Câu hỏi của Hoàng Tử Của Mej

Question

Giải Phương Trình sau

$\left(\sqrt{x+1}-\sqrt{x-2}\right).\left(1+\sqrt{x^2-x-2}\right)=3$

Lightning Farron · Accepted Answer

ĐK:$x\ge 2$

$pt\Leftrightarrow\sqrt{\left(x+1\right)\left(x^2-x-2\right)}-\sqrt{\left(x-2\right)\left(x^2-x-2\right)}+\sqrt{x+1}-\sqrt{x-2}-3=0$

$\Leftrightarrow\left(\sqrt{\left(x+1\right)\left(x^2-x-2\right)}-4\right)-\left(\sqrt{\left(x-2\right)\left(x^2-x-2\right)}-2\right)+\sqrt{x+1}-2-\left(\sqrt{x-2}-1\right)=0$

$\Leftrightarrow\dfrac{\left(x+1\right)\left(x^2-x-2\right)-16}{\sqrt{\left(x+1\right)\left(x^2-x-2\right)}+4}-\dfrac{\left(x-2\right)\left(x^2-x-2\right)-4}{\sqrt{\left(x-2\right)\left(x^2-x-2\right)}+2}+\dfrac{x+1-4}{\sqrt{x+1}+2}-\dfrac{x-2-1}{\sqrt{x-2}+1}=0$

$\Leftrightarrow\dfrac{\left(x-3\right)\left(x^2+3x+6\right)}{\sqrt{\left(x+1\right)\left(x^2-x-2\right)}+4}-\dfrac{\left(x-3\right)x^2}{\sqrt{\left(x-2\right)\left(x^2-x-2\right)}+2}+\dfrac{x-3}{\sqrt{x+1}+2}-\dfrac{x-3}{\sqrt{x-2}+1}=0$

$\Leftrightarrow\left(x-3\right)\left(\dfrac{x^2+3x+6}{\sqrt{\left(x+1\right)\left(x^2-x-2\right)}+4}-\dfrac{x^2}{\sqrt{\left(x-2\right)\left(x^2-x-2\right)}+2}+\dfrac{1}{\sqrt{x+1}+2}-\dfrac{1}{\sqrt{x-2}+1}\right)=0$

$\Rightarrow x-3=0\Rightarrow x=3$Câu trả lời: ĐK:$x\ge 2$

$pt\Leftrightarrow\sqrt{\left(x+1\right)\left(x^2-x-2\right)}-\sqrt{\left(x-2\right)\left(x^2-x-2\right)}+\sqrt{x+1}-\sqrt{x-2}-3=0$

$\Leftrightarrow\left(\sqrt{\left(x+1\right)\left(x^2-x-2\right)}-4\right)-\left(\sqrt{\left(x-2\right)\left(x^2-x-2\right)}-2\right)+\sqrt{x+1}-2-\left(\sqrt{x-2}-1\right)=0$

$\Leftrightarrow\dfrac{\left(x+1\right)\left(x^2-x-2\right)-16}{\sqrt{\left(x+1\right)\left(x^2-x-2\right)}+4}-\dfrac{\left(x-2\right)\left(x^2-x-2\right)-4}{\sqrt{\left(x-2\right)\left(x^2-x-2\right)}+2}+\dfrac{x+1-4}{\sqrt{x+1}+2}-\dfrac{x-2-1}{\sqrt{x-2}+1}=0$

$\Leftrightarrow\dfrac{\left(x-3\right)\left(x^2+3x+6\right)}{\sqrt{\left(x+1\right)\left(x^2-x-2\right)}+4}-\dfrac{\left(x-3\right)x^2}{\sqrt{\left(x-2\right)\left(x^2-x-2\right)}+2}+\dfrac{x-3}{\sqrt{x+1}+2}-\dfrac{x-3}{\sqrt{x-2}+1}=0$

$\Leftrightarrow\left(x-3\right)\left(\dfrac{x^2+3x+6}{\sqrt{\left(x+1\right)\left(x^2-x-2\right)}+4}-\dfrac{x^2}{\sqrt{\left(x-2\right)\left(x^2-x-2\right)}+2}+\dfrac{1}{\sqrt{x+1}+2}-\dfrac{1}{\sqrt{x-2}+1}\right)=0$

$\Rightarrow x-3=0\Rightarrow x=3$

Ly Hoàng · Answer

ĐK:$x\ge 2$

$pt\Leftrightarrow\sqrt{\left(x+1\right)\left(x^2-x-2\right)}-\sqrt{\left(x-2\right)\left(x^2-
x-2\right)}+\sqrt{x+1}-\sqrt{x-2}-3=0$

$\Leftrightarrow\left(\sqrt{\left(x+1\right)\left(x^2-x-2\right)}-4\right)-\left(\sqrt{\left(x-
2\right)\left(x^2-x-2\right)}-2\right)+\sqrt{x+1}-2-\left(\sqrt{x-2}-1\right)=0$

$\Leftrightarrow\dfrac{\left(x+1\right)\left(x^2-x-2\right)-16}
{\sqrt{\left(x+1\right)\left(x^2-x-2\right)}+4}-\dfrac{\left(x-2\right)\left(x^2-x-2\right)-4}
{\sqrt{\left(x-2\right)\left(x^2-x-2\right)}+2}+\dfrac{x+1-4}{\sqrt{x+1}+2}-\dfrac{x-2-1}
{\sqrt{x-2}+1}=0$

$\Leftrightarrow\dfrac{\left(x-3\right)\left(x^2+3x+6\right)}
{\sqrt{\left(x+1\right)\left(x^2-x-2\right)}+4}-\dfrac{\left(x-3\right)x^2}{\sqrt{\left(x-
2\right)\left(x^2-x-2\right)}+2}+\dfrac{x-3}{\sqrt{x+1}+2}-\dfrac{x-3}{\sqrt{x-2}+1}=0$

$\Leftrightarrow\left(x-3\right)\left(\dfrac{x^2+3x+6}{\sqrt{\left(x+1\right)\left(x^2-x-
2\right)}+4}-\dfrac{x^2}{\sqrt{\left(x-2\right)\left(x^2-x-2\right)}+2}+\dfrac{1}
{\sqrt{x+1}+2}-\dfrac{1}{\sqrt{x-2}+1}\right)=0$