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Mình giải thử xem đúng ko nha:
M=1/22+1/32+1/42+....+1/202
=1/2.2+1/3.3+1/4.4+.....+1/20.20
mà M<1/1.2+1/2.3+1/3.4+....+1/19.20
M<1/1-1/2+1/2-1/3+...+1/19-1/20
M<1-1/20
M<19/20<1
=> M<1
B1: để x là số nguyên thì: 5 chia hết cho 2x+1
=> \(2x+1\in U\left(5\right)\)
+> \(2x+1\in\left\{1;-1;5;-5\right\}\)
=> \(x\in\left\{0;-1;2;-3\right\}\)
a, (sửa đề )
\(1+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+.....+\frac{1}{x.\left(x+1\right)}=\frac{1999}{2000}\)
=\(1+\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{x.\left(x+1\right)}\right)=\frac{1999}{2000}\)
=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{x+\left(x+1\right)}=1-\frac{1999}{2000}=\frac{1}{2000}\)
=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{2000}\)
=\(\frac{1}{1}-\frac{1}{x+1}=\frac{1}{2000}\)
=\(\frac{1}{x+1}=\frac{1}{1}-\frac{1}{2000}=\frac{1999}{2000}\)
=> \(x+1=1:\frac{1999}{2000}=\frac{2000}{1999}\)
=>\(x=\frac{2000}{1999}-1=\frac{1}{1999}\)
Vậy x ∈{ \(\frac{1}{1999}\)}
b, \(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+.....+\frac{2}{x+\left(x+1\right)}=\frac{2}{9}\)
=> \(\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+.....+\frac{2}{x+\left(x+1\right)}=\frac{2}{9}\)
=>\(\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+.....+\frac{2}{x+\left(x+1\right)}=\frac{2}{9}\)
=>2.(\(\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+....+\frac{1}{x.\left(x+1\right)}\))=\(\frac{2}{9}\)
=>\(\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+....+\frac{1}{x+\left(x+1\right)}=\frac{2}{9}:2=\frac{1}{9}\)
=>\(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{9}\)
=>\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
=>\(\frac{1}{x+1}=\frac{1}{6}-\frac{1}{9}=\frac{1}{18}\)
=>\(x+1=18\)
=>\(x=18-1=17\)
=>x∈{17}
1, x - [42 + (-28)] = -8
=> x - 14 = -8
=> x = -8 + 14
=> x = 6
Vậy x = 6
3, 4x = 64
=> 4x = 43
=> x = 3 (vì cùng cơ số 4 > 1)
Vậy x = 3
4, 9x - 1 = 9
=> 9x - 1 = 91
=> x - 1 = 1 (vì cùng cơ sộ̣́ 9 > 1)
=> x = 1 + 1
=> x = 2
Vậy x = 2
