Câu hỏi của Nguyễn Minh Dương

Question

So sánh bt: $M=\dfrac{100^{100}+1}{100^{99}+1};N=\dfrac{100^{101}+1}{100^{100}+1}$

HT.Phong (9A5) · Accepted Answer

Ta có:$M=\dfrac{100^{100}+1}{100^{99}+1}$$\Rightarrow\dfrac{M}{100}=\dfrac{100^{100}+1}{100\cdot\left(100^{99}+1\right)}$$\Rightarrow\dfrac{M}{100}=\dfrac{100^{100}+1}{100^{100}+100}$$\Rightarrow\dfrac{M}{100}=1-\dfrac{99}{100^{100}+100}$ $N=\dfrac{100^{101}+1}{100^{100}+1}$$\Rightarrow\dfrac{N}{100}=\dfrac{100^{101}+1}{100\cdot\left(100^{100}+1\right)}$$\Rightarrow\dfrac{N}{100}=\dfrac{100^{101}+1}{100^{101}+100}$$\Rightarrow\dfrac{N}{100}=1-\dfrac{99}{100^{101}+100}$Mà: $100^{101}>100^{100}$$\Rightarrow100^{101}+100>100^{100}+100$$\Rightarrow\dfrac{99}{100^{101}+100}< \dfrac{99}{100^{100}+100}$$\Rightarrow1-\dfrac{99}{101^{101}+100}< 1-\dfrac{99}{100^{100}+100}$$\Rightarrow\dfrac{N}{100}< \dfrac{M}{100}$$\Rightarrow N< M$Câu trả lời: Ta có:$M=\dfrac{100^{100}+1}{100^{99}+1}$$\Rightarrow\dfrac{M}{100}=\dfrac{100^{100}+1}{100\cdot\left(100^{99}+1\right)}$$\Rightarrow\dfrac{M}{100}=\dfrac{100^{100}+1}{100^{100}+100}$$\Rightarrow\dfrac{M}{100}=1-\dfrac{99}{100^{100}+100}$ $N=\dfrac{100^{101}+1}{100^{100}+1}$$\Rightarrow\dfrac{N}{100}=\dfrac{100^{101}+1}{100\cdot\left(100^{100}+1\right)}$$\Rightarrow\dfrac{N}{100}=\dfrac{100^{101}+1}{100^{101}+100}$$\Rightarrow\dfrac{N}{100}=1-\dfrac{99}{100^{101}+100}$Mà: $100^{101}>100^{100}$$\Rightarrow100^{101}+100>100^{100}+100$$\Rightarrow\dfrac{99}{100^{101}+100}< \dfrac{99}{100^{100}+100}$$\Rightarrow1-\dfrac{99}{101^{101}+100}< 1-\dfrac{99}{100^{100}+100}$$\Rightarrow\dfrac{N}{100}< \dfrac{M}{100}$$\Rightarrow N< M$

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