Toán Lớp 11: So Sánh Phân Số :
A = $\dfrac{8^10 + 1 }{8^11 + 1 }$ Và B = $\dfrac{8^11 + 1}{8^12 + 1 }$
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Toán Lớp 11: So Sánh Phân Số :
A = $\dfrac{8^10 + 1 }{8^11 + 1 }$ Và B = $\dfrac{8^11 + 1}{8^12 + 1 }$
Comments ( 1 )
$A=\dfrac{8^{10}+1}{8^{11}+1};B=\dfrac{8^{11}+1}{8^{12}+1}\\
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A-B=\dfrac{8^{10}+1}{8^{11}+1}-\dfrac{8^{11}+1}{8^{12}+1}=\dfrac{\left ( 8^{10}+1 \right )\left ( 8^{12}+1 \right )-\left ( 8^{11}+1 \right )^2}{\left ( 8^{11}+1 \right )\left ( 8^{12}+1 \right )}$
Xét $\left ( 8^{10}+1 \right )\left ( 8^{12}+1 \right )-\left ( 8^{11}+1 \right )^2$
$=8^{22}+65.8^{10}+1-\left ( 8^{22}+2.8^{11}+1 \right )\\
=65.8^{10}-2.8^{11}\\
=65.8^{10}-16.8^{10}=49.8^{10}>0$
Do đó $A-B>0$ hay $A>B$