Giair phương trình: ∛x-1 + ∛x+8 = 1+x ³ 27/07/2021 Bởi Kinsley Giair phương trình: ∛x-1 + ∛x+8 = 1+x ³
Đáp án: x=0 Giải thích các bước giải: \(\begin{array}{l} \sqrt[3]{{x – 1}} + \sqrt[3]{{x + 8}} = 1 + x^3 \\ \Leftrightarrow \sqrt[3]{{x – 1}} + 1 – 2 + \sqrt[3]{{x + 8}} = x^3 \\ \Leftrightarrow \frac{{x – 1 + 1}}{{\sqrt[3]{{(x – 1)^2 }} – \sqrt[3]{{x – 1}} + 1}} + \frac{{x + 8 – 8}}{{\sqrt[3]{{(x + 8)^2 }} + 2\sqrt[3]{{x + 8}} + 4}} = x^3 \\ \Leftrightarrow \frac{x}{{\sqrt[3]{{(x – 1)^2 }} – \sqrt[3]{{x – 1}} + 1}} + \frac{x}{{\sqrt[3]{{(x + 8)^2 }} + 2\sqrt[3]{{x + 8}} + 4}} – x^3 = 0 \\ \Leftrightarrow x(\frac{1}{{\sqrt[3]{{(x – 1)^2 }} – \sqrt[3]{{x – 1}} + 1}} + \frac{1}{{\sqrt[3]{{(x + 8)^2 }} + 2\sqrt[3]{{x + 8}} + 4}} – x^2 ) = 0 \\ \Leftrightarrow x = 0 \\ \end{array}\) Bình luận
Đáp án:
x=0
Giải thích các bước giải:
\(
\begin{array}{l}
\sqrt[3]{{x – 1}} + \sqrt[3]{{x + 8}} = 1 + x^3 \\
\Leftrightarrow \sqrt[3]{{x – 1}} + 1 – 2 + \sqrt[3]{{x + 8}} = x^3 \\
\Leftrightarrow \frac{{x – 1 + 1}}{{\sqrt[3]{{(x – 1)^2 }} – \sqrt[3]{{x – 1}} + 1}} + \frac{{x + 8 – 8}}{{\sqrt[3]{{(x + 8)^2 }} + 2\sqrt[3]{{x + 8}} + 4}} = x^3 \\
\Leftrightarrow \frac{x}{{\sqrt[3]{{(x – 1)^2 }} – \sqrt[3]{{x – 1}} + 1}} + \frac{x}{{\sqrt[3]{{(x + 8)^2 }} + 2\sqrt[3]{{x + 8}} + 4}} – x^3 = 0 \\
\Leftrightarrow x(\frac{1}{{\sqrt[3]{{(x – 1)^2 }} – \sqrt[3]{{x – 1}} + 1}} + \frac{1}{{\sqrt[3]{{(x + 8)^2 }} + 2\sqrt[3]{{x + 8}} + 4}} – x^2 ) = 0 \\
\Leftrightarrow x = 0 \\
\end{array}
\)