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\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+....+\frac{1}{98\times99}+\frac{1}{99\times100}\)
\(=\frac{2-1}{1\times2}+\frac{3-2}{2\times3}+\frac{4-3}{3\times4}+....+\frac{99-98}{98\times99}+\frac{100-99}{99\times100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{98\cdot99}+\frac{1}{99\cdot100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
\(\frac{2}{1X2}+\frac{2}{2X3}+\frac{2}{3X4}+...+\frac{2}{98X99}+\frac{2}{99X100}\)
\(2X\left(\cdot\frac{1}{1X2}+\frac{1}{2X3}+...+\frac{1}{98X99}+\frac{1}{99X100}\right)\)
\(2X\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(2X\left(1-\frac{1}{100}\right)\)
\(2X\frac{99}{100}\)
\(\frac{99}{50}\)
\(a,\frac{3\times4\times7}{12\times8\times9}\)
\(=\frac{3\times4\times7}{3\times4\times8\times9}\)
\(=\frac{7}{72}\)
\(b,\frac{4\times5\times6}{12\times10\times8}\)
\(=\frac{4\times5\times3\times2}{4\times3\times5\times2\times8}\)
\(=\frac{1}{8}\)
\(c,\frac{5\times6\times7\times9}{12\times7\times27}\)
\(=\frac{5\times6\times7\times9}{6\times2\times7\times9\times3}\)
\(=\frac{5}{6}\)
\(d,\frac{5\times6\times7}{12\times14\times15}\)
\(=\frac{5\times6\times7}{6\times2\times7\times2\times5\times3}\)
\(=\frac{1}{12}\)
a) \(\frac{3\times4\times7}{12\times8\times9}=\frac{3\times4\times7}{3\times4\times8\times9}=\frac{7}{8\times9}=\frac{7}{72}\)
b) \(\frac{4\times5\times6}{12\times10\times8}=\frac{4\times5\times6}{6\times2\times2\times5\times4\times2}=\frac{1}{2\times2\times2}=\frac{1}{8}\)
c) \(\frac{5\times6\times7\times9}{12\times7\times27}=\frac{5\times6\times9}{12\times27}=\frac{5\times6\times9}{2\times6\times3\times9}=\frac{5}{2\times3}=\frac{5}{6}\)
d) \(\frac{5\times6\times7}{12\times14\times15}=\frac{5\times6\times7}{2\times6\times2\times7\times3\times5}=\frac{1}{2\times2\times3}=\frac{1}{12}\)
Ta thấy :
1/1x2 = 1/1 - 1/2
1/2x3 = 1/2 - 1/3
....
=>( 1/1x2 + 1/2x3 + 1/3x4 + 1/5x6 ) x 10 - x = ( 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 ) x 10 - x
= ( 1/1 - 1/6 ) x 10 - x =0
5/6 x 10 - x = 0
25/3 - x = 0
=> x = 25/3
câu 2:
A = 1/1*2 + 1/2*3 + 1/3*4 + ... + 1/98*99 + 1/99*100
A = 2-1/1*2 + 3-2/2*3 + 4-3/3*4 + ... + 99-98/98*99 + 100-99/99*100
A = 2/1*2 - 1/ 1*2 + 3/2*3 - 2/2*3 + 4/ 3*4 -3/3*4 +...+ 99/98*99 - 98/98*99 + 100/99*100 - 99/99*100
A = 1 - 1/ 100
A = 99 / 100
phần 2 mk ko =bít
bài 1, 3 mk ko bít
\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2004\cdot2005}+\frac{1}{2005\cdot2006}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2004}-\frac{1}{2005}+\frac{1}{2005}-\frac{1}{2006}\)
\(A=1-\frac{1}{2006}=\frac{2005}{2006}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2005.2006}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2005}-\frac{1}{2006}\)
\(\Rightarrow A=1-\frac{1}{2006}\)
\(\Rightarrow A=\frac{2005}{2006}\)
nhân biểu thức vs 2. đcj bao nhiêu chia cho 2