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\(=\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x.\left(x+1\right)}=\frac{2018}{2019}\)
\(\Rightarrow2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{x.\left(x+1\right)}\right)=\frac{2018}{2019}\)
\(\Rightarrow2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2018}{2019}\)
\(\Rightarrow2.\left(1-\frac{1}{x+1}\right)=\frac{2018}{2019}\)
\(\Rightarrow1-\frac{1}{x+1}=\frac{2018}{2019.2}\)
Tự làm nốt
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)}=\frac{2019}{2021}\)
<=> \(2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2019}{2021}\)
<=> \(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2019}{2021}\)
<=> \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2019}{4042}\)
<=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2019}{2042}\)
<=> \(\frac{1}{x+1}=\frac{1}{2021}\)
<=> x + 1 = 2021
<=> x = 2020
Có phải là bình 6a3 học trường THCS Nguyễn Trãi đúng không
# Giải :
|x - 2| - 4 = 6
|x - 2| = 6 + 4
|x - 2| = 12
=> x - 2 = 12 hoặc x - 2 = -12
+) x - 2 = 12
=> x = 14
+) x - 2 = -12
=> x = 10
Vậy x = 14 hoặc x = 10
401 . ( x - 3 ) = 20052019 : 20052018
401 . (x - 3) = 2005
x - 3 = 2005 : 401
x - 3 = 5
x = 5 + 3
x = 8
Vậy x = 8
#By_Ami
1, ( x - 3 )100 - 1 = 0 => ( x - 3 )100 = 0 + 1 = 1
Mà 1 = 1100 => x - 3 = 1 => x = 1 + 3 = 4 hoặc 1 = (-1)100 => x - 3 = -1 => x = -1 + 3 = 2
2, ( 9 - 5x )2019 + 1 = 0 => ( 9 - 5x )2019 = 0 - 1 = -1
Mà -1 = (-1)2019 => 9 - 5x = -1 => 5x = 9 - ( -1 ) = 10 => x = 10 : 5 = 2
3, ( 4x - 3 )5 = ( 2 - x )5 => 4x - 3 = 2 - x
=> 4x + x = 3 + 2 => 5x = 5 => x = 5 : 5 = 1
4, ( 2x - 9 )2 = ( 5x - 6 )2 => 2x - 9 = 5x - 6 ... ( tự làm )
5, ( 11 - 4x )6 - ( 2 - 5x )6 = 0 => ( 11- 4x )6 = ( 2 - 5x )6
=> 11 - 4x = 2 - 5x _ Đến đây làm tương tự 2 câu trên
6, ( x - 9 )9 = ( x - 9 )7 mà cơ số bằng nhau ( = x - 9 )
=> x - 9 = 1 hoặc -1 vì 19 = 17 và ( -1 )9 = ( -1 )7
TH1: x - 9 = 1 => x = 1 + 9 = 10
TH2: x - 9 = -1 => x = -1 + 9 = 8
7, 8, 9 tương tự 6 ( kết quả của cơ số đều = 1 hoặc -1 )
Ta có : \(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{x\left(x+1\right)}=\frac{2018}{2019}\)
\(\Rightarrow\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+...+\left(\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2018}{2019}\)
\(\Rightarrow1-\frac{1}{x+1}=\frac{2018}{2019}\)
\(\Rightarrow\frac{1}{x+1}=1-\frac{2018}{2019}=\frac{1}{2019}\)
\(\Rightarrow x+1=2019\)
\(\Rightarrow x=2018\)
Vậy x = 2018
Nhớ t.i.c.k cho mình nha!
Chỗ \(x(x+1)\Rightarrow\frac{1}{x(x+1)}\) nhé
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x(x+1)}=\frac{2018}{2019}\)
\(\Rightarrow\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x(x+1)}=\frac{2018}{2019}\)
\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2018}{2019}\)
\(\Rightarrow1-\frac{1}{x+1}=\frac{2018}{2019}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2019}\Leftrightarrow x+1=2019\Leftrightarrow x=2018\)
\(x\cdot\left(6-x\right)^{2019}=\left(6-x\right)^{2019}\\ \Leftrightarrow x\cdot\left(6-x\right)^{2019}-\left(6-x\right)^{2019}=0\\ \Leftrightarrow\left(x-1\right)\left(6-x\right)^{2019}=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\\left(6-x\right)^{2019}=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=6\end{matrix}\right.\)
Vậy x ∊ {1; 6}
\(=>x\left(6-x\right)^{2019}-\left(6-x\right)^{2019}=0\)
\(=>\left(x-1\right)\left(6-x\right)^{2019}=0\)
\(=>\left[{}\begin{matrix}x-1=0\\6-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=6\end{matrix}\right.\)