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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) \(\frac{3}{2}\times\frac{1}{7}\times\frac{5}{4}+\frac{15}{2}\times\frac{6}{7}\times\frac{1}{4}==\frac{15}{56}+\frac{80}{56}=\frac{95}{56}\)
\(a)\) \(2x-5=21\)
\(\Leftrightarrow\) \(2x=21+5\)
\(\Leftrightarrow\) \(2x=26\)
\(\Leftrightarrow\) \(x=26:2\)
\(\Leftrightarrow\) \(=13\)
\(b)\) \(\frac{3}{4}+\frac{1}{4}x=\frac{5}{6}\)
\(\Leftrightarrow\) \(\frac{1}{4}x=\frac{5}{6}-\frac{3}{4}\)
\(\Leftrightarrow\) \(\frac{1}{4}x=\frac{1}{12}\)
\(\Leftrightarrow\) \(x=\frac{1}{3}\)
\(G=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(G=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^5}\)
\(3G=3+1+\frac{1}{3}+...+\frac{1}{3^4}\)
\(3G-G=\left(3+1+...+\frac{1}{3^4}\right)-\left(1+\frac{1}{3}+...+\frac{1}{3^5}\right)\)
\(2G=3-\frac{1}{3^5}\)
\(2G=3-\frac{1}{243}\)
\(2G=\frac{729}{243}-\frac{1}{243}\)
\(G=\frac{728}{243}:2\)
\(G=\frac{364}{243}\)
\(\frac{3}{1.2}+\frac{3}{2.3}+...+\frac{3}{x.\left(x+1\right)}=\frac{6042}{2015}\)
\(3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{6042}{2015}\)
\(1-\frac{1}{x+1}=\frac{6042}{2015}:3\)
\(1-\frac{1}{x-1}=\frac{2014}{2015}\)
\(\frac{1}{x-1}=1-\frac{2014}{2015}\)
\(\frac{1}{x-1}=\frac{1}{2015}\)
\(\Rightarrow x-1=2015\)
\(\Rightarrow x=2016\)
1/1*2 + 1/2*3 + 1/3*4 + .... + 1/99 * 100
= 1- 1/100
= 99/100
a)
Số số hạng của dãy trên là;
(n - 1) : 1 + 1 = n(số hạng)
Tổng dãy trên là:
(n + 1) x n : 2 = ? (tùy giá trị n)
b) Đặt A = 1x2 + 2x3 + 3x4 + ... + 99 x 100
3A= 3 x ( 1x2 + 2x3 + 3x4 + ... + 99 x 100)
3A = 1 x 2 x (3 - 0) + 2 x 3 x(4-1) + .....+99.100.(101 - 98)
3A = 1 x 2 x 3 - 1 x 2 x 3 + 2 x 3 x 4 - 2 x 3 x 4 + .......+ 99.100.101
3A = 99.100.101
A = \(\frac{\text{99.100.101}}{3}=333300\)
a, 1 + 2 + 3 + ... + n
= ( 1 + n) × n : 2
b, 1×2 + 2×3 + 3×4 + ... + 99×100
= 1/3 × ( 1×2×3 + 2×3×3 + 3×4×3 + ... + 99×100×3)
= 1/3 × [ 1×2×(3-0) + 2×3×(4-1) + 3×4×(5-2) + ... + 99×100×(101-98) ]
= 1/3 × ( 1×2×3 - 0×1×2 + 2×3×4 - 1×2×3 + 3×4×5 - 2×3×4 + ... + 99×100×101 - 98×99×100 )
= 1/3 × [ ( 1×2×3 + 2×3×4 + 3×4×5 + ... + 99×100×101) - ( 0×1×2 + 1×2×3 + 2×3×4 + ... + 98×99×100) ]
= 1/3 × ( 99×100×101 - 0×1×2)
= 1/3 × ( 99×100×101 - 0)
= 1/3 × 99×100×101
= 333 300