[Toán 7]Tính nhanh; Tìm x

[thaoanh110]

[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
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Bài 1: Thực hiện phép tính 1 cách thuận tiện :

a) $\dfrac{1}{90}- \dfrac{1}{72}- \dfrac{1}{56}- \dfrac{1}{42}- \dfrac{1}{30}- \dfrac{1}{20}- \dfrac{1}{12}- \dfrac{1}{6}- \dfrac{1}{2}$

b) $0,5+ \dfrac{1}{3} +0,4 + \dfrac{5}{7}+ \dfrac{1}{6} - \dfrac{4}{35} + \dfrac{1}{41}$

c) $\dfrac{1}{2}-(-\dfrac{2}{5})+ \dfrac{1}{3}+ \dfrac{5}{7}-(-\dfrac{1}{6})+(-\dfrac{4}{35})+ \dfrac{1}{41}$

d) $\dfrac{1}{100.99} - \dfrac{1}{99.98} - \dfrac{1}{98.97} - ... - \dfrac{1}{3.2} - \dfrac{1}{2.1}$

Bài 2: Tìm số hữu tỉ x biết rằng :

a) $\dfrac{x+10}{100} + \dfrac{x+10}{101} + \dfrac{x+10}{102}= \dfrac{x+ 10}{103}$

b) $\dfrac{x+4}{2010} + \dfrac{x+3}{2011} =\dfrac{x+2}{2012}+ \dfrac{x+ 1}{2013}$

Chú ý: Tiêu đề

 

[hiensau99]

Bài 1:
a, $\dfrac{1}{90} - \dfrac{1}{72} - \dfrac{1}{56} - \dfrac{1}{42} - \dfrac{1}{30} - \dfrac{1}{20} - \dfrac{1}{12} - \dfrac{1}{6} - \dfrac{1}{2} $

$= \dfrac{1}{10.9} - \dfrac{1}{9.8} - \dfrac{1}{8.7} - \dfrac{1}{7.6} - ... - \dfrac{1}{3.2} - \dfrac{1}{2.1} $

$= \dfrac{1}{10}- \dfrac{1}{9}- \dfrac{1}{9} + \dfrac{1}{8}- \dfrac{1}{8} + \dfrac{1}{7} - ... - \dfrac{1}{4} + \dfrac{1}{3} - \dfrac{1}{3} + \dfrac{1}{2}- \dfrac{1}{2} +1$

$= \dfrac{1}{10}- \dfrac{1}{9}- \dfrac{1}{9}+1 = \dfrac{79}{90} $

b, $= 0,5+ \dfrac{1}{3} +0,4 + \dfrac{5}{7}+ \dfrac{1}{6} - \dfrac{4}{35} + \dfrac{1}{41}$

$= \dfrac{1}{2}+ \dfrac{1}{3} +\dfrac{2}{5}+ \dfrac{5}{7}+ \dfrac{1}{6} - \dfrac{4}{35} + \dfrac{1}{41}$

$= (\dfrac{2}{5}+ \dfrac{5}{7} - \dfrac{4}{35})+ (\dfrac{1}{2}+ \dfrac{1}{3}+ \dfrac{1}{6} )+ \dfrac{1}{41} $

$= 1+ 1 + \dfrac{1}{41} = 2 \dfrac{1}{41}$

c, $\dfrac{1}{2}-(-\dfrac{2}{5})+ \dfrac{1}{3}+ \dfrac{5}{7}-(-\dfrac{1}{6})+(-\dfrac{4}{35})+ \dfrac{1}{41}$

$=\dfrac{1}{2}+ \dfrac{2}{5}+ \dfrac{1}{3}+ \dfrac{5}{7}+ \dfrac{1}{6}-\dfrac{4}{35}+ \dfrac{1}{41}$

$= 2 \dfrac{1}{41}$ (theo phần b)

d, $= \dfrac{1}{100.99} - \dfrac{1}{99.98} - \dfrac{1}{98.97} - ... - \dfrac{1}{3.2} - \dfrac{1}{2.1} $

$= \dfrac{1}{100}- \dfrac{1}{99}- \dfrac{1}{99} + \dfrac{1}{98}- \dfrac{1}{98} + \dfrac{1}{97} - ... - \dfrac{1}{4} + \dfrac{1}{3} - \dfrac{1}{3} + \dfrac{1}{2}- \dfrac{1}{2} +1$

$= \dfrac{1}{100}- \dfrac{1}{99}- \dfrac{1}{99} +1 = \dfrac{9799}{9900}$

Bài 2:

a, $ \dfrac{x+10}{100} + \dfrac{x+10}{101} + \dfrac{x+10}{102}= \dfrac{x+ 10}{103}$

$\to \dfrac{x+10}{100} + \dfrac{x+10}{101} + \dfrac{x+10}{102}- \dfrac{x+ 10}{103}=0$

$\to (x+10).(\dfrac{1}{100} + \dfrac{1}{101} + \dfrac{1}{102}- \dfrac{1}{103})=0$

$\to x+10= 0 \to x= -10$

b, $\dfrac{x+4}{2010} + \dfrac{x+3}{2011} =\dfrac{x+2}{2012}+ \dfrac{x+ 1}{2013}$

$\to \dfrac{x+2014- 2010}{2010} + \dfrac{x+2014- 2011}{2011} -\dfrac{x+2014-2012}{2012}- \dfrac{x+ 2014- 2013}{2013}=0$

$\to \dfrac{x+2014}{2010} -1 + \dfrac{x+2014}{2011}-1 -\dfrac{x+2014}{2012}+1- \dfrac{x+ 2014}{2013}+1=0$

$\to (x+2014). (\dfrac{1}{2010}+ \dfrac{1}{2011} -\dfrac{1}{2012}- \dfrac{1}{2013})=0$

$\to x+ 2014 = 0 \to x= -2014$