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\(\frac{1313}{1212}:x=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5}\)
\(\frac{1313}{1212}:x=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}\)
\(\frac{1313}{1212}:x=\frac{1}{1}-\frac{1}{5}+\frac{1}{5}\)
\(\frac{1313}{1212}:x=\frac{4}{5}+\frac{1}{5}\)
\(\frac{1313}{1212}:x=1\)
\(x=\frac{1313}{1212}:1\)
\(x=\frac{13}{12}\)
Lời giải
\(\frac{1313}{1212}:x=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}\)
\(\frac{1313}{1212}:x=\frac{1}{1}-\frac{1}{5}+\frac{1}{5}\)
\(\frac{1313}{1212}:x=\frac{4}{5}+\frac{1}{5}\)
\(x=\frac{1313}{1212}:1\)
\(x=\frac{13}{12}\)
\(\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+...+\frac{1}{99x100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}\)
\(=\frac{49}{100}\)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(=\frac{1}{1}-\frac{1}{6}\)
\(=\frac{5}{6}\)
\(\frac{1}{1.2}\)\(+\)\(\frac{1}{2.3}\)\(+\)\(\frac{1}{3.4}\)\(+\)\(\frac{1}{4.5}\)\(+\)\(\frac{1}{5.6}\)
\(=\)\(\frac{1}{1}\)\(-\)\(\frac{1}{2}\)\(+\)\(\frac{1}{2}\)\(-\)\(\frac{1}{3}\)\(+\)\(\frac{1}{3}\)\(-\)\(\frac{1}{4}\)\(+\)\(\frac{1}{4}\)\(-\)\(\frac{1}{5}\)\(+\)\(\frac{1}{5}\)\(-\)\(\frac{1}{6}\)
\(=\)\(\frac{1}{1}\)\(-\)\(\frac{1}{6}\)
\(=\)\(\frac{5}{6}\)
Hok tốt
Đặt \(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(\Leftrightarrow A=\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+...+\left(\frac{1}{99}-\frac{1}{100}\right)\)
\(\Leftrightarrow A=\frac{1}{2}-\frac{1}{100}\)
\(\Leftrightarrow A=\frac{49}{100}\)
\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+....+\frac{1}{11\cdot12}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{11}-\frac{1}{12}\)
\(=\frac{1}{2}-\frac{1}{12}=\frac{5}{12}\)
+Câu a:
A = 1/1.2 + 1/2.3 + ...+ 1/5.6 + 1
A = 1/1 - 1/2 + 1/2 - 1/3 + ...+ 1/5 - 1/6 + 1
A = 1/1 - 1/6 + 1
A = 6/6 - 1/6 + 6/6
A = 5/6 + 6/6
A = 11/6
Câu b:
B = 1/1.2.3 + 1/2.3.4 + 1/3.4.5 + ...+ 1/98.99.100
B = 1/2. (2/1.2.3 + 2/2.3.4 + ...+ 2/98.99.100)
B = 1/2.(1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 + ...+ 1/98.99 - 1/99.100)
B = 1/2.(1/2 - 1/9900)
B = 1/2.4949/9900
B = 4949/19800
Câu a:
1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1
= 1/1 - 1/2 + 1/2 - 1/3 + 1/4 - 1/5 + 1/5 - 1/6 + 1
= 1/1 - 1/6 + 1
= 6/6 - 1/6 + 6/6
= 5/6 + 1
= 11/6
Câu b:
Câu b:
B = 1/1.2.3 + 1/2.3.4 + 1/3.4.5 + ...+ 1/98.99.100
B = 1/2. (2/1.2.3 + 2/2.3.4 + ...+ 2/98.99.100)
B = 1/2.(1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 + ...+ 1/98.99 - 1/99.100)
B = 1/2.(1/2 - 1/9900)
B = 1/2.4949/9900
B = 4949/19800
\(B=\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{2022.2023}\)
\(B=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\)
\(B=\dfrac{1}{2}-\dfrac{1}{2023}=\dfrac{2021}{4046}\)