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Ta có:
\(a=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{n}=\frac{39}{40}\)
Coi n=a.(a+1)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{a.\left(a+1\right)}\)
Ta thấy:
\(\frac{1}{1.2}=1-\frac{1}{2};\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3};...\)
\(\Rightarrow a=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{a}-\frac{1}{a+1}\)
\(=1+\frac{-1}{2}+\frac{1}{2}+\frac{-1}{3}+\frac{1}{3}+...+\frac{-1}{a}+\frac{1}{a}-\frac{1}{a+1}\)
\(=1+\left(\frac{-1}{2}+\frac{1}{2}\right)+\left(\frac{-1}{3}+\frac{1}{3}\right)+...-\frac{1}{a+1}\)
\(=1+0+0+...+0-\frac{1}{a+1}\)
\(\Rightarrow1-\frac{1}{a+1}=\frac{39}{40}\)
\(\Rightarrow a+1=40\Rightarrow a=39\)
\(\Rightarrow n=39.40=1560\)
Bài 1 :
a) A= (1;2;3;4;5)
b) B= ( 63;64;65;66;67;68;69;70)
Bài 2 :
a) 10x-5 = 11.5-10
10x-5 = 55-10
10x=45+5
10x=50
x=5
b) 27-3x=9.2-3
27-3x = 18-3
27-3x=15
3x=27-15
3x=12
x=4
c) 4x-15=12:12
4x-15=1
4x=16
x=4
d) 2+13x=14.2
13x=28-2
13x=26
x=2
a) \(10x-5=45\)
\(10x=40\)
\(x=4\)
b) \(27-3x=15\)
\(3x=27-15=12\)
\(x=\dfrac{12}{3}=4\)
c) \(4x-15=1\)
\(4x=16\)
\(x=\dfrac{16}{4}=4\)
d) \(2+13x=28\)
\(13x=26\)
\(x=\dfrac{26}{13}=2\)
A=1826+−527+−2286+1239+−32431826+−527+−2286+1239+−3243
A=913+−527+−1143+413+−3243913+−527+−1143+413+−3243
A=(913+413)+(−1143+−3243)+−527(913+413)+(−1143+−3243)+−527
A= 1+(-1)+−527−527
A=0+−527−527
A=−527−527
B=−1012+815+−1956+−318+2860−1012+815+−1956+−318+2860
B=−56+815+−1956+−16+715−56+815+−1956+−16+715
B=(−56+−16)+(815+715)+−1956(−56+−16)+(815+715)+−1956
B= -1+1+−1956−1956
B=0+−1956−1956
B=−1956
mình chỉ biết làm nhiêu đó thôi! Chúc bạn học tốt!
Bài 3:
Để A là số nguyên thì \(n-2+5⋮n-2\)
\(\Leftrightarrow n-2\in\left\{1;-1;5;-5\right\}\)
hay \(n\in\left\{3;1;7;-3\right\}\)
Bài 3:
\(a,\dfrac{x-1}{10}+\dfrac{x-1}{11}=\dfrac{x-1}{12}+\dfrac{x-1}{13}\)
\(\Rightarrow\dfrac{x-1}{10}+\dfrac{x-1}{11}-\dfrac{x-1}{12}-\dfrac{x-1}{13}=0\)
\(\Rightarrow\left(x-1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}-\dfrac{1}{12}-\dfrac{1}{13}\right)=0\)
Mà \(\dfrac{1}{10}+\dfrac{1}{11}-\dfrac{1}{12}-\dfrac{1}{13}\ne0\)
\(\Rightarrow x-1=0\Rightarrow x=1\)
Vậy x = 1
b, \(\dfrac{x-2000}{10}+\dfrac{x-1999}{9}=\dfrac{x-1998}{8}+\dfrac{x-1997}{7}\)
\(\Rightarrow\dfrac{x-2000}{10}+1+\dfrac{x-1999}{9}+1=\dfrac{x-1998}{8}+\dfrac{x-1997}{7}+1\)
\(\Rightarrow\dfrac{x-1990}{10}+\dfrac{x-1990}{9}-\dfrac{x-1990}{8}-\dfrac{x-1990}{7}=0\)
\(\Rightarrow\left(x-1990\right)\left(\dfrac{1}{10}+\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{7}\right)=0\)
Mà \(\dfrac{1}{10}+\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{7}\ne0\)
\(\Rightarrow x-1990=0\Rightarrow x=1990\)
1/3^2<1/2.3
1/4^2<1/3.4
1/5^2<1/4.5
…………...
1/25^2<1/24.25
=>A=1/3^2+1/4^2+1/5^2+…+1/25^2<1/2.3+1/3.4+1/4.5+…+1/24.25
=>A<23/50
Mà 11/39<23/50<12/25
=>11/39<A<12/25(đpcm)
Theo tôi A <23/50 chưa chắc đã nhỏ hơn 11/39.Xin giải thích.
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{n}=\frac{39}{40}\)
Đặt \(n=x\left(x+1\right)\);ta được :
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{x\left(x+1\right)}=\frac{39}{40}\)
\(\Leftrightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{39}{40}\)
\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{39}{40}\)
\(\Leftrightarrow1-\frac{1}{x+1}=\frac{39}{40}\)
\(\Leftrightarrow\frac{1}{x+1}=1-\frac{39}{40}=\frac{1}{40}\)
\(\text{Vậy }:x+1=40\Rightarrow x=39\)
\(\Rightarrow n=39.\left(39+1\right)=39.40=1560\)
tớ khuyên bạn nên lấy máy tính bỏ túi ra mà giải
tich nha bạn