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\(B=\frac{2^3}{3.5}+\frac{2^3}{5.7}+....+\frac{2^3}{101.103}\)
\(\Rightarrow\frac{1}{2^2}.B=\frac{2}{3.5}+\frac{2}{4.7}+....+\frac{2}{101.103}\)
\(\Rightarrow\frac{1}{4}.B=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{101}-\frac{1}{103}\)
\(\Rightarrow\frac{1}{4}.B=\frac{1}{3}-\frac{1}{103}=\frac{100}{309}\)
\(\Rightarrow B=\frac{100}{309}:\frac{1}{4}=\frac{400}{309}\)
\(=2^2\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{101.103}\right)\)
\(=4\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{101}-\frac{1}{103}\right)\)
\(=4\left(\frac{1}{3}-\frac{1}{103}\right)\)
\(=4\cdot\frac{100}{309}=\frac{400}{309}\)
\(A=1-\frac{2}{3.5}-\frac{2}{5.7}-\frac{2}{7.9}-...-\frac{2}{63.65}\)
\(A=1-\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{63-65}\right)\)
\(A=1-\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{63}-\frac{1}{65}\right)\)
\(A=1-\left(\frac{1}{3}-\frac{1}{65}\right)\)
\(A=1-\frac{62}{195}\)
\(A=\frac{133}{195}\)
a)Ta có:
\(A=4\frac{25}{16}+25\left(\frac{9}{16}:\frac{125}{64}\right):\frac{-27}{8}\)
\(\Rightarrow A=\frac{89}{16}+25.\frac{36}{125}:\frac{-27}{8}\)
\(\Rightarrow A=\frac{89}{16}+\frac{36}{5}:\frac{-27}{8}\)
\(\Rightarrow A=\frac{89}{16}+\frac{-32}{15}\)
\(\Rightarrow A=\frac{823}{240}\)
Vậy A=.....
b)Ta có:
\(C=\frac{2^3}{3.5}+\frac{2^3}{5.7}+\frac{2^3}{7.9}+...+\frac{2^3}{101.103}\)
\(\Rightarrow C=\frac{2^2.2}{3.5}+\frac{2^2.2}{5.7}+\frac{2^2.2}{7.9}+...+\frac{2^2.2}{101.103}\)
\(\Rightarrow C=2^2\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{101.103}\right)\)
\(\Rightarrow C=4\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{101}-\frac{1}{103}\right)\)
\(\Rightarrow C=4\left(\frac{1}{3}-\frac{1}{103}\right)\)
\(\Rightarrow C=4.\frac{100}{309}\)
\(\Rightarrow C=\frac{400}{309}\)
Vậy C=.....
\(\dfrac{2^3}{3\cdot5}+\dfrac{2^3}{5\cdot7}+...+\dfrac{2^3}{101\cdot103}\)
\(=2^2\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{101\cdot103}\right)\)
\(=4\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{101}-\dfrac{1}{103}\right)\)
\(=4\cdot\left(\dfrac{1}{3}-\dfrac{1}{103}\right)\)
\(=4\cdot\dfrac{100}{309}=\dfrac{400}{309}\)
\(\Leftrightarrow N=\frac{\left(2.3.4....50\right)\left(2.3.4...........50\right)}{\left(1.2.3.........49\right)\left(3.4.5...........51\right)}=\frac{50.2}{51}=\frac{100}{51}\)
\(\frac{2^2}{1.3}+\frac{3^2}{2.4}+\frac{4^2}{3.5}+....+\frac{50^2}{49.51}\)
\(=\frac{2^2-1}{1.3}+\frac{3^2-1}{2.4}+....+\frac{50^2-1}{49.51}+\frac{1}{1.3}+\frac{1}{2.4}+....+\frac{1}{49.51}\)
\(=\frac{1}{2}.\left(1+1+...+1\right)+\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{49}-\frac{1}{51}\)
Tự làm tiếp :))
tớ nhầm đoạn này tí :((
\(=\left(1+1+....+1\right)+\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{51}\right)\)(49 chữ số 1)
\(=49+\frac{1}{2}.\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}\right)-\left(\frac{1}{3}+\frac{1}{4}+...+\frac{1}{51}\right)\right]\)
\(=49+\left(\frac{3}{2}-\frac{1}{50}-\frac{1}{51}\right):2\)Tự tính
\(\frac{2^7.9^3}{6^5.8^2}=\frac{2^7.\left(3^2\right)^3}{\left(2.3\right)^5.\left(2^3\right)^2}=\frac{2^7.3^6}{2^5.3^5.2^6}=\frac{2^7.3^6}{2^{11}.3^5}=\frac{3}{2^4}=\)\(\frac{3}{16}\)
\(\frac{2^7.9^3}{6^5.8^2}=\frac{2^7.\left(3^2\right)^3}{\left(3.2\right)^5.\left(2^3\right)^2}=\frac{2^7.3^6}{3^5.2^5.2^6}=\frac{2^7.3^5.3}{3^5.2^{11}}=\frac{2^7.3}{2^7.2^4}=\frac{3}{2^4}=\frac{3}{16}\)
Bài này cx là BTVN của mk, mk làm giống vậy đấy