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Question

Giair phuong trinh$\sqrt{x^2-\frac{1}{4}-\sqrt{x^2+x+\frac{1}{4}}}=\frac{1}{2}\left(2x^3+x^2+2x+1\right)$

Dương Lam Hàng · Accepted Answer

$\sqrt{x^2-\frac{1}{4}-\sqrt{x^2+x+\frac{1}{4}}}=\frac{1}{2}\left(2x^3+x^2+2x+1\right)$    (ĐK: $x\ge\frac{-1}{2}$ )$\Leftrightarrow\sqrt{x^2-\frac{1}{4}-\sqrt{\left(x+\frac{1}{2}\right)^2}}=\frac{1}{2}\left[2x\left(x^2+1\right)+\left(x^2+1\right)\right]$$\Leftrightarrow\sqrt{x^2-\frac{1}{4}-x-\frac{1}{2}}=\frac{1}{2}\left(x^2+1\right)\left(2x+1\right)$$\Leftrightarrow\sqrt{\left(x+\frac{1}{2}\right)^2}=\frac{1}{2}\left(x^2+1\right)\left(2x+1\right)$$\Leftrightarrow x+\frac{1}{2}=\frac{1}{2}\left(x^2+1\right)\left(2x+1\right)$$\Leftrightarrow2x+1=\left(x^2+1\right)\left(2x+1\right)$$\Leftrightarrow\left(x^2+1\right)\left(2x+1\right)-\left(2x+1\right)=0$$\Leftrightarrow\left(2x+1\right)\left(x^2+1-1\right)=0$$\Leftrightarrow x^2\left(2x+1\right)=0$$\Leftrightarrow\orbr{\begin{cases}2x+1=0\x^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\x=0\end{cases}}$ (nhận)Vậy .....Câu trả lời: $\sqrt{x^2-\frac{1}{4}-\sqrt{x^2+x+\frac{1}{4}}}=\frac{1}{2}\left(2x^3+x^2+2x+1\right)$    (ĐK: $x\ge\frac{-1}{2}$ )$\Leftrightarrow\sqrt{x^2-\frac{1}{4}-\sqrt{\left(x+\frac{1}{2}\right)^2}}=\frac{1}{2}\left[2x\left(x^2+1\right)+\left(x^2+1\right)\right]$$\Leftrightarrow\sqrt{x^2-\frac{1}{4}-x-\frac{1}{2}}=\frac{1}{2}\left(x^2+1\right)\left(2x+1\right)$$\Leftrightarrow\sqrt{\left(x+\frac{1}{2}\right)^2}=\frac{1}{2}\left(x^2+1\right)\left(2x+1\right)$$\Leftrightarrow x+\frac{1}{2}=\frac{1}{2}\left(x^2+1\right)\left(2x+1\right)$$\Leftrightarrow2x+1=\left(x^2+1\right)\left(2x+1\right)$$\Leftrightarrow\left(x^2+1\right)\left(2x+1\right)-\left(2x+1\right)=0$$\Leftrightarrow\left(2x+1\right)\left(x^2+1-1\right)=0$$\Leftrightarrow x^2\left(2x+1\right)=0$$\Leftrightarrow\orbr{\begin{cases}2x+1=0\x^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\x=0\end{cases}}$ (nhận)Vậy .....

Incursion_03 · Answer

$\sqrt{x^2-\frac{1}{4}-\sqrt{x^2+x+\frac{1}{4}}}=\frac{1}{2}\left(2x^3+x^2+2x+1\right)$$\Leftrightarrow\sqrt{x^2-\frac{1}{4}-\sqrt{\left(x+\frac{1}{2}\right)^2}}=\frac{1}{2}\left[x^2\left(2x+1\right)+2x+1\right]$$\Leftrightarrow\sqrt{x^2-\frac{1}{4}-\left|x+\frac{1}{2}\right|}=\frac{1}{2}\left(x^2+1\right)\left(2x+1\right)$(1) Vì VT > 0 nên VP >0$\Leftrightarrow\frac{1}{2}\left(x^2+1\right)\left(2x+1\right)\ge0$$\Leftrightarrow x\ge-\frac{1}{2}$Khi đó $\left(1\right)\Leftrightarrow\sqrt{x^2-\frac{1}{4}-x-\frac{1}{2}}=\frac{1}{2}\left(x^2+1\right)\left(2x+1\right)$                    $\Leftrightarrow\sqrt{x^2-x-\frac{3}{4}}=\frac{1}{2}\left(x^2+1\right)\left(2x+1\right)$                    $\Leftrightarrow x^2-x-\frac{3}{4}=\frac{1}{4}\left(x^2+1\right)^2\left(2x+1\right)^2$                   $\Leftrightarrow\left(2x-3\right)\left(2x+1\right)-\frac{1}{4}\left(x^2+1\right)^2\left(2x+1\right)^2=0$                 $\Leftrightarrow\left(2x+1\right)\left(2x-3-\frac{1}{4}\left(x^2+1\right)^2\left(2x+1\right)\right)=0$                $\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\2x-3=\frac{1}{4}\left(x^2+1\right)^2\left(2x+1\right)\end{cases}}$ Cần cù bù thông minh , phá tung pt dưới ra được cái phương trình bậc 5, sau đó dùng Wolfram|Alpha: Computational Intelligence để tính nghiệm rồi phân tích nhân tử =))Câu trả lời: $\sqrt{x^2-\frac{1}{4}-\sqrt{x^2+x+\frac{1}{4}}}=\frac{1}{2}\left(2x^3+x^2+2x+1\right)$$\Leftrightarrow\sqrt{x^2-\frac{1}{4}-\sqrt{\left(x+\frac{1}{2}\right)^2}}=\frac{1}{2}\left[x^2\left(2x+1\right)+2x+1\right]$$\Leftrightarrow\sqrt{x^2-\frac{1}{4}-\left|x+\frac{1}{2}\right|}=\frac{1}{2}\left(x^2+1\right)\left(2x+1\right)$(1) Vì VT > 0 nên VP >0$\Leftrightarrow\frac{1}{2}\left(x^2+1\right)\left(2x+1\right)\ge0$$\Leftrightarrow x\ge-\frac{1}{2}$Khi đó $\left(1\right)\Leftrightarrow\sqrt{x^2-\frac{1}{4}-x-\frac{1}{2}}=\frac{1}{2}\left(x^2+1\right)\left(2x+1\right)$                    $\Leftrightarrow\sqrt{x^2-x-\frac{3}{4}}=\frac{1}{2}\left(x^2+1\right)\left(2x+1\right)$                    $\Leftrightarrow x^2-x-\frac{3}{4}=\frac{1}{4}\left(x^2+1\right)^2\left(2x+1\right)^2$                   $\Leftrightarrow\left(2x-3\right)\left(2x+1\right)-\frac{1}{4}\left(x^2+1\right)^2\left(2x+1\right)^2=0$                 $\Leftrightarrow\left(2x+1\right)\left(2x-3-\frac{1}{4}\left(x^2+1\right)^2\left(2x+1\right)\right)=0$                $\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\2x-3=\frac{1}{4}\left(x^2+1\right)^2\left(2x+1\right)\end{cases}}$ Cần cù bù thông minh , phá tung pt dưới ra được cái phương trình bậc 5, sau đó dùng Wolfram|Alpha: Computational Intelligence để tính nghiệm rồi phân tích nhân tử =))

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