a) Đặt $A = \dfrac{3^{10}-1}{3^{12}-1}$ $\to 9A = \dfrac{3^{12}-9}{3^{12}-1}$ $ = 1-\dfrac{8}{3^{12}-1}$ $B = \dfrac{3^{12}-1}{3^{14}-1}$ $\to 9B = 1- \dfrac{8}{3^{14}-1}$ Vì $\dfrac{8}{3^{12}-1} > \dfrac{8}{3^{14}-1}$ $to 1- \dfrac{8}{3^{12}-1} < 1-\dfrac{8}{3^{14}-1}$ $\to A < B$ b) Tương tự phần a) chỉ ra được ngược lại $A>B$
Đáp án:
$\begin{array}{l}
a)\frac{{{3^{12}} – 1}}{{{3^{14}} – 1}} < 1\\
\Rightarrow \frac{{{3^{12}} – 1}}{{{3^{14}} – 1}} > \frac{{{3^{12}} – 1 – 8}}{{{3^{14}} – 1 – 8}}\\
\Rightarrow \frac{{{3^{12}} – 1}}{{{3^{14}} – 1}} > \frac{{{3^{12}} – {3^2}}}{{{3^{14}} – {3^2}}}\\
\Rightarrow \frac{{{3^{12}} – 1}}{{{3^{14}} – 1}} > \frac{{{3^{10}} – 1}}{{{3^{12}} – 1}}\\
b)\frac{{{7^{22}} – 1}}{{1 – {7^{22}}}} = – 1\\
\frac{{{7^{20}} – 1}}{{1 – {7^{22}}}} = – \frac{{{7^{20}} – 1}}{{{7^{22}} – 1}}\\
Do:\frac{{{7^{20}} – 1}}{{{7^{22}} – 1}} < 1\\
\Rightarrow – \frac{{{7^{20}} – 1}}{{{7^{22}} – 1}} > – 1\\
\Rightarrow \frac{{{7^{20}} – 1}}{{1 – {7^{22}}}} > \frac{{{7^{22}} – 1}}{{1 – {7^{22}}}}
\end{array}$
a) Đặt $A = \dfrac{3^{10}-1}{3^{12}-1}$
$\to 9A = \dfrac{3^{12}-9}{3^{12}-1}$
$ = 1-\dfrac{8}{3^{12}-1}$
$B = \dfrac{3^{12}-1}{3^{14}-1}$
$\to 9B = 1- \dfrac{8}{3^{14}-1}$
Vì $\dfrac{8}{3^{12}-1} > \dfrac{8}{3^{14}-1}$
$to 1- \dfrac{8}{3^{12}-1} < 1-\dfrac{8}{3^{14}-1}$
$\to A < B$
b) Tương tự phần a) chỉ ra được ngược lại $A>B$