Bài 1:
Tính \(\dfrac{1}{99.97}-\dfrac{1}{99.95}-\dfrac{1}{95.93}-\dfrac{1}{5.3}-\dfrac{1}{3.1}=...\)
Bài 2: Cho tam giác vuông ABC vuông tại C, biết AC = 6cm; AB = 4cm. N là trung điểm của AB. Bình phương độ dài CN = ... cm.
Bài 3:
Cho \(\dfrac{x+16}{9}=\dfrac{y-25}{-16}=\dfrac{z+49}{25}\) và \(4x^2-3=29\). Gía trị biểu thức A = x+2y+3z là
Bài 1:
\(\dfrac{1}{99.97}-\dfrac{1}{97.95}-\dfrac{1}{95.93}-...-\dfrac{1}{5.3}-\dfrac{1}{3.1}\)
\(=\dfrac{1}{99.97}-\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{93.95}+\dfrac{1}{95.97}\right)\)
\(=\dfrac{1}{99.97}-\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{93.95}+\dfrac{2}{95.97}\right)\)
\(=\dfrac{1}{97.99}-\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{93}-\dfrac{1}{95}+\dfrac{1}{95}-\dfrac{1}{97}\right)\)
\(=\dfrac{1}{97.99}-\dfrac{1}{2}\left(1-\dfrac{1}{97}\right)\)
\(=\dfrac{1}{97.99}-\dfrac{1}{2}.\dfrac{96}{97}\)
\(=\dfrac{1}{97.99}-\dfrac{48}{97}\)
Bạn tính nốt nhé
Bài 2, 3 bạn kiểm tra lại đề giúp mk
Bài 1 :
\(\dfrac{1}{99.97}-\dfrac{1}{99.95}-\dfrac{1}{95.93}-......-\dfrac{1}{5.3}-\dfrac{1}{3.1}\)
\(=\dfrac{1}{97.99}-\left(\dfrac{1}{97.95}+\dfrac{1}{95.93}+...+\dfrac{1}{5.3}-\dfrac{1}{3.1}\right)\)
\(=\dfrac{1}{97.99}-\dfrac{1}{2}\left(\dfrac{1}{95}-\dfrac{1}{97}+\dfrac{1}{93}-\dfrac{1}{95}+...+\dfrac{1}{3}-\dfrac{1}{5}+1-\dfrac{1}{3}\right)\)
\(=\dfrac{1}{97.99}-\dfrac{1}{2}\left(1-\dfrac{1}{97}\right)\)
\(=\dfrac{1}{97.99}-\dfrac{48}{97}\)
\(=\dfrac{51}{97}\)