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a,
13[x-9] = 169
=> x - 9 = 169/13
=> x - 9 = 13
=> x = 13+9
=> x = 22
b,
Viết lại đề:
7x+3 = 343
<=> 7x+3 = 73
=> x + 3 = 3
=> x = 3-3
=> x = 0
c,
230 + [16 + [x-5]] = 315 . 23
=> 230 + [16 + x - 5] = 315 . 8
=> 230 + 16 + x - 5 = 2520
=> 230 + 16 + x = 2520 + 5 = 2525
=> x = 2525 - 230 - 16 = 2279
d,
13.x - 32.x = 20171 - 12018
=> 13x - 9x = 2017 - 1
=> 4x = 2016
=> x = 504
a) 13 ( x-9 )=169
=> x-9 =169 : 13 =13
=> x=13+9 =22
b)\(7^{x+3}=343\)
\(7^x.7^3=343\)
\(7^x=343:7^3\)
\(7^x=1\Rightarrow x=1\)
c)230 + 16 +x -5 =315.8
241 +x =2520
x=2520-241=2279
d) 13x -\(3^2.x\)=2017-1
x(13-9)=2016
x.4=2016
x=2016:4
x=504
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
\(E=1-2+3-4+5-6+...2015-2016-2017\)
\(=\left(1-2\right)+\left(3-4\right)+\left(5-6\right)+...\left(2015-2016\right)-2017\)
Mỗi nhóm có kết quả = -1.
\(=>\left[\left(2016-1\right)+1\right]:2=1008\)
\(=-1.1008+2017=-1008+2017=1009\)