#)Giải :

\(123-5\left(x+4\right)=38\)

\(\Leftrightarrow5\left(x+4\right)=85\)

\(\Leftrightarrow x+4=17\)

\(\Leftrightarrow x=13\)

\(\left(3x-16\right)343=2.2401\)

\(\Leftrightarrow\left(3x-16\right)343=4802\)

\(\Leftrightarrow3x-16=14\)

\(\Leftrightarrow3x=30\)

\(\Leftrightarrow x=10\)

\(\left(2600+6400\right)-3x=1200\)

\(\Leftrightarrow9000-3x=1200\)

\(\Leftrightarrow3x=7800\)

\(\Leftrightarrow x=2600\)

P/s : Câu cuối cùng lỗi chỗ b4 ph k ? hay là tìm cả b và x thế ???

\(123-5\left(x+4\right)=38\)

\(=>123-5x-20=38\)

\(=>-5x=-123+20+38\)

\(=>-5x=-65\)

\(=>x=13\)

b, \(\left(3x-16\right)343=2.2401\)

=>1029x -5488= 4802

=> 1029x=5488+4802

=> 1029x= 10290

=> x=10

c, (2600+6400)-3x=1200

=> 9000-3x=1200

=> -3x= -9000+1200

=> -3x= -7800

=> x=2600

d,

k hiểu đề

xin lỗi b4 sữa lại 84 nha

3:

a: 3^x*3=243

=>3^x=81

=>x=4

b; 2^x*16^2=1024

=>2^x=4

=>x=2

c: 64*4^x=16^8

=>4^x=4^16/4^3=4^13

=>x=13

d: 2^x=16

=>2^x=2^4

=>x=4

\(a,2^x.4=128\\2^x.2^2=2^7\\ 2^x=\dfrac{2^7}{2^2}=2^{7-2}=2^5\\ Vậy:x=5\\ ----\\ b,\left(2x+1\right)^3=125=5^3\\ \Rightarrow 2x+1=5\\ 2x=5-1=4\\ x=\dfrac{4}{2}=2\\ ----\\ c,2x-2^6=6\\ 2x=6+2^6=6+64\\ 2x=70\\ x=\dfrac{70}{2}=35\\ ----\\ d,49.7^x=2401\\ 7^x=\dfrac{2401}{49}=49=7^2\\ Vậy:x=2\)

a.  64.4^x=4⁵

4³.4^x=4⁵

4^x=4⁵:4³

4^x=4²

x=2

có ai biết hoặc là fan gacha life không làm ơn kết bạn với mình đi team gacha life của mình đang thiếu người

\(A=3^x+3^{x+1}+...+3^{x+100}\)

=>\(3A=3^{x+1}+3^{x+2}+...+3^{x+101}\)

=>\(2A=3^{x+101}-3^x\)

=>\(A=\dfrac{3^{x+101}-3^x}{2}\)

=>\(3^{x+101}-3^x=3^{105}-3^4\)

=>x=4

x=4

Bài 1 :

a) 7^2x-1 = 343

=> 7^2x-1 = 7³

=> 2x - 1 = 3 => 2x = 4 => x = 2

b) (7x - 11)³ = 2⁵.3² + 200

=> (7x - 11)³ = 32.9 + 200

=> (7x - 11)³ = 488

xem kĩ lại đề này :vvv

c) 174 - (2x - 1)² = 5³

=> (2x - 1)² = 174 - 5³

=> (2x - 1)² = 174 - 125 = 49

=> (2x - 1)² = (\(\pm\)7)²

=> \(\orbr{\begin{cases}2x-1=7\\2x-1=-7\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=-3\end{cases}}\)

Mà x \(\in\)N nên x = 4( thỏa mãn điều kiện)

Bài 2 :

a) x⁵ = 32 => x⁵ = 2⁵ => x = 2

b) (x + 2)³ = 27

=> (x + 2)³ = 3³

=> x + 2 = 3 => x = 3 - 2 = 1

c) (x - 1)⁴ = 16

=> (x - 1)⁴ = 2⁴

=> x - 1 = 2 => x = 3 ( vì đề bài cho x thuộc N nên thỏa mãn)

d) (x - 1)⁸ = (x - 1)⁶

=> (x - 1)⁸ - (x - 1)⁶ = 0

=> (x - 1)⁶ [(x - 1)² - 1] = 0

=> \(\orbr{\begin{cases}\left(x-1\right)^6=0\\\left(x-1\right)^2-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\\left(x-1\right)^2=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\\left(x-1\right)^2=\left(\pm1\right)^2\end{cases}}\)

+) x - 1 = 1 => x = 2 ( tm)

+) x - 1 = -1 => x = 0 ( tm)

Vậy x = 1,x = 2,x = 0

a) \(64^x:16^x=256\)

\(\Rightarrow\left(2^6\right)^x:\left(2^4\right)^x=2^8\)

\(\Rightarrow2^{6x}:2^{4x}=2^8\)

\(\Rightarrow2^{6x-4x}=2^8\)

\(\Rightarrow2^{2x}=2^8\)

\(\Rightarrow2x=8\)

\(\Rightarrow x=4\)

b) \(\dfrac{-2401}{7^x}=-7\)

\(\Rightarrow\dfrac{-7^4}{7^x}=-7\)

\(\Rightarrow-7^{4-x}=-7\)

\(\Rightarrow7^{4-x}=7\)

\(\Rightarrow4-x=1\)

\(\Rightarrow x=4-1\)

\(\Rightarrow x=3\)

c) \(\dfrac{64}{\left(-4\right)^x}=-256\)

\(\Rightarrow\left(-4\right)^x=\dfrac{64}{-256}\)

\(\Rightarrow\left(-4\right)^x=-4\)

\(\Rightarrow\left(-4\right)^x=\left(-4\right)^1\)

\(\Rightarrow x=1\)

\(a) 64^x:16^x=256\\\Rightarrow (64:16)^x=256\\\Rightarrow 4^x=4^4\\\Rightarrow x=4\\---\)

\(b,\dfrac{-2401}{7^x}=-7\)

\(\Rightarrow7^x=-2401:\left(-7\right)\)

\(\Rightarrow7^x=343\)

\(\Rightarrow7^x=7^3\)

\(\Rightarrow x=3\)

\(c,\dfrac{64}{\left(-4\right)^x}=-256\)

\(\Rightarrow\left(-4\right)^x=64:\left(-256\right)\)

\(\Rightarrow\left(-4\right)^x=-\dfrac{1}{4}\)

\(\Rightarrow\left(-4\right)^x=\left(-4\right)^{-1}\)

\(\Rightarrow x=-1\)

#\(Toru\)

A, (2x+1)³=343

=> (2x+1)³=7³

=> 2x + 1 = 7

=> 2x = 6

=> x = 3

B, 2^x+2^x+3=144

=> 2^x+2^x . 2³ =144

=> 2^x ( 1 + 2³ ) =144

=> 2^x ( 1 + 8 ) =144

=> 2^x . 9 =144

=> 2 ^x = 16

=> 2 ^x = 2 ⁴

=> x = 4

C, 3^x+3^x+2=2430

=> 3^x+3^x . 3² =2430

=> 3^x . ( 1 + 3² ) =2430

=> 3^x . ( 1 + 9 ) =2430

=> 3^x . 10 =2430

=> 3^x = 243

=> 3^x = 3 ⁵

=> x = 5