Question

Giải phương trình \(\sqrt{x+3}+\sqrt{3-x}=x^3+x^2-4x-1\)

Answer 1

\(\sqrt{x+2}+\sqrt{3-x}=x^3+x^2-4x-1\left(-2\le x\le3\right)\\ \Leftrightarrow\left(\sqrt{x+2}-1\right)+\left(\sqrt{3-x}-2\right)=x^3+x^2-4x-4\\ \Leftrightarrow\dfrac{x+1}{\sqrt{x+2}-1}-\dfrac{x+1}{\sqrt{3-x}+2}=\left(x+1\right)\left(x-2\right)\left(x+2\right)\\ \Leftrightarrow\left(x+1\right)\left(\dfrac{1}{\sqrt{x+2}-1}-\dfrac{1}{\sqrt{3-x}+2}\right)=\left(x+1\right)\left(x^2-4\right)\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\left(tm\right)\\\dfrac{1}{\sqrt{x+2}-1}-\dfrac{1}{\sqrt{3-x}+2}=x^2-4\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\dfrac{1}{\sqrt{x+2}-1}-1-\left(\dfrac{1}{\sqrt{3-x}+2}-1\right)-x^2+4=0\\ \Leftrightarrow\dfrac{2-\sqrt{x+2}}{\sqrt{x+2}-1}+\dfrac{\sqrt{3-x}+1}{\sqrt{3-x}+2}-x^2+4=0\\ \Leftrightarrow\dfrac{2-x}{\left(\sqrt{x+2}-1\right)\left(2+\sqrt{x+2}\right)}+\dfrac{2-x}{\left(\sqrt{3-x}+2\right)\left(\sqrt{3-x}-1\right)}+\left(2-x\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2-x=0\Rightarrow x=2\left(tm\right)\\\dfrac{1}{\left(\sqrt{x+2}-1\right)\left(2+\sqrt{x+2}\right)}+\dfrac{1}{\left(\sqrt{3-x}+2\right)\left(\sqrt{3-x}-1\right)}+x+2=0\left(1\right)\end{matrix}\right.\)

Với \(x\ge-2\Leftrightarrow\left(1\right)>0\left(\text{vô nghiệm}\right)\)

Vậy PT có nghiệm \(x\in\left\{2;-1\right\}\)

Answer 2

ờ ha, thảo nào bấm máy xong thay vô thấy sai sai :D

Sửa từ chỗ PT (1) đi

\(\left(1\right)\Leftrightarrow\dfrac{1}{\sqrt{x+2}+1}-\dfrac{1}{3}-\left(\dfrac{1}{\sqrt{3-x}+2}-\dfrac{1}{3}\right)-x^2+4=0\\ \Leftrightarrow\dfrac{2-\sqrt{x+2}}{3\left(\sqrt{x+2}+1\right)}-\dfrac{1-\sqrt{3-x}}{3\left(\sqrt{3-x}+2\right)}-\left(x-2\right)\left(x+2\right)=0\\ \Leftrightarrow\dfrac{2-x}{3\left(\sqrt{x+2}+1\right)\left(2+\sqrt{x+2}\right)}+\dfrac{2-x}{3\left(\sqrt{3-x}+2\right)\left(\sqrt{3-x}+1\right)}+\left(2-x\right)\left(2+x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\\dfrac{1}{3\left(\sqrt{x+2}+1\right)\left(2+\sqrt{x+2}\right)}+\dfrac{1}{3\left(\sqrt{3-x}+2\right)\left(1+\sqrt{3-x}\right)}+x+2=0\left(2\right)\end{matrix}\right.\)

Với \(x\ge-2\Leftrightarrow\left(2\right)>0\)

Vậy ...