\(2^{1+2+3+...+x}=2014\)

\(1+2+3+...+x=10\)

x=4

\(2\times2^2\times2^3\times2^4\times...\times2^x=\left(2^3\right)^{12}\)

\(\Leftrightarrow2^{1+2+3+4+...+x}=2^{3\times12}\)

\(\Leftrightarrow2^{1+2+3+4+...+x}=2^{36}\)

\(\Leftrightarrow1+2+3+4+...+x=36\)

Ta có : Số số hạng = \(\frac{x-1}{1}+1=x\)

Tổng = \(\frac{\left(x+1\right)\times x}{2}=36\)

\(\Leftrightarrow\left(x+1\right)\times x=72\)

\(\Leftrightarrow x^2+x-72=0\)

\(\Leftrightarrow x^2-8x+9x-72=0\)

\(\Leftrightarrow x\times\left(x-8\right)+9\times\left(x-8\right)=0\)

\(\Leftrightarrow\left(x-8\right)\times\left(x+9\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-8=0\\x+9=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=8\\x=-9\end{cases}}\)

=> x = 8 ( do x là số nguyên dương )

x = 0

* Trả lời:

Ta có: \(1024=2^{10}\)

Lại có: \(2.2^2.2^3.2^4=2^{10}\)

\(\Rightarrow2^x=1\)

\(\Leftrightarrow x=0\) ( vì số nào mũ 0 cũng bằng 1)

x=0

2.2².2³....2ⁿ = 1024

2.2².2³....2ⁿ = 2¹⁰

=> 1+2+3+...+n = 10

(n+1).n : 2 = 10

(n+1).n = 10.2

(n+1).n = 20

(n+1).n = 5.4

=> n = 4

Ta có: \(2.2^2.2^3.....2^n=1024\)

\(\Rightarrow2.2^2.2^3......2^n=2^{10}\)

\(\Rightarrow1+2+3+...+n=10\)

\(\Rightarrow n=4\)

2.2^2.2^3.2^4......2^x = 1024

2^(1+2+...+x) = 2^10

=> 1+2 + 3+...+ x = 10

=>x.(1+x):2 = 10

=>x.(1+x) = 20

=> x = 4 (vì 4.5 =20)

dung haha

Đúng cái gì cơ

1) \(7.4^x=7.4^3\Leftrightarrow4^x=4^3;x=3\)

2) \(\frac{3}{2.5^x}=\frac{3}{2.5^{12}}\Leftrightarrow5^x=5^{12};x=12\)

\(2^x=2.2^8=2^9;x=9\)

4) \(5.3^x=7.3^5-2.3^5\Leftrightarrow5.3^x=3^5.\left(7-2\right)\)

\(\Leftrightarrow3^5.x=3^5.5;x=5\)

Bài 1 )

a ) \(2.2^2.2^3.....2^x=1024\Leftrightarrow2^{1+2+....+x}=2^{10}\Leftrightarrow1+2+....+x=10\)

\(\Leftrightarrow\frac{x\left(x+1\right)}{2}=10\Leftrightarrow\left(x+1\right)x=20=4.5\Rightarrow x=4\)

b ) \(\frac{37-x}{x+13}=\frac{3}{7}\Leftrightarrow3x+39=259-7x\Leftrightarrow3x+7x=259-39\Leftrightarrow10x=220\Rightarrow x=22\)

Bài 2 ) \(\frac{1}{2}\sqrt{64}-\sqrt{\frac{4}{25}}+\left(\frac{50^2-15.125}{5^4}\right)^{2014}=\frac{1}{2}.8-\frac{2}{5}+\left(\frac{5^4.2^2-3.5^4}{5^4}\right)^{2014}\)

\(=4-\frac{2}{5}+\left[\frac{5^4\left(4-3\right)}{5^4}\right]^{2014}=\frac{18}{5}+1=\frac{23}{5}\)

Mình làm bài 1 thui nha, còn bài 2 thì còn tự tính là được thôi mừ !!!

Bài 1:

a) \(2.2^2.2^3...2^x=1024\)

\(=>2^{1+2+3+...+x}=2^{10}\)

\(< =>1+2+3+...+x=10\)

\(=>6+x=10\)

\(=>x=10-6\)

\(=>x=4.\)

Nếu đúng thì k cho mình nhá

Mink ko biết kks

b./ \(\Leftrightarrow\frac{x+1}{2009}+1+\frac{x+2}{2008}+1+\frac{x+3}{2007}+1=\frac{x+10}{2000}+1+\frac{x+11}{1999}+1+\frac{x+12}{1998}+1.\)

\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}=0\)

\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0\)(b)

Mà \(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}< 0\)

(b) \(\Leftrightarrow x+2010=0\Leftrightarrow x=-2010\)

a./

\(\Leftrightarrow\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}-\frac{x+1}{5}-\frac{x+1}{6}=0.\)

\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\right)=0\)(a)

Mà \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}>0\)

(a) \(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)

c./ \(\Leftrightarrow2\cdot2^2\cdot2^3\cdot...\cdot2^x=2^{10}\Leftrightarrow2^{1+2+3+...+x}=2^{10}\Leftrightarrow1+2+3+...+x=10\)

\(\Leftrightarrow\frac{x\left(x+1\right)}{2}=10\Leftrightarrow x\left(x+1\right)=20=4\cdot5\Rightarrow x=4\)