(1-1/2)(1-1/3)(1-1/4)….(1-1/2002).x=1-1/1x2-1/2x3-1/3x4-...1/2002x2003 ae ghi lời giải jup mình nhé. Tìm x

Gọi \(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)....\left(1-\frac{1}{2002}\right).x\)

\(\Rightarrow A=\frac{1}{2}.\frac{2}{3}....\frac{2001}{2002}.x=\frac{x}{2002}\)

Gọi \(B=1-\frac{1}{1.2}-\frac{1}{2.3}-...-\frac{1}{2002.2003}\)

=>\(B=1-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2002.2003}\right)\)

\(\Rightarrow B=1-\left(1-\frac{1}{2003}\right)=1-\frac{2002}{2003}=\frac{1}{2003}\)

\(\Rightarrow\frac{x}{2002}=\frac{1}{2003}\Rightarrow x=\frac{2002}{2003}\)

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

A=1²/1.2 .2²/2.3 .3²/3.4 .4²/4.5

=1/2. 2.2/2.3 .3.3/3.4 .4.4/4.5

=1/2.2/3.3.4.4./5

=1/5

b) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2013.2015}\)

\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\right)\)

\(=\frac{1}{2}\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{2015-2013}{2013.2015}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{2015}\right)=\frac{1007}{2015}\)

Phương trình tương đương với: 

\(\frac{1007X}{2015}=\frac{4}{2015}\Leftrightarrow X=\frac{4}{1007}\)

c) \(\frac{x+1}{2015}+\frac{x+2}{2016}=\frac{x+3}{2017}+\frac{x+4}{2018}\)

\(\Leftrightarrow\frac{x+1}{2015}-1+\frac{x+2}{2016}-1=\frac{x+3}{2017}-1+\frac{x+4}{2018}-1\)

\(\Leftrightarrow\frac{x-2014}{2015}+\frac{x-2014}{2016}=\frac{x-2014}{2017}+\frac{x-2014}{2018}\)

\(\Leftrightarrow x-2014=0\)

\(\Leftrightarrow x=2014\)

làm ơn hãy giúp mình

\(A=1\times2+2\times3+3\times4+...+19\times20\)

\(A\times3=3\times\left(1\times2+2\times3+3\times4+...+19\times20\right)\)

\(A\times3=1\times2\times3+2\times3\times3+3\times4\times3+...+19\times20\times3\)

\(A\times3=1\times2\times3+2\times3\times\left(4-1\right)+3\times4\times\left(5-2\right)+....+19\times20\times\left(21-18\right)\)

\(A\times3=1\times2\times3-1\times2\times3+2\times3\times4-2\times3\times4+3\times4\times5+...+19\times20\times21\)

\(A\times3=\left(1\times2\times3-1\times2\times3\right)+\left(2\times3\times4-2\times3\times4\right)+...+\left(18\times19\times20-18\times19\times20\right)+19\times20\times21\)

\(A\times3=19\times20\times21\)

\(A\times3=7980\)

cảm ơn bạn nhiều nhé

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/\(\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}\right)\cdot10-x=0\)

\(\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\right)\cdot10-x=0\)

\(\left(\frac{30}{60}+\frac{10}{60}+\frac{5}{60}+\frac{2}{30}\right)\cdot10-x=0\)

\(\frac{47}{60}\cdot10-x=0\)

\(\frac{47}{6}-x=0\)

\(x=\frac{47}{6}-0\)

\(x=\frac{47}{6}\)