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a ) 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256
Đạt A = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256
A x 2 = 2 x ( 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256)
A x 2 = 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128
Lấy A x 2 - A ta có :
A x 2 - A = 1 + 1/2 + ..... + 1/128 - 1/2 + 1/4 + ........ + 1/256
A x ( 2 - 1 ) = 1 - 1/ 256
A = 255/256
b) 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729
Đặt A = 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729
A x 3 = 3 x ( 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729)
= 1 + 1/ 3 + 1/9 + 1/27 + 1/81 + 1/243
Lấy A x 3 - A ta có :
A x 3 - A = 1 + 1/3 + 1/9 +..... + 1/243 - 1/3 + 1/9 +........+ 1/243 + 1/29
A x ( 3 - 1 ) = 1 - 1/29
A x2 = 28/29
A = 28/29 : 2 ( tự tính
Đặt A = \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+.....+\frac{1}{256}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{128}\)
\(\Rightarrow2A-A=1-\frac{1}{256}\)
\(\Rightarrow A=\frac{255}{256}\)
Bài 3 :
b) Ta có 1+ 2 + 3 +4 + ...+ x =15
Nên \(\frac{x\left(x+1\right)}{2}=15\)
\(x\left(x+1\right)=30\)
=> \(x\left(x+1\right)=5.6\)
=> x = 5
Bài 2:
h; \(\dfrac{2}{3}\)\(x\) + 50% + \(x\) = \(\dfrac{1}{10}\)
\(\dfrac{2}{3}\)\(x\) + \(\dfrac{1}{2}\) + \(x\) = \(\dfrac{1}{10}\)
(\(\dfrac{2}{3}\)\(x\) + \(x\)) + \(\dfrac{1}{2}\) = \(\dfrac{1}{10}\)
\(x\) \(\times\) (\(\dfrac{2}{3}\) + 1) + \(\dfrac{1}{2}\) = \(\dfrac{1}{10}\)
\(x\) \(\times\) \(\dfrac{5}{3}\) + \(\dfrac{1}{2}\) = \(\dfrac{1}{10}\)
\(x\) \(\times\) \(\dfrac{5}{3}\) = \(\dfrac{1}{10}\) - \(\dfrac{1}{2}\)
\(x\) \(\times\) \(\dfrac{5}{3}\) = \(\dfrac{-2}{5}\)
\(x\) = \(\dfrac{-2}{5}\): \(\dfrac{5}{3}\)
\(x\) = - \(\dfrac{6}{25}\)
Lớp 5 chưa học số âm em nhé.
a) 2006/2005 = 1/2005 ; 2005/2006 =1/2006
Vì: 1/2005 > 1/2006 ; nên 2006/2005 < 2005/2006
b) 99/1000 = 990/10000
Vì: 990/10000 <999/10000 ; nên 99/1000 < 999/10000
c) 13/15 =1- 2/15 ;1333/1555 = 1- 222/555 => 1- 2/5
Vì: 2/15 < 2/5 ; nên 13/15 >1333/1555
d) 650650/480480 = 650/480 =65/48 ; 222222/144144 = 222/144
Vì: 65/48 < 222/144 ; nên 650650/480480 < 222222/144144
(câu a với câu c là tính theo kiểu phần bù các bạn nhớ nhé!)
Bài làm:
\(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right).y=\frac{2}{3}\)
\(\Leftrightarrow\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right).y=\frac{4}{3}\)
\(\Leftrightarrow\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+\frac{11-9}{9.11}\right).y=\frac{4}{3}\)
\(\Leftrightarrow\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right).y=\frac{4}{3}\)
\(\Leftrightarrow\left(1-\frac{1}{11}\right).y=\frac{4}{3}\)
\(\Leftrightarrow\frac{10}{11}.y=\frac{4}{3}\)
\(\Leftrightarrow y=\frac{4}{3}:\frac{10}{11}=\frac{4}{3}.\frac{11}{10}=\frac{22}{15}\)
Chú ý dấu \(\left(.\right)\)là dấu \(\left(\times\right)\)
Vậy \(y=\frac{22}{15}\)
\(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right).y=\frac{2}{3}\)
\(\Leftrightarrow\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right).y=\frac{2}{3}\)
\(\Leftrightarrow\frac{1}{2}.\left(1-\frac{1}{11}\right).y=\frac{2}{3}\)
\(\Leftrightarrow\frac{1}{2}.\frac{10}{11}.y=\frac{2}{3}\)
\(\Leftrightarrow\frac{5}{11}.y=\frac{2}{3}\)
\(\Rightarrow y=\frac{2}{3}:\frac{5}{11}=\frac{22}{15}\)
LƯU Ý:các dấu chấm(.) là dấu nhân ^^.
Bạn tham khảo theo đường link này nhé có 1 bài tương tự đó : https://olm.vn/hoi-dap/question/1042256.html
Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{64}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{64}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\right)\)
\(A=1-\frac{1}{128}=\frac{127}{128}\)
Ủng hộ mk nha ^_-
\(2\cdot A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{64}\)
\(A=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{64}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\right)\)
\(A=1-\frac{1}{128}=\frac{128}{128}-\frac{1}{128}=\frac{127}{128}\)