Toán Bài tập đại số 9

Q

quynhnhung81

Bài 2: Cho
A = 1/(căn1 + căn2) + 1/(căn2 + căn3) + ... + 1/(căn120 + căn121):confused:
B = 1/căn1 + 1/căn2 +.....+ 1/căn35
chứng minh rằng A<B
[TEX]A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+ \sqrt{3}}+...+\frac{1}{\sqrt{120}+\sqrt{121}}[/TEX]

[TEX]=\frac{\sqrt{2}-\sqrt{1}}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+...+\frac{\sqrt{121}-\sqrt{120}}{121-120}[/TEX]

[TEX]=\sqrt{121}-1=10[/TEX]

[TEX]B=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{35}}[/TEX]

[TEX]=\frac{2}{\sqrt{1}+\sqrt{1}}+\frac{2}{\sqrt{2}+\sqrt{2}}+...+\frac{2}{\sqrt{35}+\sqrt{35}}[/TEX]

[TEX]B < 2(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+ \sqrt{3}}+...+\frac{1}{\sqrt{35}+\sqrt{36}})[/TEX]

[TEX]=2(\sqrt{35}-1) \approx 9,83[/TEX]

\Rightarrow A <B
 
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C

camnhivd4

pạn ơi

[TEX]B > 2(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+ \sqrt{3}}+...+\sqrt{1}{\sqrt{35}+\sqrt{36}})[/TEX \Rightarrow B > căn 36 - 1 \Rightarrow B lớn hơn 10 \Rightarrow A<B[/TEX]
 
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