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\(\frac{1}{1.3}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{n\left(n+3\right)}=\frac{2018}{6057}\)
\(\Rightarrow\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{n\left(n+3\right)}\right)=\frac{2018}{6057}\)
\(\Rightarrow\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}=\frac{2018}{6057}.3\)
\(\Rightarrow1-\frac{1}{n+3}=\frac{2018}{2019}\)
\(\Rightarrow\frac{1}{n+3}=1-\frac{2018}{2019}\)
\(\Rightarrow\frac{1}{n+3}=\frac{1}{2019}\)
\(\Rightarrow n+3=2019\)
\(\Rightarrow n=2016\)
Vậy n = 2016
ta có 3A = 3/1.4 + 3/4.7 + ... + 3/(3n-2).(3n+1)
3A = 1-1/4 + 1/4 - 1/7 +....+ 1/(3n-2) - 1/(3n+1)
3A = 1- 1/(3n+1)
Mà 1/(3n+1) > 0 suy ra 3A < 1 suy ra A<1/3
tk giúp mình nha
Cộng 1 vào từng phân số ta sẽ đc
\(\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}=\frac{x+100}{101}+\frac{x+100}{102}+\frac{x+100}{103}\)
\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}-\frac{1}{101}-\frac{1}{102}-\frac{1}{103}\right)=0\)
\(\Rightarrow x=-100\)
\(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}=\frac{x-1}{101}+\frac{x-2}{102}+\frac{x-3}{103}\)
<=> \(\frac{x+1}{99}+1+\frac{x+2}{98}+1+\frac{x+3}{97}+1=\frac{x-1}{101}+1+\frac{x-2}{102}+1+\frac{x-3}{103}+1\)
<=> \(\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}=\frac{x+100}{101}+\frac{x+100}{102}+\frac{x+100}{103}\)
<=> \(\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}-\frac{1}{101}-\frac{1}{102}-\frac{1}{103}\right)=0\)
<=> x + 100 = 0 (vì \(\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}-\frac{1}{101}-\frac{1}{102}-\frac{1}{103}\right)\ne0\))
<=> x = -100
\(\frac{x-5}{100}+\frac{x-4}{101}+\frac{x-3}{102}=\frac{x-100}{5}+\frac{x-101}{4}+\frac{x-102}{3}\)
<=> \(\frac{x-5}{100}-1+\frac{x-4}{101}-1+\frac{x-3}{102}-1=\frac{x-100}{5}-1+\frac{x-101}{4}-1+\frac{x-102}{3}-1\)
<=> \(\frac{x-105}{100}+\frac{x-105}{101}+\frac{x-105}{102}=\frac{x-105}{5}+\frac{x-105}{4}+\frac{x-105}{3}\)
<=> \(\left(x-105\right)\left(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\right)=0\)
Nhận thấy: \(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}+\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\ne0\)
=> \(x-105=0\)
<=> \(x=105\)
\(\frac{x-5}{100}+\frac{x-4}{101}+\frac{x-3}{102}=\frac{x-100}{5}+\frac{x-101}{4}+\frac{x-102}{3}\)
\(\Leftrightarrow\frac{x-5}{100}+\frac{x-4}{101}+\frac{x-3}{102}-\frac{x-100}{5}-\frac{x-101}{4}-\frac{x-102}{3}=0\)
\(\Leftrightarrow\left(\frac{x-5}{100}-1\right)+\left(\frac{x-4}{101}-1\right)+\left(\frac{x-3}{102}-1\right)-\left(\frac{x-100}{5}-1\right)-\left(\frac{x-101}{4}-1\right)-\left(\frac{x-102}{3}-1\right)=0\)
\(\Leftrightarrow\frac{x-105}{100}+\frac{x-105}{101}+\frac{x-105}{102}-\frac{x-105}{5}-\frac{x-105}{4}-\frac{x-105}{3}=0\)
\(\Leftrightarrow\left(x-105\right)\left(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\right)=0\)
\(\Leftrightarrow x-105=0\left(Vì\frac{1}{100}+\frac{1}{101}+\frac{1}{102}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\ne0\right)\)
\(\Leftrightarrow x=105\)
+ \(n\left(n+3\right)+2\) \(=n^2+3n+2\)
\(=n^2+2n+n+2=\left(n+1\right)\left(n+2\right)\)
\(A=\frac{1\cdot4+2}{1\cdot4}\cdot\frac{2\cdot5+2}{2\cdot5}\cdot...\cdot\frac{2019\cdot2022+2}{2019\cdot2022}\)
\(=\frac{2\cdot3}{1\cdot4}\cdot\frac{3\cdot4}{2\cdot5}\cdot...\cdot\frac{2020\cdot2021}{2019\cdot2022}\)
\(=\frac{2\cdot3\cdot..\cdot2020}{1\cdot2\cdot...\cdot2019}\cdot\frac{3\cdot4\cdot...\cdot2021}{4\cdot5\cdot...\cdot2022}\)
\(=2020\cdot\frac{3}{2022}=\frac{1010}{337}\)
a) <=>(x - 3/4)(x-3/4 +x-1/2)=0
<=>(x-3/4)(2x-5/4)=0
<=>x-3/4=0 hoặc 2x-5/4=0
<=>x=3/4 hoặc x=5/8
Vậy tập nghiệm của phương trình trên là S={3/4;5/8}
b)<=>140x/35 - 7(4x-3)/35 - 10(x+3)/35=0
<=>140x-28x+21-10x-30=0
<=>102x=9
<=>x=3/34
Vậy tập nghiệm của phương trình trên là S={3/34}
=> 3x/4+3/4.7+3/7.10+...+3/100.103=306/103(nhân cả 2 vế của đt lên 2)
=>3x/4+(1/4-1/7)+(1/7-1/10)+...+(1/100-1/103)=306/103
=>3x/4+1/4-1/103+=306/103
=>3x/4+99/412=306/103
=>3x/4=306/103-99/412=1125/412
=>x=1125/412:3/4
=>x=1125/309
( nếu thấy đúng thì tick cho mk nha