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\(\Rightarrow\dfrac{1}{5\times6}+\dfrac{1}{6\times7}+\dfrac{1}{7\times8}+...+\dfrac{1}{11\times12}=\dfrac{21}{x}\\ \Rightarrow\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{11}-\dfrac{1}{12}=\dfrac{21}{x}\\ \Rightarrow\dfrac{1}{5}-\dfrac{1}{12}=\dfrac{21}{x}\\ \Rightarrow\dfrac{21}{x}=\dfrac{7}{60}\Rightarrow x=\dfrac{21\cdot60}{7}=180\)
\(\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}+\dfrac{1}{132}=\dfrac{21}{x}\)
\(\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}+\dfrac{1}{10.11}+\dfrac{1}{11.12}=\dfrac{21}{x}\)
\(\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}=\dfrac{21}{x}\)
\(\dfrac{1}{5}-\dfrac{1}{12}=\dfrac{21}{x}\)
Còn lại bạn tự tính
\(\frac{1}{6x7}+\frac{1}{7x8}+\frac{1}{8x9}+...+\frac{1}{Xx\left(X+1\right)}=\frac{1}{9}\)
\(\frac{7-6}{6x7}+\frac{8-7}{7x8}+\frac{9-8}{8x9}+...+\frac{\left(X+1\right)-X}{Xx\left(X+1\right)}=\frac{1}{9}\)
\(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{X}-\frac{1}{X+1}=\frac{1}{9}\)
\(\frac{1}{6}-\frac{1}{X+1}=\frac{1}{9}\)
\(\frac{1}{X+1}=\frac{1}{6}-\frac{1}{9}=\frac{1}{18}\)
X+1=18
X=17
a)12,8 x 195 + 12,8 x 804 + 12,8
= 12,8 x (195+804+1)
=12,8 x 1000
=12800
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 - \(\frac{1}{10}\)
= \(\frac{9}{10}\)
rút gọn thành
1/2 x 2/3 x 3/4 x 4/5 x 5/6 x 6/7 x 7/8 x 8/9
1x2x3x4x5x6x7x8
__________________
2x3x4x5x6x7x8x9
gạch các số giống nhau ở tử và mẫu và bằng 1/9
1/20 + 1/30 + 1/42 + 1/56 + 1/72 + 1/90
= 1/4.5 + 1/5.6 + 1/6.7 + 1/7.8 + 1/8.9 + 1/9.10
= 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8 + 1/8 - 1/9 + 1/9 - 1/10
= 1/4 - 1/10
= 3/20
\(A=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\)
\(A=\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\)
\(A=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-...+\frac{1}{9}-\frac{1}{10}\)
\(A=\frac{1}{4}-\frac{1}{10}\)
\(A=\frac{3}{20}\)
\(A=\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{9.10}\)
\(A=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}\)
\(A=\frac{1}{4}-\frac{1}{10}=\frac{5}{20}-\frac{2}{20}=\frac{3}{20}\)
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)\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{9}{10}\)
\(\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{9}{10}\)
\(\Rightarrow\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{9}{10}\)
\(\Rightarrow\frac{1}{1}-\frac{1}{x+1}=\frac{9}{10}\)
\(\Leftrightarrow\frac{x+1-1}{x+1}=\frac{9}{10}\)
\(\Rightarrow\frac{x}{x+1}=\frac{9}{10}\)
\(\Rightarrow x=9\)
b) \(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}+x=\frac{1}{3}\)
\(\Leftrightarrow\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{15-13}{13.15}+x=\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}+x=\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{3}-\frac{1}{15}+x=\frac{1}{3}\)
\(\Leftrightarrow x=\frac{1}{15}\)
Ta có: \(B=\dfrac{1}{6}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\)
\(\Leftrightarrow B=\dfrac{1}{6}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
\(\Leftrightarrow B=\dfrac{1}{6}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)
\(\Leftrightarrow B=\dfrac{1}{6}+\dfrac{1}{4}-\dfrac{1}{10}=\dfrac{19}{60}\)
\(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\)
\(=\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=\dfrac{1}{2}-\dfrac{1}{10}\)
\(=\dfrac{2}{5}\)
\(B=\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}+\dfrac{1}{132}+\dfrac{1}{156}\)
\(B=\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}+\dfrac{1}{10\cdot11}+\dfrac{1}{11\cdot12}+\dfrac{1}{12\cdot13}\)
\(B=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{13}\)
\(B=\dfrac{1}{5}-\dfrac{1}{13}=\dfrac{8}{65}\)
B=\(\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}+\dfrac{1}{10.11}+\dfrac{1}{11.12}+\dfrac{1}{12.13}\)
B=\(\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-.......-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}\)
B=\(\dfrac{1}{5}-\dfrac{1}{12}=\dfrac{7}{60}\)
(sở dĩ có như trên vì \(\dfrac{1}{x\left(x+1\right)}=\dfrac{1}{x}-\dfrac{1}{x+1}\) nên cứ áp dụng vào là ra nhé)