1) [tex]\sqrt{x^2 - x + 5} = x+4[/tex]
2)[tex]\sqrt{x^2 - x+5 } = \sqrt{x+5}[/tex]
3)[tex]\sqrt{x+3} - \sqrt{x} = 1[/tex]
4)[tex]\sqrt{2x-1} = x-2[/tex]
5)[tex]\sqrt{2x-1}-\sqrt{x-1}=\sqrt{x-4}[/tex]
6)[tex](4x-1)\sqrt{x^2 + 1} = 2x^2 + 2x +1[/tex]
1) ĐK: $x\geq -4$
pt $\Leftrightarrow x^2-x+5=x^2+8x+16$
$\Leftrightarrow 9x=-11$
$\Leftrightarrow x=\dfrac{-11}9$ (TM)
Vậy...
2) ĐK: $x\geq -5$
pt $\Leftrightarrow x^2-x+5=x+5$
$\Leftrightarrow x^2-2x=0
\\\Leftrightarrow x(x-2)=0$
$\Leftrightarrow x=0 \ or \ x=2$ (TM)
Vậy...
3) ĐK: $x\geq 0$
pt $\Leftrightarrow \sqrt{x+3}=\sqrt{x}+1$
$\Leftrightarrow x+3=x+2\sqrt{x}+1
\\\Leftrightarrow 2\sqrt{x}=2
\\\Leftrightarrow \sqrt{x}=1$
$\Leftrightarrow x=1$ (TM)
Vậy...
4) ĐK: $x\geq 2$
pt $\Leftrightarrow 2x-1=x^2-4x+4$
$\Leftrightarrow x^2-6x+5=0
\\\Leftrightarrow (x-1)(x-5)=0$
$\Leftrightarrow x=5$ (TM) or $x=1$ (KTM)
Vậy...
5) ĐK: $x\geq 4$
pt $\Leftrightarrow \sqrt{2x-1}=\sqrt{x-4}+\sqrt{x-1}$
$\Leftrightarrow 2x-1=x-4+x-1+2\sqrt{(x-4)(x-1)}
\\\Leftrightarrow \sqrt{x^2-5x+4}=2
\\\Leftrightarrow x^2-5x+4=4
\\\Leftrightarrow x(x-5)=0$
$\Leftrightarrow x=0$ (KTM) or $x=5$ (TM)
Vậy...
6) pt $\Leftrightarrow 2x^2+2-\sqrt{x^2+1}(4x-2+1)+2x-1$
$\Leftrightarrow 2(x^2+1)-\sqrt{x^2+1}(4x-2)-\sqrt{x^2+1}+2x-1
\\\Leftrightarrow 2(x^2+1)-2\sqrt{x^2+1}(2x-1)-\sqrt{x^2+1}+2x-1=0
\\\Leftrightarrow 2\sqrt{x^2+1}(\sqrt{x^2+1}-2x+1)-(\sqrt{x^2+1}-2x+1)=0
\\\Leftrightarrow (\sqrt{x^2+1}-2x+1)(2\sqrt{x^2+1}-1)=0$
$\Leftrightarrow \sqrt{x^2+1}-2x+1=0$ (vì $2\sqrt{x^2+1}-1\geq 1$)
$\Leftrightarrow \sqrt{x^2+1}=2x-1 \ \ \ \ \ (x\geq \dfrac12)
\\\Leftrightarrow x^2+1=4x^2-4x+1
\\\Leftrightarrow 3x^2-4x=0
\\\Leftrightarrow x(3x-4)=0$
$\Leftrightarrow x=0$ (KTM) or $x=\dfrac{4}3$ (TM)
Vậy...
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
