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\(\left(1+\frac{1}{4}\right).\left(1+\frac{1}{8}\right).\left(1+\frac{1}{15}\right).\left(1+\frac{1}{24}\right)...\left(1+\frac{1}{9999}\right)\)
\(=\frac{5}{4}.\frac{9}{8}.\frac{16}{15}.\frac{25}{24}...\frac{10000}{9999}=\frac{5.9.16.25...10000}{4.8.15.24...9999}=\frac{5.3^2.4^2.5^2...100^2}{4.2.4.3.5.4.6...99.101}\)
\(=\frac{5.3.4.5...100.3.4.5...100}{4.2.3.4...99.4.5.6...101}=\frac{5.100.3}{4.2.101}=\frac{5.25.3}{2.101}=\frac{375}{202}.\)
a) 4/ 3x7 + 4/7x11+ 4/11x15+...+ 4/107x111
=1/3-1/7+ 1/7-1/11+ 1/11- 1/15+...+1/107 - 1/111
= 1/3-1/111
=12/37
\(b,\frac{3^2}{8\cdot11}+\frac{3^2}{11\cdot14}+\frac{3^2}{14\cdot17}+...+\frac{3^2}{197\cdot200}\)
\(=3\left(\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+\frac{3}{14\cdot17}+...+\frac{3}{197\cdot200}\right)\)
\(=3\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+...+\frac{1}{197}-\frac{1}{200}\right)\)
\(=3\left(\frac{1}{8}-\frac{1}{200}\right)\)
\(=3\cdot\frac{3}{25}=\frac{9}{25}\)
a) \(B=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{302\cdot305}\)
\(B=\frac{1}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{302\cdot305}\right)\)
\(B=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{302}-\frac{1}{305}\right)\)
\(B=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{305}\right)=\frac{1}{3}\cdot\frac{303}{610}=\frac{101}{610}\)
b) \(C=\frac{6}{1\cdot4}+\frac{6}{4\cdot7}+....+\frac{6}{202\cdot205}\)
\(C=2\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{202\cdot205}\right)=2\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{202}-\frac{1}{205}\right)\)
\(=2\left(1-\frac{1}{205}\right)=2\cdot\frac{204}{205}=\frac{408}{205}\)
c) \(D=\frac{5^2}{1\cdot6}+\frac{5^2}{6\cdot11}+...+\frac{5^2}{266\cdot271}\)
\(D=5\left(\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+...+\frac{5}{266\cdot271}\right)\)
\(D=5\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{266}-\frac{1}{271}\right)=5\left(1-\frac{1}{271}\right)=5\cdot\frac{270}{271}=\frac{1350}{271}\)
d) \(E=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{5}{16}\cdot...\cdot\frac{9999}{10000}=\frac{3\cdot8\cdot15\cdot...\cdot9999}{4\cdot9\cdot16\cdot...\cdot10000}=\frac{3}{10000}\)
e) \(F=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{50^2}\right)\)
\(F=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right)...\left(1-\frac{1}{2500}\right)\)
\(F=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot...\cdot\frac{2499}{2500}=\frac{3\cdot8\cdot15\cdot...\cdot2499}{4\cdot9\cdot16\cdot...\cdot2500}=\frac{3}{2500}\)
a. \(B=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{302.305}\)
\(\Rightarrow3B=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{302.305}\)
\(\Rightarrow3B=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{302}-\frac{1}{305}\)
\(\Rightarrow3B=\frac{1}{2}-\frac{1}{305}\)
\(\Rightarrow3B=\frac{303}{610}\)
\(\Rightarrow B=\frac{101}{610}\)
b. \(C=\frac{6}{1.4}+\frac{6}{4.7}+...+\frac{6}{202.205}\)
\(\Rightarrow\frac{1}{2}C=\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{202.205}\)
\(\Rightarrow\frac{1}{2}C=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{202}-\frac{1}{205}\)
\(\Rightarrow\frac{1}{2}C=1-\frac{1}{205}\)
\(\Rightarrow\frac{1}{2}C=\frac{204}{205}\)
\(\Rightarrow C=\frac{408}{205}\)
c. \(D=\frac{5^2}{1.6}+\frac{5^2}{6.11}+...+\frac{5^2}{266.271}\)
\(\Rightarrow\frac{1}{5}D=\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{266.271}\)
\(\Rightarrow\frac{1}{5}D=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{266}-\frac{1}{271}\)
\(\Rightarrow\frac{1}{5}D=1-\frac{1}{271}\)
\(\Rightarrow\frac{1}{5}D=\frac{270}{271}\)
\(\Rightarrow D=\frac{1350}{271}\)
- A ở trên giữa các phân số là dấu " + " nha mấy bạn !
\(A=\frac{1}{2.2}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\)
\(A=\frac{1}{4}+\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)\)
\(A=\frac{1}{4}+\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\)
( gạch bỏ các phân số giống nhau)
\(A=\frac{1}{4}+\left(\frac{1}{3}-\frac{1}{9}\right)\)
\(A=\frac{1}{4}+\frac{2}{9}\)
\(A=\frac{17}{36}\)
phần b, c bn lm tương tự như phần a nha
🔹 Bài 1a
\(\frac{4}{5} : \left(\right. \frac{7}{13} - \frac{1}{3} \left.\right) \times \frac{13}{2}\)
Bước 1: Tính trong ngoặc
\(\frac{7}{13} - \frac{1}{3} = \frac{21 - 13}{39} = \frac{8}{39}\)
Bước 2: Thực hiện phép chia
\(\frac{4}{5} : \frac{8}{39} = \frac{4}{5} \times \frac{39}{8}\)
Rút gọn:
- \(4\) và \(8\) → còn \(1\) và \(2\)
\(= \frac{39}{10}\)
Bước 3: Nhân tiếp
\(\frac{39}{10} \times \frac{13}{2} = \frac{507}{20}\)
✅ Kết quả:
\(\boxed{\frac{507}{20}}\)
🔹 Bài 2
a)
\(12 : x = 2015 - 2011 = 4\) \(12 : x = 4 \Rightarrow x = 3\)
b)
\(13 - x = \frac{15}{4} - 0,75\)
👉 \(0,75 = \frac{3}{4}\)
\(= \frac{15}{4} - \frac{3}{4} = 3\) \(13 - x = 3 \Rightarrow x = 10\)
c)
\(6,37 x + 3,63 x = 14\) \(\left(\right. 6,37 + 3,63 \left.\right) x = 10 x = 14 \Rightarrow x = 1,4\)
d)
\(\frac{1}{2} + \frac{3}{4} x = \frac{3}{2}\)
Bước 1:
\(\frac{3}{4} x = \frac{3}{2} - \frac{1}{2} = 1\)
Bước 2:
\(x = \frac{1}{\frac{3}{4}} = \frac{4}{3}\)
✅ Kết quả bài 2:
- a) \(x = 3\)
- b) \(x = 10\)
- c) \(x = 1,4\)
- d) \(x = \frac{4}{3}\)
🔹 Bài 3
a) Cách vẽ hình
- Vẽ hình vuông \(A B C D\)
- Lấy điểm \(E\) trên cạnh \(B C\)
- Nối \(A\) với \(E\) → được tam giác \(A E D\)
- Từ \(E\) hạ đường vuông góc xuống \(A D\), cắt tại \(M\)
- Khi đó \(M E C D\) là hình chữ nhật
b) Tính diện tích tam giác \(A E D\)
👉 Diện tích hình vuông:
\(36 \Rightarrow c ạ n h = 6 \&\text{nbsp};(\text{cm})\)
👉 Nhận xét quan trọng (kiểu lớp 6):
- Tam giác \(A E D\) có đáy \(A D = 6\)
- Chiều cao chính là \(E M\)
- Mà \(E M = D C = 6\) (do tạo hình chữ nhật)
Diện tích:
\(S_{A E D} = \frac{1}{2} \times 6 \times 6 = 18 \&\text{nbsp};\text{cm}^{2}\)
✅ Kết luận:
\(\boxed{18 \&\text{nbsp};\text{cm}^{2}}\)
\(a,A=\frac{1}{25\cdot27}+\frac{1}{27\cdot29}+...+\frac{1}{73\cdot75}\)
\(A=\frac{1}{2}\left[\frac{2}{25\cdot27}+\frac{2}{27\cdot29}+...+\frac{2}{73\cdot75}\right]\)
\(A=\frac{1}{2}\left[\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\right]\)
\(A=\frac{1}{2}\left[\frac{1}{25}-\frac{1}{75}\right]=\frac{1}{2}\cdot\frac{2}{75}=\frac{1}{75}\)
\(b,B=\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+\frac{1}{14\cdot17}+...+\frac{1}{197\cdot200}\)
\(3B=\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+\frac{3}{14\cdot17}+...+\frac{3}{197\cdot200}\)
\(3B=\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200}\)
\(3B=\frac{1}{8}-\frac{1}{200}\)
\(3B=\frac{3}{25}\)
\(B=\frac{3}{25}:3=\frac{1}{25}\)
#)Giải :
a, \(A=\frac{1}{25.27}+\frac{1}{27.29}+...+\frac{1}{73.75}\)
\(A=\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\)
\(A=\frac{1}{25}-\frac{1}{75}\)
\(A=\frac{2}{75}\)
b, \(B=\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+...+\frac{1}{197.200}\)
\(B=\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+...+\frac{1}{197}-\frac{1}{200}\)
\(B=\frac{1}{8}-\frac{1}{200}\)
\(B=\frac{3}{25}\)
#~Will~be~Pens~#
\(A=\frac{1}{25.27}+\frac{1}{27.29}+...+\frac{1}{73.75}\)
\(2A=\frac{2}{25.27}+\frac{2}{27.29}+...+\frac{2}{73.75}\)
\(2A=\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\)
\(2A=\frac{1}{25}-\frac{1}{75}\)
\(2A=\frac{2}{75}\)
\(A=\frac{2}{75}:2=\frac{1}{75}\)
\(B=\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+...+\frac{1}{197.200}\)
\(3B=\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+...+\frac{3}{197.200}\)
\(3B=\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+...+\frac{1}{197}-\frac{1}{200}\)
\(3B=\frac{1}{8}-\frac{1}{200}\)
\(3B=\frac{3}{25}\)
\(B=\frac{3}{25}:3=\frac{1}{25}\)
#)Ui chết sorry bn, mk ghi thiếu nhé @@
ph là 2A và 3B ms đúng :v
Mấy bn kia lm đúng rùi !
a) \(2A=\frac{2}{25.27}+\frac{2}{27.29}+...+\frac{2}{73.75}=\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\)
\(\Leftrightarrow2A=\frac{1}{25}-\frac{1}{75}=\frac{2}{75}\Rightarrow A=\frac{1}{75}.\)
b)\(3B=\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+...+\frac{3}{197.200}=\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200}\)
\(\Leftrightarrow3B=\frac{1}{8}-\frac{1}{200}=\frac{3}{25}\Leftrightarrow B=\frac{1}{25}.\)
A = \(\frac{1}{25.27}+\frac{1}{27.29}+...+\)\(\frac{1}{73.75}\)
2A = \(\frac{2}{25.27}+\frac{2}{27.29}+...+\frac{2}{73.75}\)
2A = \(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\)
2A = \(\frac{1}{25}-\frac{1}{75}\)
2A = \(\frac{3}{75}-\frac{1}{75}\)
=>A = \(\frac{2}{75}:2\)=\(\frac{2}{75}.\frac{1}{2}\)
=> A =\(\frac{1}{75}\)
B = \(\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+...+\frac{1}{197.200}\)
3B = \(\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+...+\frac{3}{197.200}\)
3B = \(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+...+\frac{1}{197}-\frac{1}{200}\)
3B = \(\frac{1}{8}-\frac{1}{200}\)
=> 3B = \(\frac{25}{200}-\frac{1}{200}\)
=> B = \(\frac{24}{200}.\frac{1}{3}\)
=> B = \(\frac{1}{25}\)
#Học'Tốt#
\(A=\frac{1}{25.27}+\frac{1}{27.29}+...+\frac{1}{73.75}\)
\(2A=\frac{2}{25.27}+\frac{2}{27.29}+...+\frac{2}{73.75}\)
\(2A=\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\)
\(2A=\frac{1}{25}+\left(\frac{1}{27}-\frac{1}{27}\right)+\left(\frac{1}{29}-\frac{1}{29}\right)+...+\left(\frac{1}{73}-\frac{1}{73}\right)-\frac{1}{75}\)
\(2A=\frac{1}{25}-\frac{1}{75}\)
\(2A=\frac{2}{75}\)
\(A=\frac{2}{75}:2\)
\(A=\frac{1}{75}\)
~ Hok tốt ~
\(B=\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+...+\frac{1}{197.200}\)
\(3B=\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+...+\frac{3}{197.200}\)
\(3B=\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+...+\frac{1}{197}-\frac{1}{200}\)
\(3B=\frac{1}{8}+\left(\frac{1}{11}-\frac{1}{11}\right)+\left(\frac{1}{14}-\frac{1}{14}\right)+...+\left(\frac{1}{197}-\frac{1}{197}\right)-\frac{1}{200}\)
\(3B=\frac{1}{8}-\frac{1}{200}\)
\(3B=\frac{3}{25}\)
\(B=\frac{3}{25}:3\)
\(B=\frac{1}{25}\)
~ Hok tốt ~