tính
\(\frac{2^2}{1.3}\).\(\frac{3^2}{2.4}\).\(\frac{4^2}{3.5}\)...\(\frac{59^2}{58.60}\)
\(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{59^2}{58.60}\)
\(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.....\frac{59^2}{58.60}\)
\(=\frac{2^2.3^2.4^2....59^2}{1.3.2.4.3.5....58.60}\)
\(=\frac{\left(2.3.4...59\right)\left(2.3.4...59\right)}{\left(2.3.4...58\right)\left(3.4.5....60\right)}\)
\(=\frac{59.2}{60}=\frac{59}{30}\)
Tính:
\(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}....\frac{59^2}{58.60}=\)
\(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.....\frac{59^2}{58.60}\)
22/1.3*32/2.4*42/3.5....592/58.60
=(2.3.4....59)(2.3....59)/(1.2.3....58)(3.4.5...60)
=59.2/60
=1 29/30
=59/30
Tính:
a) \(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}...\frac{99^2}{99.100}.\frac{100^2}{100.101}\)
b)\(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{59^2}{58.60}\)
a,1^2/1.2 . 2^2/2.3 . 3^2/3.4 ... 99^2/99.100 . 100^2/100.101
= 1/2 . 2/3 . 3/4 ... 99/100 . 100/101
=( 2.3.4....100/2.3.4...100) . 1/101
= 1 . 1/101
=1/101
ý b tương tự nhé !
Bạn kia đéo biết thì thôi xen vào làm gì?
tính
a) \(\frac{1^2}{1.2}\).\(\frac{2^2}{2.3}\).\(\frac{3^2}{3.4}\).....\(\frac{99^2}{99.100}\).\(\frac{100^2}{100.101}\)
b) \(\frac{2^2}{1.3}\).\(\frac{3^2}{2.4}\).\(\frac{4^2}{3.5}\).....\(\frac{59^2}{58.60}\)
a,\(\frac{1.1.2.2.3.3....99.99.100.100}{1.2.2.3.3.4...99.100.100.101}\)=\(\frac{\left(1.2.3...99.100\right)\left(1.2.3...99.100\right)}{\left(1.2.3.4...99.100\right)\left(2.3.4...100.101\right)}\)=\(\frac{1.1}{101}\)=\(\frac{1}{101}\). ý b làm tương tự nha
giup mik voi help
bai 1 thuc hien phep tinh
\(\frac{1^2}{1.2}\).\(\frac{2^2}{2.3}\).\(\frac{3^2}{3.4}\)....\(\frac{99^2}{99.100}\).\(\frac{100^2}{100.101}\)
\(\frac{2^2}{1.3}\). \(\frac{3^2}{2.4}\). \(\frac{4^2}{3.5}\)... \(\frac{59^2}{58.60}\)
Bg
a)\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.....\frac{99^2}{99.100}.\frac{100^2}{100.101}\)
\(=\frac{1^2.2^2.3^2.....99^2.100^2}{1.2.2.3.3.4.....99.100.100.101}\)
\(=\frac{1^2}{101}\)
\(=\frac{1}{101}\)
Ghi chú: \(=\frac{1^2.2^2.3^2.....99^2.100^2}{1.2.2.3.3.4.....99.100.100.101}\)--> 22 chịt tiêu 2.2 (trên và dưới) làm thế này mãi đến khi còn \(\frac{1^2}{101}\).
b) \(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.....\frac{59^2}{58.60}\)
=\(\frac{2^2.3^2.4^2.....59^2}{1.3.2.4.3.5.....58.60}\)
= \(\frac{2}{1}.\frac{59}{60}\)
= \(\frac{59}{30}\)
Ghi chú: \(\frac{2^2.3^2.4^2.....59^2}{1.3.2.4.3.5.....58.60}\)--> chịt tiêu liên tục, còn \(\frac{2}{1}.\frac{59}{60}\).
TÍNH:\(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4.6}\).
\(A=\frac{2^2}{1.3}\cdot\frac{2^2}{2.4}\cdot\frac{2^2}{3.5}\cdot\frac{2^2}{4.6}\)
\(A=\frac{4}{3}\cdot\frac{1}{2}\cdot\frac{4}{15}\cdot\frac{1}{6}\)
\(A=\frac{4.1.4.1}{3.2.15.6}\)
\(A=\frac{4}{135}\)
\(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4.6}\)
\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}.\frac{5.5}{4.6}\)
\(=\frac{2.3.4.5}{1.2.3.4}.\frac{2.3.4.5}{3.4.5.6}\)
\(=\frac{5}{1}.\frac{2}{6}\)
\(=\frac{5}{1}.\frac{1}{3}\)
\(=\frac{5}{3}\)
Tính
\(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.......\frac{50^2}{49.51}\)
\(\text{= 2/1 . 2/3 . 3/2 . 3/4 . 4/3 . 4/5 ....... 50/49.50/51 }\)
Dùng phương pháp khử liên tiếp ta có
\(=\frac{2}{1}-\frac{50}{51}=\frac{52}{51}\)
Tính a=\(\frac{1.3}{2^2}.\frac{2.4}{3^2}\frac{3.5}{4^2}...\frac{2016.2018}{2017^2}\)
bạn làm ntn
ta có
\(\frac{1.2.3.....2016}{2.3.4.....2017}.\frac{3.4.5.....2018}{2.3.4.....2017}\)
và rút gọn