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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 \(K=1+2+2^2+...+2^{2015}\)
\(\Rightarrow2^x.K=2^{2019}-8\)
\(\Rightarrow2K=2.\left(1+2+2^2+...+2^{2015}\right)\)
\(\Rightarrow2K=2+2^2+2^3+...+2^{2015}+2^{2016}\)
\(\Rightarrow2K-K=\left(2+2^2+2^3+...+2^{2015}+2^{2016}\right)-\left(1+2+2^2+...+2^{2015}\right)\)
\(\Rightarrow K=2^{2016}-1\)
\(\Rightarrow2^x.\left(2^{2016}-1\right)=2^{2019}-8\)
\(\Rightarrow2^{x+2016}-2^x=2^{2019}-2^3\)
\(\Rightarrow\hept{\begin{cases}x+2016=2019\\x=3\end{cases}}\Rightarrow x=3\)
Bài 10:
a: 2x-3 là bội của x+1
=>\(2x-3⋮x+1\)
=>\(2x+2-5⋮x+1\)
=>\(-5⋮x+1\)
=>\(x+1\in\left\{1;-1;5;-5\right\}\)
=>\(x\in\left\{0;-2;4;-6\right\}\)
b: x-2 là ước của 3x-2
=>\(3x-2⋮x-2\)
=>\(3x-6+4⋮x-2\)
=>\(4⋮x-2\)
=>\(x-2\inƯ\left(4\right)\)
=>\(x-2\in\left\{1;-1;2;-2;4;-4\right\}\)
=>\(x\in\left\{3;1;4;0;6;-2\right\}\)
Bài 14:
a: \(4n-5⋮2n-1\)
=>\(4n-2-3⋮2n-1\)
=>\(-3⋮2n-1\)
=>\(2n-1\inƯ\left(-3\right)\)
=>\(2n-1\in\left\{1;-1;3;-3\right\}\)
=>\(2n\in\left\{2;0;4;-2\right\}\)
=>\(n\in\left\{1;0;2;-1\right\}\)
mà n>=0
nên \(n\in\left\{1;0;2\right\}\)
b: \(n^2+3n+1⋮n+1\)
=>\(n^2+n+2n+2-1⋮n+1\)
=>\(n\left(n+1\right)+2\left(n+1\right)-1⋮n+1\)
=>\(-1⋮n+1\)
=>\(n+1\in\left\{1;-1\right\}\)
=>\(n\in\left\{0;-2\right\}\)
mà n là số tự nhiên
nên n=0
Ta có :
a) \(1+3+5+...+\left(2x-1\right)=\frac{\left(2x-1\right)+1}{2}\left(\frac{\left(2x-1\right)-1}{2}+1\right)=x^2\)
\(\Leftrightarrow x^2=225\Rightarrow x=15\)
b) \(2^x+2^{x+1}+...+2^{x+2015}=2^x\left(2^0+2^1+...+2^{2015}\right)\)
Đặt A = 20 + 21 + ... + 22015 . Ta có :
2A = 21 + 22 + ... + 22016
⇒ A = 2A - A = (21 +22 +...+22016 )-(20 + 21 + ... +22015 )
⇒ A = 22016 - 1
⇔ 2x.A = 22019 - 8
⇔ 2x( 22016 - 1 ) = 23 ( 22016 - 1 )
⇔ x = 3
Đề bài c) chưa đủ ý nên o làm đc
a)
\(\begin{array}{l}\left( {9x - {2^3}} \right):5 = 2\\9x - {2^3} = 2.5\\9x - 8 = 10\\9x = 18\\x = 2\end{array}\)
Vậy \(x = 2\)
b)
\(\begin{array}{l}\left[ {{3^4} - \left( {{8^2} + 14} \right):13} \right]x = {5^3} + {10^2}\\\left[ {81 - \left( {64 + 14} \right):13} \right]x = 125 + 100\\\left[ {81 - 78:13} \right]x = 125 + 100\\\left[ {81 - 6} \right]x = 225\\75x = 225\\x = 3\end{array}\)
Vậy \(x = 3\)
\(a.x+\dfrac{1}{6}=-\dfrac{3}{8}\)
\(\Leftrightarrow x=-\dfrac{13}{24}\)
\(b.2-\left(\dfrac{3}{4}-x\right)=\dfrac{7}{12}\)
\(\Leftrightarrow2-\dfrac{3}{4}+x=\dfrac{7}{12}\)
\(\Leftrightarrow x=-\dfrac{2}{3}\)
\(c.\dfrac{1}{2}x+\dfrac{1}{8}x=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{5}{8}x=\dfrac{3}{4}\)
\(\Leftrightarrow x=\dfrac{6}{5}\)
\(d.75\%-1\dfrac{1}{2}+0,5:\dfrac{5}{12}-\left(\dfrac{-1}{2}\right)^2\)
\(=\dfrac{75}{100}-\dfrac{3}{2}+\dfrac{1}{2}:\dfrac{5}{12}-\dfrac{1}{4}\)
\(=-\dfrac{3}{4}+\dfrac{6}{5}-\dfrac{1}{4}\)
\(=\dfrac{1}{5}\)
a) \(x+\dfrac{1}{6}=\dfrac{-3}{8}\)
\(x=\dfrac{-3}{8}-\dfrac{1}{6}\)
\(x=\dfrac{-13}{24}\)
vậy x =....
b) \(2-\left(\dfrac{3}{4}-x\right)=\dfrac{7}{12}\)
\(\dfrac{3}{4}-x=2-\dfrac{7}{12}\)
\(\dfrac{3}{4}-x=\dfrac{17}{12}\)
\(x=\dfrac{3}{4}-\dfrac{17}{12}\)
\(x=\dfrac{-2}{3}\)
vậy x =....
c: Ta có: \(\dfrac{1}{3}-\dfrac{7}{8}x=\dfrac{1}{4}\)
\(\Leftrightarrow x\cdot\dfrac{7}{8}=\dfrac{1}{12}\)
\(\Leftrightarrow x=\dfrac{1}{12}\cdot\dfrac{8}{7}=\dfrac{2}{21}\)
d: Ta có: \(\dfrac{3}{2}x+\dfrac{1}{7}=\dfrac{7}{8}\cdot\dfrac{64}{49}\)
\(\Leftrightarrow x\cdot\dfrac{3}{2}=1\)
hay \(x=\dfrac{2}{3}\)
a) Ta có: 1+3+5+7+9+...+(2x-1)=225 (1)
Số số hạng của vế trái là:
(2x-1-1):2+1=(2x-2):2+1=(x-1)+1=x (số hạng)
Vậy khi đó từ (1) ta có:
[(1+2x-1).x]:2=225
=> (2x.x):2=225
=> (2.x2):2=225
=> x2=225
=> x2=152
=> x=15
Vậy x=15
b) 2x+2x+1+2x+2+2x+3+...+2x+2015=22019-8
=> 2x.1+2x.2+2x.22+2x.23+...+2x.22015=22019-23
=> 2x.(1+2+22+23+...+22015)=22019-23 (1)
Đặt A=1+2+22+23+...+22015
=> 2A=2+22+23+24...+22016
=> 2A-A=(2+22+23+24...+22016)-(1+2+22+23+...+22015)
=> A=22016-1 (2)
Từ (1), (2) => 2x.(22016-1)=22019-23
=> 2x.(22016-1)=23.(22016-1)
=> 2x=23
=> x=3
Vậy x=3
Bài giải