Bài 1: a) \(A=\frac{5}{11.16}+\frac{5}{16.21}+\frac{5}{21.26}+...+\frac{5}{61.66}\) b) \(B=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\) c) \(C=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{1989.1990}\)Bài 2: a. Tính tổng: \(M=\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}\) b. Cho: \(S=\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}\) chứng minh rằng 1 < S < 2Bài 3: Tính giá trị của biểu thức...
Đọc tiếp
Bài 1: a) \(A=\frac{5}{11.16}+\frac{5}{16.21}+\frac{5}{21.26}+...+\frac{5}{61.66}\)
b) \(B=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)
c) \(C=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{1989.1990}\)
Bài 2: a. Tính tổng: \(M=\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}\)
b. Cho: \(S=\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}\) chứng minh rằng 1 < S < 2
Bài 3: Tính giá trị của biểu thức sau:
\(A=\left(\frac{1}{7}+\frac{1}{23}-\frac{1}{1009}\right):\left(\frac{1}{23}+\frac{1}{7}-\frac{2}{2009}+\frac{1}{7}.\frac{1}{23}.\frac{1}{2009}\right)+1:\left(30.1009-160\right)\)
Bài 4: Tính nhanh:
\(\text{a) 35 . 34 + 35 . 86 + 67 . 75 + 65 . 45}\)
\(\text{b) 21 . }7^2-11.7^2+90.7^2+49.125.16\)
Bài 5: Thực hiện phép tinh sau:
a. \(\frac{2181.729+243.81.27}{3^2.9^2.234+18.54+162.9+723.729}\)
b. \(\frac{1}{1.2+}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
c. \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1\)
d. \(\frac{5.4^{15}-9^9-4.3^{20}}{5.2^{19}.6^{19}-7.2^{29}.27^6}\)
giúp mk nha! nhớ viết cách làm nha!
*)\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\)
=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
=\(1-\frac{1}{6}\)
=\(\frac{6}{6}-\frac{1}{6}\)
\(=\frac{5}{6}\)
*)\(\frac{2003}{1.2}+\frac{2003}{2.3}+\frac{2003}{3.4}+...+\frac{2003}{2002.2003}\)
\(=\frac{2003}{1}-\frac{2003}{2}+\frac{2003}{2}-\frac{2003}{3}+\frac{2003}{3}-\frac{2003}{4}+...+\frac{2003}{2002}-\frac{2003}{2003}\)
\(=2003-1\)
\(=2002\)
Thanks bạn nha (Tuy thiếu câu 2)