a: \(\Leftrightarrow x^3=\dfrac{539}{64}\)

hay \(x=\dfrac{7\sqrt{11}}{4}\)

c: \(\Leftrightarrow2^{2x-1}=2^9\cdot2^2=2^{11}\)

=>2x-1=11

hay x=6

d: \(\Leftrightarrow x^{17}-x=0\)

\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\)

hay \(x\in\left\{0;1;-1\right\}\)

2^x=8 hay 2.2.2=8

=>x=3

x^3=64 

=>x =4

a) \(2^{4x+1}-8^{x+2}=0\)\(\Leftrightarrow2^{4x+1}-2^{3\left(x+2\right)}=0\)

\(\Leftrightarrow2^{4x+1}-2^{3x+6}=0\)\(\Leftrightarrow2^{4x+1}=2^{3x+6}\)

\(\Leftrightarrow4x+1=3x+6\)\(\Leftrightarrow4x-3x=6-1\)\(\Leftrightarrow x=5\)

Vậy \(x=5\)

b) \(3^2.9^{2x}=27^{x+3}\)\(\Leftrightarrow3^2.3^{2.2x}=3^{3\left(x+3\right)}\)\(\Leftrightarrow3^2.3^{4x}=3^{3x+9}\)

\(\Leftrightarrow3^{2+4x}=3^{3x+9}\)\(\Leftrightarrow2+4x=3x+9\)\(\Leftrightarrow4x-3x=9-2\)\(\Leftrightarrow x=7\)

Vậy \(x=7\)

c) \(8^{2x}.64^2=16^{x+4}\)\(\Leftrightarrow2^{3.2x}.2^{6.2}=2^{4\left(x+4\right)}\)\(\Leftrightarrow2^{6x}.2^{12}=2^{4\left(x+4\right)}\)

\(\Leftrightarrow2^{6x+12}=2^{4x+16}\)\(\Leftrightarrow6x+12=4x+16\)\(\Leftrightarrow6x-4x=16-12\)

\(\Leftrightarrow2x=4\)\(\Leftrightarrow x=2\)

Vậy \(x=2\)

D, x = 1

C, x = 6

màu giải như cái lồnn vậy

Thèo đề bài, ta có:

\(\frac{x^3}{2^3}=\frac{y^3}{4^3}=\frac{z^3}{6^3}=\frac{x}{2}=\frac{y}{4}=\frac{z}{6}=\frac{x^2}{4}=\frac{y^2}{16}=\frac{z^2}{36}=\frac{x^2+y^2+z^2}{4+16+36}=\frac{14}{56}=\frac{1}{4}\)

x ; y ; z thì bạn tự tìm nhé , chắc cái này không khó đâu nhỉ ??

\(\frac{x^3}{8}=\frac{y^3}{64}=\frac{z^3}{216}\Rightarrow\frac{x}{2}=\frac{y}{4}=\frac{z}{6}\Rightarrow\frac{x^2}{4}=\frac{y^2}{16}=\frac{z^2}{36}\) \(=\frac{x^2+y^2+z^2}{4+16+36}=\frac{14}{56}=\frac{1}{4}\)

\(\frac{x}{2}=\frac{1}{4}\Rightarrow x=\frac{1}{2}\)

\(\frac{y}{4}=\frac{1}{4}\Rightarrow y=1\)

\(\frac{z}{6}=\frac{1}{4}\Rightarrow z=\frac{3}{2}\)

a, 4ⁿ=(4³)³.(4⁴)³

=>4ⁿ=4⁹.4¹²

=>4ⁿ=4²¹

=>n=21

b, 8ⁿ.25³=125²

=>8ⁿ=125²:25³

=>8ⁿ=(5³)²:(5²)³

=>8ⁿ=5⁶:5⁶

=>8ⁿ=1=8⁰

=>n=0

Chúc bạn học giỏi nha!!!

\(4^n=64^3.256^3\)

\(\Leftrightarrow4^n=\left(4^3\right)^3.\left(4^4\right)^3\)

\(\Leftrightarrow4^n=4^9.4^{12}=4^{21}\)

\(\Leftrightarrow n=21\)

     Vậy \(n=21\)

1/4 . 2/6 . 3/8 . ... .30/62 .31/64 = 2^x

(1/2 . 1/2).(2/3 . 1/2).(3/4 . 1/2). ... .(30/31 . 1/2).(31/32 . 1/2) = 2^x

(1/2.1/2. ... .1/2).(1/2 . 2/3 . 3/4. ... .30/31 . 31/32) = 2^x

   (31 số 1/2) 

(1/2)^31.  = 2^x

=> 0=x+36

      x=0-36

      x=-36

Vậy x=-36

Theo mk nghĩ,mk làm đúng nha .Tk cho mk

Để mk sửa phần này một chút

\((\frac{1}{2})^{31}\cdot\frac{1\cdot2\cdot3.....30\cdot31}{2\cdot3\cdot4.....31\cdot32}=2^x\)

\(\frac{1^{31}}{2^{31}}\cdot\frac{1}{32}=2^x\)

\(\frac{1}{2^{31}}\cdot\frac{1}{2^5}=2^x\)

\(\frac{1}{2^{36}}=2^x\)

\(1=2^x\cdot2^{36}\)

\(2^0=2^x+36\)

Rồi bn tự suy luận nha

a) \(3^{x-2}=27\cdot9\)

\(3^{x-2}=3^3\cdot3^2=3^5\)

\(\Rightarrow\)\(x-2=5\Rightarrow x=7\)

b) \(2^{x+1}+2^{x+3}=80\)

\(\Rightarrow2^{x+1}\left(1+2^2\right)=80\)

\(\Rightarrow2^{x+1}\cdot5=80\)

\(\Rightarrow2^{x+1}=16=2^4\)

\(\Rightarrow x+1=4\Rightarrow x=3\)

c) \(2^{2x-3}=16\cdot8\)

\(2^{2x-3}=2^4\cdot2^3=2^7\)

\(\Rightarrow2x-3=7\)

\(\Rightarrow2x=4\Rightarrow x=2\)

d) \(2^{x-2}\cdot2^x=64\)

\(\Rightarrow2^{x-2+x}=64=2^6\)

\(\Rightarrow x-2+x=6\)

\(\Rightarrow2x-2=6\)

\(\Rightarrow2x=8\Rightarrow x=4\)

Giải:

a) \(3^{x-2}=27.9\)

\(\Leftrightarrow3^{x-2}=3^3.3^2\)

\(\Leftrightarrow3^{x-2}=3^5\)

Vì \(3=3\)

Nên \(x-2=5\)

\(\Leftrightarrow x=5+2\)

\(\Leftrightarrow x=7\)

Vậy x = 7.

b) \(2^{x+1}+2^{x+3}=80\)

\(\Leftrightarrow2^{x+1}\left(1+2^2\right)=80\)

\(\Leftrightarrow2^{x+1}.5=80\)

\(\Leftrightarrow2^{x+1}=\dfrac{80}{5}=16\)

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

Vì \(2=2\)

Nên \(x+1=4\)

\(\Leftrightarrow x=4-1\)

\(\Leftrightarrow x=3\)

Vậy x = 3.

c) \(2^{2x-3}=16.8\)

\(\Leftrightarrow2^{2x-3}=2^4.2^3\)

\(\Leftrightarrow2^{2x-3}=2^7\)

Vì \(2=2\)

Nên \(2x-3=7\)

\(\Leftrightarrow2x=7+3=10\)

\(\Leftrightarrow x=\dfrac{10}{2}=5\)

Vậy x = 5.

d) \(2^{x-2}.2^x=64\)

\(2^{2x-2}=2^6\)

Vì \(2=2\)

Nên \(2x-2=6\)

\(\Leftrightarrow2x=6+2=8\)

\(\Leftrightarrow x=\dfrac{8}{2}=4\)

Vậy x = 4.

Chúc bạn học tốt!

\(2^{x+9}-2^{x+8}-2^{x+7}-..........-2^{x+1}-2^x=1024\)

\(\Leftrightarrow2^x.2^9-2^x.2^8-.........-2^x.2-2^x=1024\)

\(\Leftrightarrow2^x\left(2^9-2^8-.........-2-1\right)=1024\)

Đặt :

\(A=2^9-2^8-......-2-1\Leftrightarrow2^x.A=1024\)

\(\Leftrightarrow A=2^9-\left(2^8+2^7+........+2+1\right)\)

\(\Leftrightarrow A=2^9-\left(2^9-1\right)\)

\(\Leftrightarrow A=1\)

\(\Leftrightarrow2^x.1=1024\)

\(\Leftrightarrow2^x=2^{10}\)

\(\Leftrightarrow x=10\)

Vậy ...

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