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\(C=\frac{5}{1.5}+\frac{5}{5.9}+\frac{5}{9.13}+...+\frac{5}{101.105}\)
\(C=5.\left(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{101.105}\right)\)
\(C=5.\frac{1}{4}.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+....+\frac{1}{101}-\frac{1}{105}\right)\)
\(C=\frac{5}{4}.\left(1-\frac{1}{105}\right)\)
\(C=\frac{5}{4}.\frac{104}{105}\)
\(C=\frac{26}{21}\)
a, \(\frac{1}{2}.B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(\frac{1}{2}.B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)
\(\frac{1}{2}.B=1-\frac{1}{101}=\frac{100}{101}\)
\(B=\frac{100}{101}.2=\frac{200}{101}\)
b, \(\frac{4}{5}.C=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{101.105}\)
\(\frac{4}{5}.C=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{101}-\frac{1}{105}\)
\(\frac{4}{5}.C=1-\frac{1}{105}=\frac{104}{105}\)
\(C=\frac{104}{105}.\frac{5}{4}=\frac{26}{21}\)
\(B=\frac{2}{2}\cdot\left(\frac{4}{1\cdot3}+\frac{4}{3\cdot5}+\frac{4}{5\cdot7}+....+\frac{4}{99\cdot101}\right)\)
\(=\frac{4}{2}\cdot\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{99\cdot101}\right)\)
\(=2\cdot\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=2\cdot\left(1-\frac{1}{101}\right)\)
\(=2\cdot\frac{100}{101}\)
\(=1\frac{99}{101}\)
\(A=3\times\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{101}-\frac{1}{105}\right)\)
\(A=3\times\left(1-\frac{1}{105}\right)\)
\(A=3\times\frac{104}{105}\)
\(A=\frac{104}{35}\)
Chỉ cần để các thừa số ra ngoài rồi nhân các số mà bằng khoảng cách của mẫu lên tử là giải được
\(4A=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{x.\left(x+4\right)}\)
\(4A=\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{x}-\frac{1}{x+4}\)
\(4A=1-\frac{1}{x+4}\)
\(4A=\frac{x+4-1}{x+4}\)
\(A=\frac{x+3}{\text{4(x+4)}}\)
Bạn tự thay rồi tính nhé
\(A=\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+........+\frac{1}{x\cdot\left(x+4\right)}\)
\(4A=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+........+\frac{4}{x\cdot\left(x+4\right)}\)
\(4A=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+.......+\frac{1}{x}-\frac{1}{x+4}\)
\(4A=1-\frac{1}{x+4}\)
\(A=\left(1-\frac{1}{x+4}\right):4\)
Khi x = 12 => \(A=\left(1-\frac{1}{12+4}\right):4\)
A = \(\left(1-\frac{1}{16}:4\right)\)
A = \(\frac{15}{16}:4=\frac{15}{64}\)
Khi x = 2 => \(A=\left(1-\frac{1}{2+4}\right):4\)
A = \(\left(1-\frac{1}{6}\right):4\)
A \(=\frac{5}{6}:4=\frac{5}{24}\)
Khi x = \(\frac{5}{6}\)=> \(A=\left(1-\frac{1}{\frac{5}{6}+4}\right):4\)
A = \(\left(1-\frac{1}{\frac{29}{6}}\right):4\)
A = \(\frac{23}{29}:4=\frac{23}{116}\)
a) \(=\frac{9}{1.4}+\frac{9}{4.7}+\frac{9}{7.10}+...+\frac{9}{61.64}\)
\(=3\left(\frac{1}{1}-\frac{1}{64}\right)\)
\(=\frac{189}{64}\)
b) \(=\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{21}-\frac{1}{25}\)
\(=\frac{1}{1}-\frac{1}{25}\)
\(=\frac{24}{25}\)
c) Chưa học tới
b)1/1.5+1/5.9+1/9.13+...+1/21.25
=1/4.(4/1.5+4/5.9+4/9.13+4/21.25)
=1/4.(4-4/5+4/5-4/9+4/9-4/13+...+4/21-4/25)
=1/4.(4-4/25)
=1/4.(100/25-4/25)
=1/4.96/25
=24/25
\(P=\frac{3}{1.5}+\frac{3}{5.9}+\frac{3}{9.13}+...+\frac{3}{197.201}\)
\(P=\frac{3}{4}.\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{197.201}\right)\)
\(P=\frac{3}{4}.\left(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}+\frac{1}{13}+...+\frac{1}{197}-\frac{1}{201}\right)\)
\(P=\frac{3}{4}.\left(\frac{1}{1}-\frac{1}{201}\right)\)
\(P=\frac{3}{4}.\left(\frac{201}{201}-\frac{1}{201}\right)\)
\(P=\frac{3}{4}.\frac{200}{201}\)
\(P=\frac{50}{67}\)
Vậy \(P=\frac{50}{67}\)
\(P=\frac{3}{1\cdot5}+\frac{3}{5\cdot9}+...+\frac{3}{197\cdot201}\)
\(=3\cdot\left(\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+...+\frac{1}{197\cdot201}\right)\)
\(=\frac{3}{4}\cdot\left(\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+...+\frac{4}{197\cdot201}\right)\)
\(=\frac{3}{4}\cdot\left(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{197}-\frac{1}{201}\right)\)
\(=\frac{3}{4}\cdot\left(\frac{1}{1}-\frac{1}{201}\right)\)
\(=\frac{3}{4}\cdot\left(\frac{201-1}{201}\right)\)
\(=\frac{3}{4}\cdot\frac{200}{201}\)
\(\Rightarrow B=\frac{50}{67}\)
\(A=\frac{12}{1.5}+\frac{12}{5.9}+\frac{12}{9.13}+.............+\frac{12}{101.105}\)
\(=3.\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+............+\frac{4}{101.105}\right)\)
\(=3\left(1-\frac{1}{105}\right)\)
\(=3.\frac{104}{105}=\frac{312}{105}\)
TẬP HỢP RA HAI NHÓM .MỘT NHÓM SỐ ÂM.CÒN NHÓM KIA LÀ SỐ DƯƠNG MÀ TÍNH
STUDY WELL
K NHA
MK XIN CẢM ƠN CÁC BẠN NHÌU
C = 24.7 −35.9 +27.10 −39.13 +...+2301.304 −3401.405
\(C=\left(\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{301.304}\right)-\left(\frac{3}{5.9}+\frac{3}{9.13}+...+\frac{3}{401.405}\right)\)
\(C=\frac{2}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{301}-\frac{1}{304}\right)-\frac{3}{4}\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{401}-\frac{1}{405}\right)\)
\(C=\frac{2}{3}\left(\frac{1}{4}-\frac{1}{304}\right)-\frac{3}{4}\left(\frac{1}{5}-\frac{1}{405}\right)\)
\(C=\frac{2}{3}.\frac{75}{304}-\frac{3}{4}.\frac{16}{81}\)
\(C=\frac{25}{152}-\frac{4}{27}\)
\(C=\frac{67}{4104}\)
Study well
\(\) Xét tổng \(S=\frac{2}{1 \cdot5}+\frac{2}{5 \cdot9}+\frac{2}{9 \cdot13}+\ldots+\frac{2}{101 \cdot105}\).
Mỗi số hạng có dạng \(\frac{2}{\left(\right. 4 n - 3 \left.\right) \left(\right. 4 n + 1 \left.\right)}\). Phân tích thành:
\(\frac{2}{\left(\right. 4 n - 3 \left.\right) \left(\right. 4 n + 1 \left.\right)} = \frac{1}{2} \left(\right. \frac{1}{4 n - 3} - \frac{1}{4 n + 1} \left.\right) .\)
Thay vào tổng:
\(S=\frac{1}{2}\left(\right.\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+..\ldots+\frac{1}{101}-\frac{1}{105}\left.\right).\)
Các số hạng giữa triệt tiêu, còn lại:
\(S = \frac{1}{2} \left(\right. 1 - \frac{1}{105} \left.\right) = \frac{1}{2} \cdot \frac{104}{105} = \frac{52}{105} .\)
Kết quả:
\(\boxed{\frac{52}{105}}\)
\(\frac{2}{1\cdot5}+\frac{2}{5\cdot9}+\cdots+\frac{2}{101\cdot105}\)
\(=\frac12\left(\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\cdots+\frac{4}{101\cdot105}\right)\)
\(=\frac12\left(1-\frac15+\frac15-\frac19+\cdots+\frac{1}{101}-\frac{1}{105}\right)\)
\(=\frac12\left(1-\frac{1}{105}\right)=\frac12\cdot\frac{104}{105}=\frac{52}{105}\)