Câu 1: $x\geq -\frac{1}{2}$Ta có với $x\geq -\frac{1}{2}$ thì $|x+\frac{1}{2}|=x+\frac{1}{2}$Pt $\Leftrightarrow \sqrt{x^2-\frac{1}{4}+\sqrt{(x+\frac{1}{2})^2}}=\frac{1}{2}(2x^3+x^2+2x+1)$$\Leftrightarrow \sqrt{x^2-\frac{1}{4}+|x+\frac{1}{2}|}=\frac{1}{2}(x+\frac{1}{2})(2x^2+2)$$\Leftrightarrow \sqrt{x^2+x+\frac{1}{2}}=\frac{1}{2}(x+\frac{1}{2})(2x^2+2)$$\Leftrightarrow |x+\frac{1}{2}|=(x+\frac{1}{2})(x^2+1)$$\Leftrightarrow (x+\frac{1}{2})x^2=0$$\Leftrightarrow x=-\frac{1}{2}\vee x= 0$
Câu 1: $x\geq -\frac{1}{2}$Ta có với $x\geq -\frac{1}{2}$ thì $|x+\frac{1}{2}|=x+\frac{1}{2}$Pt $\Leftrightarrow \sqrt{x^2-\frac{1}{4}+\sqrt{(x+\frac{1}{2})^2}}=\frac{1}{2}(2x^3+x^2+2x+1)$$\Leftrightarrow \sqrt{x^2-\frac{1}{4}+|x+\frac{1}{2}|}=\frac{1}{4}(2x+1)(2x^2+2)$$\Leftrightarrow \sqrt{x^2+x+\frac{1}{4}}=\frac{1}{4}(2x+1)(2x^2+2)$$\Leftrightarrow |x+\frac{1}{2}|=\frac{1}{4}(2x+1)(2x^2+2)$$\Leftrightarrow (2x+1)(\frac{x^2}{2}-\frac{1}{2})=0$$\Leftrightarrow x=-\frac{1}{2}\vee x= 1$
Câu 1: $x\geq -\frac{1}{2}$Ta có với $x\geq -\frac{1}{2}$ thì $|x+\frac{1}{2}|=x+\frac{1}{2}$Pt $\Leftrightarrow \sqrt{x^2-\frac{1}{4}+\sqrt{(x+\frac{1}{2})^2}}=\frac{1}{2}(2x^3+x^2+2x+1)$$\Leftrightarrow \sqrt{x^2-\frac{1}{4}+|x+\frac{1}{2}|}=\frac{1}{
2}(x+
\frac{1
}{2})(2x^2+2)$$\Leftrightarrow \sqrt{x^2+x+\frac{1}{
2}}=\frac{1}{
2}(x+
\frac{1
}{2})(2x^2+2)$$\Leftrightarrow |x+\frac{1}{2}|=
(x+\frac{1}{2
})(x^2+
1)$$\Leftrightarrow (x+\frac{1}{2})
x^2=0$$\Leftrightarrow x=-\frac{1}{2}\vee x=
0$