giải pt (√(x+1)- √(x-2) )(1+ √(x^2-x-2))=3 13/08/2021 Bởi Elliana giải pt (√(x+1)- √(x-2) )(1+ √(x^2-x-2))=3
Đáp án: $x=3$ Giải thích các bước giải: Đkxđ: $x\ge 2$ $(\sqrt{x+1}-\sqrt{x-2})(1+\sqrt{x^2-x-2})=3$ $\rightarrow (\sqrt{x+1}-\sqrt{x-2})(1+\sqrt{x^2-x-2})=(x+1)-(x-2)$ $\rightarrow (\sqrt{x+1}-\sqrt{x-2})(1+\sqrt{x^2-x-2})=(\sqrt{x+1}+\sqrt{x-2})(\sqrt{x+1}-\sqrt{x-2})$ $\rightarrow 1+\sqrt{x^2-x-2}=\sqrt{x+1}+\sqrt{x-2}$ $\rightarrow (1+\sqrt{x^2-x-2})^2=(\sqrt{x+1}+\sqrt{x-2})^2$ $\rightarrow 1+2\sqrt{x^2-x-2}+x^2-x-2=x+1+x-2+2\sqrt{(x+1)(x-2)}$ $\rightarrow x^2-3x=0$ $\rightarrow x(x-3)=0$ $\rightarrow x\in\{0,3\}$ $\rightarrow x=3$ Bình luận
Đáp án: $x=3$
Giải thích các bước giải:
Đkxđ: $x\ge 2$
$(\sqrt{x+1}-\sqrt{x-2})(1+\sqrt{x^2-x-2})=3$
$\rightarrow (\sqrt{x+1}-\sqrt{x-2})(1+\sqrt{x^2-x-2})=(x+1)-(x-2)$
$\rightarrow (\sqrt{x+1}-\sqrt{x-2})(1+\sqrt{x^2-x-2})=(\sqrt{x+1}+\sqrt{x-2})(\sqrt{x+1}-\sqrt{x-2})$
$\rightarrow 1+\sqrt{x^2-x-2}=\sqrt{x+1}+\sqrt{x-2}$
$\rightarrow (1+\sqrt{x^2-x-2})^2=(\sqrt{x+1}+\sqrt{x-2})^2$
$\rightarrow 1+2\sqrt{x^2-x-2}+x^2-x-2=x+1+x-2+2\sqrt{(x+1)(x-2)}$
$\rightarrow x^2-3x=0$
$\rightarrow x(x-3)=0$
$\rightarrow x\in\{0,3\}$
$\rightarrow x=3$