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cách này ngon hơn nè
\(2A=2\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\right)\)
\(2A=2+1+\frac{1}{2}+...+\frac{1}{2^{2011}}\)
\(2A-A=\left(2+1+\frac{1}{2}+...+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\right)\)
\(A=2-\frac{1}{2^{2012}}\)

a, Ta có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{2017^2}< \frac{1}{2016.2017}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2017^2}>\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2016.2017}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2016}-\frac{1}{2017}=1-\frac{1}{2017}< 1\)Vậy...
b, Đặt A = \(\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+...+\frac{1}{10000}\)
\(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)
\(A=\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)
Đặt B = \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)
Ta có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};.....;\frac{1}{50^2}< \frac{1}{49.50}\)
\(\Rightarrow B< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}=1-\frac{1}{50}< 1\)
Thay B vào A ta được:
\(A< \frac{1}{4}\left(1+1\right)=\frac{1}{4}.2=\frac{1}{2}\)
Vậy....
c, Ta có: \(\frac{1}{2^2}>\frac{1}{2.3};\frac{1}{3^2}>\frac{1}{3.4};....;\frac{1}{9^2}>\frac{1}{9.10}\)
\(\Rightarrow A>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)(1)
Lại có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};....;\frac{1}{9^2}< \frac{1}{8.9}\)
\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{8.9}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}=1-\frac{1}{9}=\frac{8}{9}\)(2)
Từ (1) và (2) suy ra \(\frac{2}{5}< A< \frac{8}{9}\)(đpcm)
d, chắc là đề sai
e, giống câu a

Lời giải chi tiết
12 1 13
12 – 02 (0 + 1)2
02 +12
22 1 + 3 23
32 – 12 (1 + 2)2
12 + 22
32 1 + 3 + 5 33
62 – 32 (2 + 3)2
22 + 32
43 102 – 62
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1.\(\dfrac{-1}{4}\) \(-\) x \(=\dfrac{-3}{2}\)
\(\Leftrightarrow\) x \(\dfrac{-3}{2}-\dfrac{-1}{4}\)
\(\Leftrightarrow\) x \(=\dfrac{-5}{4}\)
1. -1/4 - x = -3/2
x = -1/4 - -3/2
x = -1/4 - -6/4
x = 5/4
2. 3/5 . (x - 1/2) : 3/5 - 1/3 = -1/2
3/5 . (x - 1/2) : 3/5 = -1/2 + 1/3
3/5 . (x - 1/2) : 3/5 = -3/6 + 2/6
3/5 . (x - 1/2) : 3/5 = -1/6
3/5 . (x - 1/2) = -1/6 . 3/5
3/5 . (x - 1/2) = -1/10
x - 1/2 = -1/10 : 3/5
x - 1/2 = -1/6
x = -1/6 + 1/2
x = -1/6 + 3/6
x = 2/6
x = 1/3
Mấy câu sau bn tự làm nha!

b)Ta có:\(A=1+\frac{1}{2.\left(1+2\right)}+\frac{1}{3.\left(1+2+3\right)}+...+\frac{1}{16.\left(1+2+3+...+16\right)}\)
\(=1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+...+\frac{1}{16}.\left(1+2+3+...+16\right)\)
\(=1+\frac{1}{2}.3+\frac{1}{3}.6+...+\frac{1}{16}.136\)
\(=1+1,5+2+...+8,5\)
\(=\frac{\left(8,5+1\right).\left[\left(8,5-1\right):0,5+1\right]}{2}=76\)
B=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}<\)
B=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
B=\(1-\frac{1}{8}=\frac{8}{8}-\frac{7}{8}=\frac{1}{8}<2\)
Vậy 1/8<2 hay 1/8<16/8

Ta có : \(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+.......+\frac{1}{2^{100}}\)
\(\Leftrightarrow2A=2+1+\frac{1}{2}+\frac{1}{2^2}+......+\frac{1}{2^{99}}\)
\(\Leftrightarrow2A-A=2-\frac{1}{2^{99}}\)
\(\Leftrightarrow A=2-\frac{1}{2^{99}}\)
B tương tự

a)Ta có: 22>1.2⇒\(\frac{1}{2^2}< \frac{1}{1.2}\)
32>2.3⇒\(\frac{1}{3^2}< \frac{1}{2.3}\)
... 1002>99.100 ⇒ \(\frac{1}{100^2}< \frac{1}{99.100}\)
VT < \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)\(=1-\frac{1}{100}< 1\)(ĐPCM)
=3 nha
1+2=3
k nha