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LH
1
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
6 tháng 7 2015
\(\frac{1}{6x7}+\frac{1}{7x8}+\frac{1}{8x9}+...+\frac{1}{Xx\left(X+1\right)}=\frac{1}{9}\)
\(\frac{7-6}{6x7}+\frac{8-7}{7x8}+\frac{9-8}{8x9}+...+\frac{\left(X+1\right)-X}{Xx\left(X+1\right)}=\frac{1}{9}\)
\(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{X}-\frac{1}{X+1}=\frac{1}{9}\)
\(\frac{1}{6}-\frac{1}{X+1}=\frac{1}{9}\)
\(\frac{1}{X+1}=\frac{1}{6}-\frac{1}{9}=\frac{1}{18}\)
X+1=18
X=17
16 tháng 8 2018
a)\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+x=\frac{3}{5}\)
\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+x=\frac{3}{5}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}+x=\frac{3}{5}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{10}+x=\frac{3}{5}\)
\(\Rightarrow\frac{2}{5}+x=\frac{3}{5}\)
\(\Rightarrow x=\frac{3}{5}-\frac{2}{5}=\frac{1}{5}\)
b)\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}+x=\frac{1}{3}\)
\(\Rightarrow\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+...+\frac{2}{13}-\frac{2}{15}+x=\frac{1}{3}\)
\(\Rightarrow\frac{2}{3}-\frac{2}{15}+x=\frac{1}{3}\)
\(\Rightarrow\frac{8}{15}+x=\frac{1}{3}\)
\(\Rightarrow x=\frac{1}{3}-\frac{8}{15}=-\frac{1}{5}\)
c)\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{9}{10}\)
\(\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{9}{10}\)
\(\Rightarrow\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{9}{10}\)
\(\Rightarrow\frac{1}{1}-\frac{1}{x+1}=\frac{9}{10}\)
\(\Leftrightarrow\frac{x+1-1}{x+1}=\frac{9}{10}\)
\(\Rightarrow\frac{x}{x+1}=\frac{9}{10}\)
\(\Rightarrow x=9\)
DD
Đoàn Đức Hà
Giáo viên
10 tháng 1 2021
b) \(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}+x=\frac{1}{3}\)
\(\Leftrightarrow\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{15-13}{13.15}+x=\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}+x=\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{3}-\frac{1}{15}+x=\frac{1}{3}\)
\(\Leftrightarrow x=\frac{1}{15}\)
23 tháng 4 2016
a)12,8 x 195 + 12,8 x 804 + 12,8
= 12,8 x (195+804+1)
=12,8 x 1000
=12800
23 tháng 4 2016
a) = 12,8 \(\times\) ( 195 + 804 +1)
= 12,8 \(\times\) 1000
= 1280
b) = \(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{9\times10}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
= 1 - \(\frac{1}{10}\)
= \(\frac{9}{10}\)
1 tháng 9 2020
\(\frac{70}{3}\left(\frac{39}{30}+\frac{39}{42}\right)-\frac{246}{7}\div\left(\frac{41}{56}+\frac{41}{72}\right)\)
\(=\frac{70}{3}\left(\frac{13}{10}+\frac{13}{14}\right)-\frac{246}{7}\div\left(\frac{41}{7\cdot8}+\frac{41}{8\cdot9}\right)\)
\(=\frac{70}{3}\left(1+\frac{3}{10}+1-\frac{1}{14}\right)-\frac{246}{7}\div\left(\frac{40+1}{7\cdot8}+\frac{40+1}{8\cdot9}\right)\)
\(=\frac{70}{3}\left[\left(1+1\right)+\left(\frac{3}{10}-\frac{1}{14}\right)\right]-\frac{246}{7}\div\left(\frac{5}{7}+\frac{1}{7\cdot8}+\frac{5}{9}+\frac{1}{8\cdot9}\right)\)
\(=\frac{70}{3}\left(2+\frac{8}{35}\right)-\frac{246}{7}\div\left[\frac{5}{7}+\frac{5}{9}+\left(\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\right)\right]\)
\(=\frac{70}{3}\cdot\frac{78}{35}-\frac{246}{7}\div\left[\frac{5}{7}+\frac{5}{9}+\left(\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\right]\)
\(=\frac{35\cdot2\cdot26\cdot3}{3\cdot35}-\frac{246}{7}\div\left(\frac{5}{7}+\frac{5}{9}+\frac{1}{7}-\frac{1}{9}\right)\)
\(=52-\frac{246}{7}\div\left[\left(\frac{5}{7}+\frac{1}{7}\right)+\left(\frac{5}{9}-\frac{1}{9}\right)\right]\)
\(=52-\frac{246}{7}\div\left(\frac{6}{7}+\frac{4}{9}\right)\)
\(=52-\frac{246}{7}\div\frac{82}{63}\)
\(=52-\frac{82\cdot3\cdot9\cdot7}{7\cdot82}\)
\(=52-27=25\)
\(\frac{57}{20}-\frac{26}{15}+\frac{139}{20}\div3\)
\(=\frac{57}{20}-\frac{26}{15}+\frac{139}{60}\)
\(=\frac{171}{60}-\frac{104}{60}+\frac{139}{60}=\frac{103}{30}\)
\(\frac{39}{4}+\frac{2}{3}\left(11-\frac{23}{4}\right)\)
\(=\frac{39}{4}+11\cdot\frac{2}{3}-\frac{23}{4}\cdot\frac{2}{3}\)
\(=\frac{39}{4}+\frac{22}{3}-\frac{56}{12}\)
\(=\frac{119}{12}+\frac{88}{12}-\frac{56}{12}=\frac{151}{12}\)
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2002}\right)\left(1-\frac{1}{2003}\right)\left(1-\frac{1}{2004}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2001}{2002}\cdot\frac{2002}{2003}\cdot\frac{2003}{2004}\)
\(=\frac{1\cdot2\cdot3\cdot...\cdot2001\cdot2002\cdot2003}{2\cdot3\cdot4\cdot...\cdot2002\cdot2003\cdot2004}=\frac{1}{2004}\)
NT
2
NT
2
VT
2
23 tháng 8 2017
\(\frac{2}{3}:\frac{x}{5}+\frac{5}{7}=\frac{2}{7}:\frac{3}{5}+\frac{10}{9}\)
\(\Rightarrow\frac{2}{3}:\frac{x}{5}+\frac{5}{7}=\frac{2}{7}.\frac{5}{3}+\frac{10}{9}\)
\(\Rightarrow\frac{2}{3}:\frac{x}{5}+\frac{5}{7}=\frac{10}{21}+\frac{10}{9}\)
\(\Rightarrow\frac{2}{3}:\frac{x}{5}+\frac{5}{7}=\frac{100}{63}\)
\(\Rightarrow\frac{2}{3}:\frac{x}{5}=\frac{100}{63}-\frac{5}{7}\)
\(\Rightarrow\frac{2}{3}:\frac{x}{5}=\frac{55}{63}\)
\(\Rightarrow\frac{x}{5}=\frac{2}{3}:\frac{55}{63}\)
\(\Rightarrow\frac{x}{5}=\frac{2}{3}.\frac{63}{55}\)
\(\Rightarrow\frac{x}{5}=\frac{42}{55}\)
\(\Rightarrow x=\frac{42.5}{55}=\frac{42}{11}\)
Nên ta chọn đáp án D là đáp án đúng!
23 tháng 8 2017
( trong bài có dấu . tức là dấu dấu nhân bn nhé )
2/3 : x/5 + 5/7 = 2/7 : 3/5 + 10/9
2/3 : x/5 + 5/7 = 10/21 + 10/9
2/3 : x/5 + 5/7 = 100/63
2/3 : x/5 = 100/63 - 5/7
2/3 : x/5 = 55/63
2/3 . 5/x= 55/63
=) 5/x = 55/63 : 2/3
=) 5/x = 55/42
ta có : a/b = c/d =) a.d=b.c
(=) 55.x = 5.42
55.x = 210
=) x = 210 : 55 = 42/11
vậy đáp án đúng là D
chúc bn học tốt