Ôn tập toán 6

Nghi Nguyen Bao

Bài 1. Chứng tỏ rằng: B=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+\dfrac{1}{8^2}< 1\)

Bài 2. so sánh : A=\(\dfrac{2011+2012}{2012+2013}\)

và B=\(\dfrac{2011}{2012}+\dfrac{2012}{2013}\)

Bài 3. Rút gọn : B= \(\left(1-\dfrac{1}{1}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{20}\right)\)

Bài 4. Rút gọn biểu thức : A= \(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2012}}\)

Bài 5. Tìm số nguyên \(\pi\) sao cho \(\pi+5\) chia hết cho \(\pi-2\)

HELP ME!!!! MÌNH TICK CHO HA

Trần Minh An
11 tháng 4 2017 lúc 21:56

Bài 1)

Ta có:

A = \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+\dfrac{1}{8^2}\)

A < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}\)

A < \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\)

A < \(1-\dfrac{1}{8}\) = \(\dfrac{7}{8}\) < 1

Vậy A < 1

Bình luận (1)
Trần Minh An
12 tháng 4 2017 lúc 11:28

Bài 2)

Ta thấy:

\(\dfrac{2011}{2012+2013}< \dfrac{2011}{2012};\dfrac{2012}{2012+2013}< \dfrac{2012}{2013}\)

\(\Rightarrow\) \(\dfrac{2011}{2012+2013}+\dfrac{2012}{2012+2013}< \dfrac{2011}{2012}+\dfrac{2012}{2013}\)

\(\Rightarrow\) \(\dfrac{2011+2012}{2012+2013}< \dfrac{2011}{2012}+\dfrac{2012}{2013}\)

\(\Rightarrow\) A < B

Bài 3)

Ta có:

B = \(\left(1-\dfrac{1}{1}\right)\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right)......\left(1-\dfrac{1}{20}\right)\)

= \(0.\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)......\left(1-\dfrac{1}{20}\right)\)

= 0

Bài 3)

Ta có:

A = \(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+.....+\dfrac{1}{2^{2012}}\)

\(\Rightarrow\) 2A = \(2\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+.....+\dfrac{1}{2^{2012}}\right)\)

\(\Rightarrow\) 2A = \(2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+.....+\dfrac{1}{2^{2011}}\)

\(\Rightarrow\) 2A - A = \(\left(2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+.....+\dfrac{1}{2^{2011}}\right)\)-\(\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+.....+\dfrac{1}{2^{2012}}\right)\)

\(\Rightarrow\) A = 2 - \(\dfrac{1}{2^{2012}}\) = \(\dfrac{2^{2013}-1}{2^{2012}}\)

Bài 5)

\(\pi\) + 5 \(⋮\) \(\pi\) - 2

\(\Leftrightarrow\) \(\pi\) - 2 + 7 \(⋮\) \(\pi\) - 2

\(\Leftrightarrow\) 7 \(⋮\) \(\pi\) - 2 (vì \(\pi\) - 2 \(⋮\) \(\pi\) - 2)

\(\Leftrightarrow\) \(\pi\) - 2 \(\in\) Ư(7)

\(\Leftrightarrow\) \(\pi\) - 2 \(\in\) \(\left\{\pm1;\pm7\right\}\)

\(\Leftrightarrow\) \(\pi\) \(\in\) \(\left\{1;3;-5;9\right\}\)

Bình luận (1)

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