Có: 1024=2^10

=> 2.2^2.2^3......2^x=2^10

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

      1+2+3+...+x=1+2+3+4

=>x=4

Vậy x=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

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

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

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

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

a)\(\left(\frac{2}{5}\right)^{2014}:\left(\frac{4}{25}\right)^{1007}=\left[\left(\frac{2}{5}\right)^2\right]^{1007}:\left(\frac{4}{25}\right)^{1007}\)

                                            \(=\left(\frac{4}{25}\right)^{1007}:\left(\frac{4}{25}\right)^{1007}\)

                                              \(=1\)

b)\(3^{n+1}:9=3^{n+1}:3^2\)

                    \(=3^{n+1-2}\)

                     \(=3^{n-1}\)

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: \(2.2^2.2^3.....2^n=1024\)

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

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

\(\Rightarrow n=4\)

3^-2 . 3⁴ . 3^x = 3⁷

<=> 3^{2 + x} = 3⁷

<=> 2 + x = 7

<=> x = 7 - 2

<=> x = 5

\(2^{-2}\cdot2^x+2\cdot2^x=9\cdot2^6\\ \Rightarrow2^{x-2}+2^{x+1}=9\cdot2^6\\ \Rightarrow2^{x-2}\left(1+2^3\right)=9\cdot2^6\\ \Rightarrow2^{x-2}\cdot9=9\cdot2^6\Rightarrow2^{x-2}=2^6\\ \Rightarrow x-2=6\Rightarrow x=8\)

\(3^{-2}\cdot3^4\cdot3^x=3^7\\ \Rightarrow3^{x+4-2}=3^7\\ \Rightarrow x+2=7\Rightarrow x=5\)

1) \(2^{-2}.2^x+2.2^x=9.2^6\)

\(\Rightarrow2^x.\left(2^{-2}+2\right)=576\)

\(\Rightarrow2^x.\dfrac{9}{4}=576\)

\(\Rightarrow2^x=256=2^8\Rightarrow x=8\)

2) \(\Rightarrow3^{x+2}=3^7\Rightarrow x+2=7\Rightarrow x=5\)