{\sqrt{3x-2}+1})+(\frac{3-\sqrt{5x-1}}{\sqrt{5x-1}+2})+(x-2)(\sqrt{x^2-2x+5} )]\\[/tex]
Mình chia làm 2 trường hợp cho đỡ rối mắt
TH1:
[tex]\frac{4-3x+2}{(\sqrt{3x-2}+1).(\sqrt{3x-2}+2)}+\frac{9-5x+1}{(\sqrt{5x-1}+2).(\sqrt{5x-1}+3)}+(x-2)\sqrt{x^2 -2x+5}=0\\ \Leftrightarrow \frac{6-3x}{(\sqrt{3x-2}+1).(\sqrt{3x-2}+2)}+\frac{10-5x}{(\sqrt{5x-1}+2).(\sqrt{5x-1}+3)}+(x-2)\sqrt{x^2 -2x+5}=0\\\Leftrightarrow (x-2)(\frac{-3}{(\sqrt{3x-2}+1).(\sqrt{3x-2}+2)}+\frac{-5}{(\sqrt{5x-1}+2).(\sqrt{5x-1}+3)}+\sqrt{x^2 -2x+5})=0[/tex]
mà:x[tex]\geq \frac{2}{3}[/tex] [tex]\frac{-3}{(\sqrt{3x-2}+1).(\sqrt{3x-2}+2)}+\frac{-5}{(\sqrt{5x-1}+2).(\sqrt{5x-1}+3)}+\sqrt{x^2 -2x+5} > 0[/tex]
Dấu đẳng thức không xãy ra. Chỉ còn x-2=0[tex]\Leftrightarrow[/tex] x=2(TMĐK)
TH2:[tex]x-1=0\Leftrightarrow x=1[/tex](TMĐK)
Vậy phương trình đã cho có 2 nghiệm x=1,x=2
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
