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\(\text{= 2/1 . 2/3 . 3/2 . 3/4 . 4/3 . 4/5 ....... 50/49.50/51 }\)
Dùng phương pháp khử liên tiếp ta có
\(=\frac{2}{1}-\frac{50}{51}=\frac{52}{51}\)
\(\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}...\frac{98.100}{99^2}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{98.100}{99.99}\)
\(=\frac{1.2.3...98}{2.3.4...99}.\frac{3.4.5...100}{2.3.4...99}\)
\(=\frac{1}{99}.\frac{100}{2}\)
\(=\frac{1}{99}.50=\frac{50}{99}\)
\(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.....\frac{50^2}{49.51}=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}.....\frac{50.50}{49.51}\)
\(=\frac{2.2.3.3.4.4......50.50}{1.3.2.4.3.5....49.51}=\frac{\left(2.3.4.....50\right).\left(2.3.4......50\right)}{\left(1.2.4.....49\right).\left(3.4.5.....51\right)}\)
\(=\frac{50.2}{1.51}=\frac{100}{51}\)
Cách làm:
tách tử thành 2.2;3.3;4.4;...;50.50
Sau đó ta nhân tử với tử,mẫu với mẫu theo thứ tự chữ số 1 trước như sau:
Tử: 2.3.4...50/1.2.3....49 . 2.3.4...50/3.4.5...51
=50.2/51=100/51
*Cho tôi biết cách viết dấu gạch ngang phân số nhé!
\(B=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}......\frac{10^2}{9.11}=\frac{\left(1.2.3.....10\right)^2}{\left(1.2.3.....9\right).\left(3.4.5....9.10.11\right)}=\frac{\left(1.2.3....10\right)^2}{\left(1.2\right)\left(3.4.5.....9\right)^2\left(10.11\right)}=\frac{\left(1.2.10\right)^2}{\left(1.2\right).\left(10.11\right)}=\frac{1.2.10}{11}=\frac{20}{11}\)
Tính giá trị biểu thức \(B=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{9.9^2}{9.8.100}\)
Ta có công thức : với n thuộc N* thì ta luôn có :
\(1+\frac{1}{n\left(n+2\right)}=\frac{n\left(n+2\right)+1}{n\left(n+2\right)}=\frac{n^2+2n+1}{n\left(n+2\right)}=\frac{\left(n+1\right)^2}{n\left(n+2\right)}\)
Áp dụng vào bài toán ta được :
\(P=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right).....\left(1+\frac{1}{49.51}\right)+\frac{2}{51}\)
\(=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.......\frac{50^2}{49.51}+\frac{2}{51}\)
\(=\frac{\left(2.3.4...50\right)\left(2.3.4...50\right)}{\left(1.2.3...49\right)\left(3.4.5....51\right)}+\frac{2}{51}\)
\(=\frac{50.2}{51}+\frac{2}{51}=\frac{102}{51}=2\)
A = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}=\frac{49}{50}\)
B = \(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4.6}=\frac{\left(2.3.4.5\right).\left(2.3.4.5\right)}{\left(1.2.3.4\right).\left(3.4.5.6\right)}=\frac{5.2}{1.6}=\frac{5}{3}\)
C = \(\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{59.61}=\frac{3}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\right)=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{61}\right)=\frac{3}{2}.\frac{56}{305}=\frac{74}{305}\)
Bài làm:
1) \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{50-49}{49.50}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=1-\frac{1}{50}=\frac{49}{50}\)
2) \(B=\frac{2^2.3^2.4^2.5^2}{1.2.3^2.4^2.5.6}=\frac{2.5}{6}=\frac{5}{3}\)
3) \(C=\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{59.61}\)
\(C=\frac{3}{2}\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)\)
\(C=\frac{3}{2}\left(\frac{7-5}{5.7}+\frac{9-7}{7.9}+...+\frac{61-59}{59.61}\right)\)
\(C=\frac{3}{2}\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\right)\)
\(C=\frac{3}{2}\left(\frac{1}{5}-\frac{1}{61}\right)\)
\(C=\frac{3}{2}.\frac{56}{305}=\frac{84}{305}\)