Toán cho a^2+b^2+c^2+1/a^2+1/b^2+1/c^2=6.Tính a^2n+b^2n+c^2n 09/07/2021 By Raelynn cho a^2+b^2+c^2+1/a^2+1/b^2+1/c^2=6.Tính a^2n+b^2n+c^2n
Đáp án: $3$ Giải thích các bước giải: $a^2 + b^2 + c^2 + \dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{c^2} = 6$ $\Leftrightarrow \left(a^2 – 2 +\dfrac{1}{a^2}\right) + \left(b^2 – 2 +\dfrac{1}{b^2}\right) + \left(c^2 – 2 +\dfrac{1}{c^2}\right) = 0$ $\Leftrightarrow \left(a^2 – 2.a.\dfrac{1}{a} + \dfrac{1}{a^2}\right) + \left(a^2 – 2.b.\dfrac{1}{b} + \dfrac{1}{a^2}\right) + \left(c^2 – 2.c.\dfrac{1}{c} + \dfrac{1}{c^2}\right) = 0$ $\Leftrightarrow \left(a – \dfrac{1}{a}\right)^2 + \left(b- \dfrac{1}{b}\right)^2 + \left(c- \dfrac{1}{c}\right)^2 = 0$ $\Leftrightarrow \begin{cases}a – \dfrac{1}{a} = 0\\b – \dfrac{1}{b} = 0\\c – \dfrac{1}{c} = 0\end{cases}$ $\Leftrightarrow \begin{cases}a = \dfrac{1}{a}\\b = \dfrac{1}{b}\\c = \dfrac{1}{c}\end{cases}$ $\Leftrightarrow \begin{cases}a^2 = 1\\b^2 = 1\\c^2 = 1\end{cases}$ Ta được: $a^{2n} + b^{2n} + c^{2n}$ $= (a^2)^n + (b^2)^n + (c^2)^n$ $= 1^n + 1^n + 1^n$ $= 3$ Trả lời
Đáp án:
$3$
Giải thích các bước giải:
$a^2 + b^2 + c^2 + \dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{c^2} = 6$
$\Leftrightarrow \left(a^2 – 2 +\dfrac{1}{a^2}\right) + \left(b^2 – 2 +\dfrac{1}{b^2}\right) + \left(c^2 – 2 +\dfrac{1}{c^2}\right) = 0$
$\Leftrightarrow \left(a^2 – 2.a.\dfrac{1}{a} + \dfrac{1}{a^2}\right) + \left(a^2 – 2.b.\dfrac{1}{b} + \dfrac{1}{a^2}\right) + \left(c^2 – 2.c.\dfrac{1}{c} + \dfrac{1}{c^2}\right) = 0$
$\Leftrightarrow \left(a – \dfrac{1}{a}\right)^2 + \left(b- \dfrac{1}{b}\right)^2 + \left(c- \dfrac{1}{c}\right)^2 = 0$
$\Leftrightarrow \begin{cases}a – \dfrac{1}{a} = 0\\b – \dfrac{1}{b} = 0\\c – \dfrac{1}{c} = 0\end{cases}$
$\Leftrightarrow \begin{cases}a = \dfrac{1}{a}\\b = \dfrac{1}{b}\\c = \dfrac{1}{c}\end{cases}$
$\Leftrightarrow \begin{cases}a^2 = 1\\b^2 = 1\\c^2 = 1\end{cases}$
Ta được:
$a^{2n} + b^{2n} + c^{2n}$
$= (a^2)^n + (b^2)^n + (c^2)^n$
$= 1^n + 1^n + 1^n$
$= 3$