Tính nhanh
\(\frac{4\cdot5\cdot6\cdot7}{14\cdot15\cdot16}\)
\(\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot\frac{5}{6}\)
giúp mình với
\(\sqrt[2]{4\cdot9\frac{8}{8}+\frac{48\cdot11+5}{1\cdot\frac{814}{5+\frac{6145}{1\cdot\frac{821}{614}}}}}2548-\frac{8452}{14\cdot\frac{58}{96\cdot\frac{41}{\frac{24}{1\cdot\frac{975545}{1421+\frac{84874}{\frac{1+2+3+4+5+6+7+8+9\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9}{2\cdot\frac{2}{1}}}}}}}}\)
Bài 1:Tìm x
\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{\left(2\cdot x+1\right)\cdot\left(2\cdot x+3\right)}=\frac{9}{19}\)
Bài 2: Tính nhanh
\(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{2016\cdot2018}\)
ai giúp mình với gấp lắm không có bài là bị phạt đó
Bài 1 :
\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}=\frac{9}{19}\)
\(\Leftrightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{9}{19}\)
\(\Leftrightarrow1-\frac{1}{2x+3}=\frac{9}{19}\)
\(\Leftrightarrow\frac{1}{2x+3}=1-\frac{9}{19}\)
\(\Leftrightarrow\frac{1}{2x+3}=\frac{10}{19}\)
\(\Leftrightarrow10.\left(2x+3\right)=19\Leftrightarrow2x+3=\frac{19}{10}\)
\(\Leftrightarrow2x=\frac{19}{10}-3\Leftrightarrow2x=-\frac{11}{10}\)
\(\Leftrightarrow x=-\frac{11}{20}=-0,55\)
Bài 2 :
\(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2016.2018}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+....+\frac{1}{2016}-\frac{1}{2018}\)
\(=\frac{1}{2}-\frac{1}{2018}=\frac{504}{1009}\)
tính \(\frac{2^2}{1.3}\cdot\frac{3^2}{2\cdot4}\cdot\frac{4^2}{3\cdot5}\cdot\frac{5^2}{4\cdot6}\)
\(=\frac{\left(2.3.4.5\right)^2}{2.3^2.4^2.5.6}=\frac{2.5}{6}=\frac{5}{3}\)
\(\frac{2^2}{1.3}=\frac{4}{3};\frac{3^2}{2.4}=\frac{9}{8};\frac{4^2}{3.5}=\frac{16}{15};\frac{5^2}{4.6}=\frac{25}{24}\)
\(\Rightarrow\frac{4}{3}.\frac{9}{8}.\frac{16}{15}.\frac{25}{24}=\frac{5}{3}\)
\(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4.6}\)
\(=\frac{2^2.3^2.4^2.5^2}{1.3.2.4.3.5.4.6}\)
\(=\frac{\left(2.3.4.5\right)\left(2.3.4.5\right)}{\left(2.3.4.5\right)\left(3.4.6\right)}\)
\(=\frac{5}{3}\)
Cho \(A=\frac{1\cdot3\cdot5\cdot...\cdot995\cdot997}{4\cdot6\cdot8\cdot...\cdot998\cdot1000};B=\frac{2\cdot4\cdot6\cdot...\cdot996\cdot998}{5\cdot7\cdot9\cdot...\cdot999\cdot1001}\)So sánh A và B
\(41\sqrt[9^1]{8\sqrt[2]{\frac{12}{2.85\frac{1\cdot2+3\cdot4+5\cdot6+7\cdot8+9\sqrt[4]{16}}{2\cdot\frac{12}{2}\sqrt{4^2}-7^2}}}4\cdot5\cdot6\cdot7\cdot8\cdot9}\)
Ô phép tính khủng. Cái này do bạn chế ra à !
a)\(\left(10\frac{2}{9}\cdot2\frac{3}{5}\right)-6\frac{2}{9}\)
b)\(\frac{6}{7}+\frac{1}{7}\cdot\frac{2}{7}+\frac{1}{7}\cdot\frac{5}{7}\)
c)\(3\cdot136\cdot8+4\cdot14\cdot6-14\cdot150\)
d)\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+...+\frac{1}{110}\)
e)\(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{37\cdot39}\)
Tính nhanh nha
\(a,\left(10\frac{2}{9}.2\frac{3}{5}\right)-6\frac{2}{9}=\frac{1196}{45}-\frac{56}{9}=\frac{1196}{45}-\frac{280}{45}=\frac{916}{45}\)
\(b,\frac{6}{7}+\frac{1}{7}.\frac{2}{7}+\frac{1}{7}.\frac{5}{7}=\frac{1}{7}\left(6+\frac{2}{7}+\frac{5}{7}\right)=\frac{1}{7}.7=1\)
\(c,3.136.8+4.14.6-14.150=3264+336-2100=1500\)
\(d,\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{110}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{10.11}\)\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\)\(=\frac{1}{2}-\frac{1}{11}=\frac{9}{22}\)
\(e,\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{37.39}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}=\frac{1}{3}-\frac{1}{39}=\frac{4}{13}\)
a, \(\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+....+\frac{1}{24\cdot25}\)
b, \(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+.....+\frac{2}{99\cdot101}\)
c, \(5\frac{2}{7}\cdot\frac{8}{11}+5\frac{2}{7}\cdot\frac{5}{11}-5\frac{2}{7}\cdot\frac{2}{11}\)
giup minh voi mai nop roi!!!!
#)Giải :
a)\(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{24.25}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{24}-\frac{1}{25}\)
\(=\frac{1}{5}-\frac{1}{25}\)
\(=\frac{4}{25}\)
b)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}\)
\(=\frac{100}{101}\)
a) \(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{24.25}\)
= \(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{24}-\frac{1}{25}\)
= \(\frac{1}{5}-\frac{1}{25}\)
= \(\frac{4}{25}\)
b) \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\)
= \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)
= \(1-\frac{1}{101}\)
= \(\frac{100}{101}\)
c) \(5\frac{2}{7}.\frac{8}{11}+5\frac{2}{7}.\frac{5}{11}-5\frac{2}{7}.\frac{2}{11}\)
= \(5\frac{2}{7}.\left(\frac{8}{11}+\frac{5}{11}-\frac{2}{11}\right)\)
= \(5\frac{2}{7}\)
= \(\frac{37}{7}\)
a)\(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{24.25}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{24}-\frac{1}{25}\)
\(=\frac{1}{5}-\frac{1}{25}\)
\(=\frac{5}{25}-\frac{1}{25}=\frac{4}{25}\)
tập hợp các số nguyên x thỏa mãn
\(x\cdot\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}\right)<1\frac{6}{7}\)
Lời giải:
$x(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7})< 1\frac{6}{7}$
$x(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7})< \frac{13}{7}$
$x(1-\frac{1}{7})< \frac{13}{7}$
$x.\frac{6}{7}< \frac{13}{7}$
$x< \frac{13}{7}: \frac{6}{7}=\frac{13}{6}$
Vì $x$ là số nguyên nên $x\leq 2$
Vậy $x$ là các số nguyên sao cho $x\leq 2$.
Tính nhanh:
\(987654321\cdot\frac{1+1\cdot2+2\cdot3+3\cdot4+4\cdot5+5\cdot6+6\cdot7+7\cdot8+8\cdot9+9\cdot...\cdot999+999\cdot1000+1000}{1000+1000\cdot1001+1001\cdot1002+1002\cdot1003+1003\cdot1004+1004\cdot1005+1005\cdot...\cdot9999+9999\cdot10000+10000}\)