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=> (1+2X-1)x (2x-1+1)/4=225
=> 2x+2x/4=225
=> 4x^2/4=225
=> x^2= 225
=> x=15
cái ^ là mũ nha bạn
chúc bn hok tốt
`Answer:`
a. Tổng: \([\left(2x-1\right)-1]:2+1=x\) số hạng
Ta có: \(1+3+5+7+9+...+\left(2x-1\right)=225\)
\(\Rightarrow x.\left(2x-1+1\right):2=225\)
\(\Leftrightarrow2x^2:2=225\)
\(\Leftrightarrow x^2=225\)
\(\Leftrightarrow x=15\)
b. Mình sửa đề nhé: \(2^x+2^{x+1}+2^{x+2}+2^{x+3}+...+2^{x+2015}=2^{2019}-8\)
\(\Rightarrow2^x.\left(1+2+2^2+...+2^{2015}\right)=2^{2019}-8\)
Ta đặt \(K=1+2+2^2+...+2^{2015}\)
\(\Rightarrow2^x.K=2^{2019}-8\)
\(\Rightarrow2K=2.\left(1+2+2^2+...+2^{2015}\right)\)
\(\Rightarrow2K=2+2^2+2^3+...+2^{2015}+2^{2016}\)
\(\Rightarrow2K-K=\left(2+2^2+2^3+...+2^{2015}+2^{2016}\right)-\left(1+2+2^2+...+2^{2015}\right)\)
\(\Rightarrow K=2^{2016}-1\)
\(\Rightarrow2^x.\left(2^{2016}-1\right)=2^{2019}-8\)
\(\Rightarrow2^{x+2016}-2^x=2^{2019}-2^3\)
\(\Rightarrow\hept{\begin{cases}x+2016=2019\\x=3\end{cases}}\Rightarrow x=3\)
a) (3x - 1)2 = 100
(3x - 1)2 = 102
=>3x - 1 = 10
=> 3x = 10 + 1
3x = 11
x = 11/3
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x-1\right)}=\)\(\frac{2017}{2019}\)
\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x-1\right)}=\frac{2017}{2019}\)
\(2\left[\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right]=\frac{2017}{2019}\)
\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\)\(\frac{2017}{2019}\)
\(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2017}{2019}:2\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2017}{4038}\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{2017}{4038}\)
\(\frac{1}{x+1}=\frac{1}{2019}\)
x + 1 =2019
x = 2019-1 =2018
Vậy x = 2018
\(2\left(\frac{1}{3}.\frac{1}{2}+\frac{1}{6}.\frac{1}{2}+\frac{1}{10}.\frac{1}{2}+....+\frac{2}{x\left(x+1\right)}.\frac{1}{2}\right)=\frac{2017}{2019}\)
=>\(2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+....+\frac{1}{x\left(x+1\right)}\right)=\frac{2017}{2019}\)
=>\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{x\left(x+1\right)}\right)\)\(=\frac{2017}{2019}\)
=>\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)
=> \(2[\frac{1}{2}+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{4}\right)+....+\left(\frac{1}{x}-\frac{1}{x}\right)-\frac{1}{x+1}]=\frac{2017}{2019}\)
=>\(2\left(\frac{1}{2}+0+0+....+0-\frac{1}{x-1}\right)=\frac{2017}{2019}\)
=>\(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)
=>\(\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2017}{4038}\)
=>\(\frac{1}{x+1}=\frac{1}{2019}\)
=> x+1=2019
=>x=2018
a) 219 - 7(x + 1) = 100 <=> 7(x+1) = 219 - 100
<=> 7(x+1) = 119 <=> x + 1 = 119 : 7 <=> x + 1 = 17
b) (3x - 6).3 = 3434 <=> 3x - 6 = 3434 : 3 <=> 3x - 6 = 3333
<=> 3x = 27 + 6 <=> 3x = 33 <=> x = 11
a) 219−7(x+1)=100
- 7(x+ 1)= 100 - 219
- 7(x+ 1)= - 119
x+1 = -119 : (-7)
x+1 = 17
x= 17-1
x= 16
b) (3x - 6). 3 = 34
(3x - 6). 3 = 81
(3x - 6)= 81: 3
(3x - 6)= 27
3x= 27+ 6
3x= 33
x= 33:3
x= 11
b) (3x−6).3=34
Ta có : \(3x+14\)\(⋮\)\(3x+1\)
\(\Rightarrow\)\(\left(3x+1\right)+13\)\(⋮\)\(3x+1\)
mà \(3x+1\)\(⋮\)\(3x+1\)
\(\Rightarrow\)\(13\)\(⋮\)\(3x+1\)
\(\Rightarrow\)\(3x+1\in\text{Ư}\left(13\right)\)
\(\Rightarrow\)\(3x+1\in\left\{1;13\right\}\)
\(\Rightarrow\)\(x\in\left\{0;4\right\}\)
\(\left|x\right|+\left|x+1\right|+2019=3x\)
Ở đây x là giá trị tuyệt đối nên \(\left|x\right|\ge0\) và \(\left|x+1\right|\ge0\)
\(\Rightarrow\left|x\right|+\left|x+1\right|+2019\ge2019\)
\(\Rightarrow\) \(x>0\) (vì vế trái là số dương nên \(3x\) cũng phải là số dương)
Ta có:
\(\left|x\right|+\left|x+1\right|+2019=3x\)
Ta đã nhận định \(x>0\), suy ra:
\(x+x+1+2019=3x\)
\(\Rightarrow2x+2020=3x\)
\(\Rightarrow2020=3x-2x\)
\(\Rightarrow2020=x\) hay \(x=2020\)
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