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Question

so sánh:A=1999^1999+1/1999^1998+1

B=1999^2000+1/1999^1999+1

Phạm Quang Lộc · Accepted Answer

$A=\dfrac{1999^{1999}+1}{1999^{1998}+1}$ $\dfrac{1}{1999}A=\dfrac{1999^{1999}+1}{1999^{1999}+1999}$ $\dfrac{1}{1999}A=\dfrac{1999^{1999}}{1999^{1999}}-\dfrac{1998}{1999^{1999}+1999}$ $\dfrac{1}{1999}A=1-\dfrac{1998}{1999^{1999}+1999}$ $B=\dfrac{1999^{2000}+1}{1999^{1999}+1}$ $\dfrac{1}{1999}B=\dfrac{1999^{2000}+1}{1999^{2000}+1999}$ $\dfrac{1}{1999}B=\dfrac{1999^{2000}}{1999^{2000}}-\dfrac{1998}{1999^{2000}+1999}$ $\dfrac{1}{1999}B=1-\dfrac{1998}{1999^{2000}+1999}$ Vì $\dfrac{1998}{1999^{1999}+1999}>\dfrac{1998}{1999^{2000}+1999}=>\dfrac{1}{1999}A< \dfrac{1}{1999}B=>A< B$ Câu trả lời: $A=\dfrac{1999^{1999}+1}{1999^{1998}+1}$ $\dfrac{1}{1999}A=\dfrac{1999^{1999}+1}{1999^{1999}+1999}$ $\dfrac{1}{1999}A=\dfrac{1999^{1999}}{1999^{1999}}-\dfrac{1998}{1999^{1999}+1999}$ $\dfrac{1}{1999}A=1-\dfrac{1998}{1999^{1999}+1999}$ $B=\dfrac{1999^{2000}+1}{1999^{1999}+1}$ $\dfrac{1}{1999}B=\dfrac{1999^{2000}+1}{1999^{2000}+1999}$ $\dfrac{1}{1999}B=\dfrac{1999^{2000}}{1999^{2000}}-\dfrac{1998}{1999^{2000}+1999}$ $\dfrac{1}{1999}B=1-\dfrac{1998}{1999^{2000}+1999}$ Vì $\dfrac{1998}{1999^{1999}+1999}>\dfrac{1998}{1999^{2000}+1999}=>\dfrac{1}{1999}A< \dfrac{1}{1999}B=>A< B$

Nguyễn Đức Trí · Answer

$A=\dfrac{1999^{1999}+1}{1999^{1998}+1}=\dfrac{\left(1999^{1999}+1\right)^2}{\left(1999^{1998}+1\right)\left(1999^{1999}+1\right)}$ $A=\dfrac{\left(1999^{1999}\right)^2+2.1999^{1999}+1}{\left(1999^{1998}+1\right)\left(1999^{1999}+1\right)}\left(1\right)$ $B=\dfrac{1999^{2000}+1}{1999^{1999}+1}=\dfrac{\left(1999^{2000}+1\right)\left(1999^{1998}+1\right)}{\left(1999^{1998}+1\right)\left(1999^{1999}+1\right)}$ $B=\dfrac{\left(1999.1999^{1999}+1\right)\left(\dfrac{1}{1999}.1999^{1999}+1\right)}{\left(1999^{1998}+1\right)\left(1999^{1999}+1\right)}$ $B=\dfrac{\left(1999^{1999}\right)^2+1999.1999^{1999}+\dfrac{1}{1999}.1999^{1999}+1}{\left(1999^{1998}+1\right)\left(1999^{1999}+1\right)}$ $B=\dfrac{\left(1999^{1999}\right)^2+\left(1999+\dfrac{1}{1999}\right).1999^{1999}+1}{\left(1999^{1998}+1\right)\left(1999^{1999}+1\right)}\left(2\right)$ mà $\left(1999+\dfrac{1}{1999}\right)>2$ $\left(1\right).\left(2\right)\Rightarrow A< B$Câu trả lời: $A=\dfrac{1999^{1999}+1}{1999^{1998}+1}=\dfrac{\left(1999^{1999}+1\right)^2}{\left(1999^{1998}+1\right)\left(1999^{1999}+1\right)}$ $A=\dfrac{\left(1999^{1999}\right)^2+2.1999^{1999}+1}{\left(1999^{1998}+1\right)\left(1999^{1999}+1\right)}\left(1\right)$ $B=\dfrac{1999^{2000}+1}{1999^{1999}+1}=\dfrac{\left(1999^{2000}+1\right)\left(1999^{1998}+1\right)}{\left(1999^{1998}+1\right)\left(1999^{1999}+1\right)}$ $B=\dfrac{\left(1999.1999^{1999}+1\right)\left(\dfrac{1}{1999}.1999^{1999}+1\right)}{\left(1999^{1998}+1\right)\left(1999^{1999}+1\right)}$ $B=\dfrac{\left(1999^{1999}\right)^2+1999.1999^{1999}+\dfrac{1}{1999}.1999^{1999}+1}{\left(1999^{1998}+1\right)\left(1999^{1999}+1\right)}$ $B=\dfrac{\left(1999^{1999}\right)^2+\left(1999+\dfrac{1}{1999}\right).1999^{1999}+1}{\left(1999^{1998}+1\right)\left(1999^{1999}+1\right)}\left(2\right)$ mà $\left(1999+\dfrac{1}{1999}\right)>2$ $\left(1\right).\left(2\right)\Rightarrow A< B$

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