Tìm số hữu tỉ y biết (3y-1)^10=(3y-1)^20 02/09/2021 Bởi Hadley Tìm số hữu tỉ y biết (3y-1)^10=(3y-1)^20
$(3y-1)^{10}=(3y-1)^{20}$ $⇔(3y-1)^{10}-(3y-1)^{20}=0$ $⇔(3y-1)^{10}-(3y-1)^{10}.(3y-1)^{10}=0$ $⇔[(3y-1)^{10}-1].(3y-1)^{10}=0$ $⇔$\(\left[ \begin{array}{l}(3y-1)^{10}-1=0\\(3y-1)^{10}=0\end{array} \right.\) $⇔$\(\left[ \begin{array}{l}(3y-1)^{10}=1\\3y-1=0\end{array} \right.\) $⇔$\(\left[ \begin{array}{l}3y-1=0\\3y-1=1\\3y-1=-1\end{array} \right.\) $⇔$\(\left[ \begin{array}{l}y=\frac{1}{3}\\y=\frac{2}{3}\\y=0\end{array} \right.\) Vậy… Bình luận
Đáp án: Giải thích các bước giải: $(3y-1)^{10}$ = $(3y-1)^{20}$ ⇒$(3y-1)^{10}$-$(3y-1)^{20}$ =$0^{}$ ⇒$(3y-1)^{10}$ . [$1-(3y-1)^{10}$ ]=$0_{}$ ⇒\(\left[ \begin{array}{l}(3y-1)^{10}=0\\1-(3y-1)^{10}\end{array} \right.\) ⇔$y^{}$ ∈{$\frac{1}{3}$ ;$\frac{2}{3}$ ;$0^{}$ } Bình luận
$(3y-1)^{10}=(3y-1)^{20}$
$⇔(3y-1)^{10}-(3y-1)^{20}=0$
$⇔(3y-1)^{10}-(3y-1)^{10}.(3y-1)^{10}=0$
$⇔[(3y-1)^{10}-1].(3y-1)^{10}=0$
$⇔$\(\left[ \begin{array}{l}(3y-1)^{10}-1=0\\(3y-1)^{10}=0\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}(3y-1)^{10}=1\\3y-1=0\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}3y-1=0\\3y-1=1\\3y-1=-1\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}y=\frac{1}{3}\\y=\frac{2}{3}\\y=0\end{array} \right.\)
Vậy…
Đáp án:
Giải thích các bước giải:
$(3y-1)^{10}$ = $(3y-1)^{20}$
⇒$(3y-1)^{10}$-$(3y-1)^{20}$ =$0^{}$
⇒$(3y-1)^{10}$ . [$1-(3y-1)^{10}$ ]=$0_{}$
⇒\(\left[ \begin{array}{l}(3y-1)^{10}=0\\1-(3y-1)^{10}\end{array} \right.\)
⇔$y^{}$ ∈{$\frac{1}{3}$ ;$\frac{2}{3}$ ;$0^{}$ }