1. Tính :
a, \(A=\dfrac{\dfrac{1}{3}-\dfrac{5}{2}}{\dfrac{3}{4}-\dfrac{1}{2}}.\dfrac{\dfrac{5}{6}+\dfrac{7}{3}}{1-\dfrac{5}{6}}.\dfrac{\dfrac{-2}{5}+1}{\dfrac{2}{5}-1}\).
b, \(B=\dfrac{\dfrac{1}{3}-\dfrac{4}{5}}{\dfrac{1}{3}+\dfrac{4}{5}}.\dfrac{\dfrac{3}{4}-\dfrac{5}{3}}{\dfrac{3}{4}+\dfrac{5}{3}}:\dfrac{\dfrac{4}{5}-1}{1-\dfrac{2}{3}}\).
Tính các tổng sau:
a) A=\(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\)
b) B=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{102^2}\)
c) C=\(\dfrac{3}{1+2}+\dfrac{3}{1+2+3}+\dfrac{3}{1+2+3+4}+...+\dfrac{3}{1+2+3+...+100}\)
1 Thực hiện phép tính:
a)\(-\dfrac{1}{2}+\dfrac{2}{3}-\dfrac{3}{4}+\dfrac{4}{5}\) b)\(\dfrac{1}{2}+\dfrac{2}{9}-\dfrac{3}{5}-\dfrac{4}{5}\)
Bài 1: Tính
a) \(\dfrac{1}{5}:\dfrac{1}{10}-\dfrac{1}{3}\left(\dfrac{6}{5}-\dfrac{9}{4}\right)\)
b) \(\dfrac{1}{2}\left(\dfrac{4}{3}+\dfrac{2}{5}\right)-\dfrac{3}{4}\left(\dfrac{8}{9}+\dfrac{16}{3}\right)\)
c) \(\dfrac{6}{7}:\left(\dfrac{3}{26}-\dfrac{3}{13}\right)+\dfrac{6}{7}\left(\dfrac{1}{10}-\dfrac{8}{5}\right)\)
Chứng minh:
a. \(A=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}< \dfrac{1}{3}\)
b.\(B=\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}< \dfrac{3}{16}\)
c. \(C=\dfrac{1}{41}+\dfrac{1}{42}+\dfrac{1}{43}+...+\dfrac{1}{79}+\dfrac{1}{80}>\dfrac{7}{12}\)
a) Chứng minh rằng: \(\dfrac{1}{6}< \dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4}\)
b) Tìm số nguyên a để: \(\dfrac{2a+9}{a+3}+\dfrac{5a+17}{a+3}-\dfrac{3a}{a+3}\) là số nguyên.
8 Chứng minh rằng :
a) \(\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+...+\dfrac{99}{100!}< 1\) ; b) \(\dfrac{1.2-1}{2!}+\dfrac{2.3-1}{3!}+\dfrac{3.4-1}{4!}+...+\dfrac{99.100}{100!}\)
c) \(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{49.50}=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+...+\dfrac{1}{50}\)
d) \(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}< \dfrac{1}{2}\)
ĐỀ 1:
1. Tính:
a. \(\dfrac{7}{23}\left[\left(\dfrac{-8}{6}\right)-\dfrac{45}{18}\right]\)
b. \(\dfrac{1}{5}:\dfrac{1}{10}-\dfrac{1}{3}\left(\dfrac{6}{5}-\dfrac{9}{4}\right)\)
c. \(\dfrac{3}{5}.\left(\dfrac{-8}{3}\right)-\dfrac{3}{5}:\left(-6\right)\)
d. \(\dfrac{1}{2}\left(\dfrac{4}{3}+\dfrac{2}{5}\right)-\dfrac{3}{4}\left(\dfrac{8}{9}+\dfrac{16}{3}\right)\)
e. \(\dfrac{6}{7}:\left(\dfrac{3}{26}-\dfrac{3}{13}\right)+\dfrac{6}{7}\left(\dfrac{1}{10}-\dfrac{8}{5}\right)\)
2. Tìm x, biết:
a. \(1\dfrac{2}{5}x+\dfrac{3}{7}=\dfrac{4}{5}\)
b. |x - 1,5| = 2
c. \(\dfrac{4}{5}-\left|x-\dfrac{1}{6}\right|=\dfrac{2}{3}\)
d. 3x . 2x = 216
e. \(\dfrac{1}{2}\left(x-\dfrac{1}{3}\right)+\dfrac{-1}{2}=\dfrac{3}{4}\)
* Lm nhanh nha
Ths you <3
Rút gọn:
\(A=\left[\left(\dfrac{3}{1+x}-\dfrac{x}{x^2+x+1}\right):\dfrac{2x^2+3x}{x+1}+\dfrac{3}{x+1}\right]\cdot\dfrac{x^2+x}{1+3x}\)
\(B=\left[\dfrac{a}{2a-6}-\dfrac{a^2}{a^2-9}+\dfrac{a}{2a-9}\cdot\left(\dfrac{3}{a}+\dfrac{1}{3-a}\right)\right]:\dfrac{a^2-5a-6}{18-2a^2}\)