Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1> a) \(\frac{5}{7}x4:\frac{5}{9}=\frac{5}{7}:\frac{5}{9}x4=\frac{5}{7}x\frac{9}{5}x4=\frac{9}{7}x4=\frac{9x4}{7}=\frac{36}{7}\)
\(b,8x\frac{2}{3}:\frac{1}{2}=8x\frac{2}{3}x\frac{2}{1}=8x2x\frac{2}{3}=16x\frac{2}{3}=\frac{32}{3}\)
\(c,6:\frac{3}{5}-\frac{7}{6}x\frac{6}{7}=6x\frac{5}{3}-1=10-1=9\)
\(\frac{21}{5}x\frac{10}{11}+\frac{57}{11}=\frac{42}{11}+\frac{57}{11}=\frac{99}{11}=9\)
2) a) \(\frac{35}{9}:x=\frac{35}{6}\)
\(x=\frac{35}{9}:\frac{35}{6}\)
\(x=\frac{35}{9}x\frac{6}{35}\)
\(x=\frac{2}{3}\)
b) \(\left(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}\right)x10-X=0\)
\(\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{5}-\frac{1}{6}\right)x10-X=0\)
\(\left(\frac{1}{1}-\frac{1}{6}\right)x10-X=10\)
\(\frac{5}{6}x10-X=0\)
\(X=\frac{5}{6}x10=\frac{25}{3}\)
Đúng nha !!!!
1/a/\(\frac{5}{7}\cdot4:\frac{5}{9}=\frac{20}{7}:\frac{5}{9}=\frac{20}{7}\cdot\frac{9}{5}=\frac{36}{7}\)
b/\(8\cdot\frac{2}{3}:\frac{1}{2}=\frac{16}{3}:\frac{1}{2}=\frac{16}{3}\cdot\frac{2}{1}=\frac{32}{3}\)
c/\(6:\frac{3}{5}-\frac{7}{6}\cdot\frac{6}{7}=6\cdot\frac{5}{3}-1=10-1=9\)
2/a/\(\frac{35}{9}:x=\frac{35}{6}\)
\(x=\frac{35}{9}:\frac{35}{6}=\frac{35}{9}\cdot\frac{6}{35}\)
\(x=\frac{2}{3}\)
b/\(\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}\right)\cdot10-x=0\)
\(\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\right)\cdot10-x=0\)
\(\left(\frac{30}{60}+\frac{10}{60}+\frac{5}{60}+\frac{2}{30}\right)\cdot10-x=0\)
\(\frac{47}{60}\cdot10-x=0\)
\(\frac{47}{6}-x=0\)
\(x=\frac{47}{6}-0\)
\(x=\frac{47}{6}\)
a) \(\left(\frac{1}{3}+\frac{1}{5}\right)+\left(\frac{1}{6}-\frac{1}{5}\right)=\left(\frac{1}{3}+\frac{1}{6}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)=\frac{1}{2}\)
b) \(\frac{3}{16}\times\frac{7}{5}+\frac{3}{5}\times\frac{9}{16}=\frac{21}{80}+\frac{27}{80}=\frac{48}{80}=\frac{3}{5}\)
c) \(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{2020\times2021}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2020}-\frac{1}{2021}\)
\(=1-\frac{1}{2021}=\frac{2020}{2021}\)
d) \(\frac{1}{1\times3}+\frac{1}{3\times5}+...+\frac{1}{2021\times2023}=\frac{1}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+...+\frac{2}{2021\times2023}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2021}-\frac{1}{2023}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{2023}\right)=\frac{1}{2}\times\frac{2022}{2023}=\frac{1011}{2023}\)
e) \(\frac{3}{2}\times\frac{1}{7}\times\frac{5}{4}+\frac{15}{2}\times\frac{6}{7}\times\frac{1}{4}==\frac{15}{56}+\frac{80}{56}=\frac{95}{56}\)
\(a)\) \(2x-5=21\)
\(\Leftrightarrow\) \(2x=21+5\)
\(\Leftrightarrow\) \(2x=26\)
\(\Leftrightarrow\) \(x=26:2\)
\(\Leftrightarrow\) \(=13\)
\(b)\) \(\frac{3}{4}+\frac{1}{4}x=\frac{5}{6}\)
\(\Leftrightarrow\) \(\frac{1}{4}x=\frac{5}{6}-\frac{3}{4}\)
\(\Leftrightarrow\) \(\frac{1}{4}x=\frac{1}{12}\)
\(\Leftrightarrow\) \(x=\frac{1}{3}\)
( 1 + 3 + 5 + 7 + ... + 2003 + 2005 ) x ( 1212 x 27 - 2727 x 12 )
= ( 1 + 3 + 5 + 7 + ... + 2003 + 2005 ) x ( 32724 -32724 )
= ( 1 + 3 + 5 + 7 + ... + 2003 + 2005 ) x 0
= 0
\(\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3.4}+....+\frac{5}{x.\left(x+1\right)}=\frac{44}{9}\)
\(5.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{x.\left(x+1\right)}\right)=\frac{44}{9}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{44}{9}:5\)
\(1-\frac{1}{x+1}=\frac{44}{45}\)
\(\frac{1}{x+1}=1-\frac{44}{45}\)
\(\frac{1}{x+1}=\frac{1}{45}\)
=> x + 1 = 45
=> x = 45 - 1
=> x = 44
Ta có : \(\left(x+1\right)+\left(x+3\right)+\left(x+5\right)+\left(x+7\right)=32\)
\(\Leftrightarrow x+x+x+x+1+3+5+7=32\)
\(\Leftrightarrow4x+16=32\Leftrightarrow4x=16\Leftrightarrow x=4.\)