2.Tìm x, biết

a,(x-1)³=3 (x-1)³=343

=>  (x-1)³=7³

=> x -1 = 7

=> x = 8

B, (X-2)⁴=4096

(X-2)⁴= 8⁴

=> x - 2 = 8

=> x = 10

C,(2x²-13)⁴=(-5)⁴

=> 2x²-13 = -5

=> 2x²      = 8

=> x²        = 4

=> x          = 2

Study well 

a. (x-1)³ = 343

=> (x-1)³ = 7³

=> x - 1 = 7

=> x   = 7 + 1

=> x   =   8

Ai đó hãy giúp giải bà này đi 

Chọn A

a ) ( x - 1 )³ = 343

7³ = 343

x + 1 = 7

x = 7 - 1

x = 6

b ) ( x - 2 )⁴ = 4096

8⁴ = 4096

x - 2 = 8

x = 8 + 2

x = 10

\(\left(x-1\right)^3=343=7^3\) => x -1 =7 => x =8

\(\left(x-2\right)^4=4096=8^4=\left(-8\right)^4\)\(\Rightarrow\orbr{\begin{cases}x-2=8\\x-2=-8\end{cases}\Rightarrow\orbr{\begin{cases}x=10\\x=-6\end{cases}}}\)

a) \(\left(x-1\right)^3=343\)

\(\left(x-1\right)=7^3\)

  \(x-1=7\)

  \(x=7+1\)        

  \(x=8\)

b) \(\left(x-2\right)^4=4094\)

    \(\left(x-2\right)^4=8^4\)

      \(x-2=8\) hoặc \(x-2=-8\)

      \(x=8+2\)         \(x=-8+2\)

      \(x=10\)               \(x=-6\)

Vậy \(x=10\) hoặc \(x=-6\)

a) \(\frac{3}{7}-\frac{1}{7}x=\frac{2}{3}\)

=> \(\frac{1}{7}x=\frac{3}{7}-\frac{2}{3}=-\frac{5}{21}\)

=> \(x=-\frac{5}{21}:\frac{1}{7}=-\frac{5}{21}\cdot7=-\frac{5}{3}\)

b) \(3x^2-2=72\)=> 3x² = 74 => x² = 74/3 => x không thỏa mãn

c) \(\left(19x+2\cdot5^2\right):14=\left(13-8\right)^2-4^2\)

=> \(\left(19x+2\cdot25\right):14=5^2-4^2=9\)

=> \(\left(19x+50\right):14=9\)

=> \(19x+50=126\)

=> \(19x=76\)

=> x = 4

d) \(x:\frac{1}{2}+x:\frac{1}{4}+x:\frac{1}{8}+x:\frac{1}{16}+x:\frac{1}{32}=343\)

=> \(x\cdot2+x\cdot4+x\cdot8+x\cdot16+x\cdot32=343\)

=> \(x\left(2+4+8+16+32\right)=343\)

=> x . 62 = 343

=> x = 343/62

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) \(\frac{14}{15}:\frac{9}{10}=x:\frac{3}{7}\Rightarrow\frac{28}{27}=x:\frac{3}{7}\Rightarrow x=\frac{4}{9}\)

b) \(\left(x-\frac{4}{7}\right)^3=343\Rightarrow\left(x-\frac{4}{7}\right)^3=7^3\Rightarrow x-\frac{4}{7}=7\Rightarrow x=\frac{53}{7}\)

c) \(x^5=x^3\Leftrightarrow\hept{\begin{cases}x=1\\x=0\end{cases}}\)

e) \(\left(x-1\right)^4=16\Leftrightarrow\orbr{\begin{cases}\left(x-1\right)^4=2^4\\\left(x-1\right)^4=\left(-2\right)^4\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x-1=2\\x-1=\left(-2\right)\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)

a: =>9x=5(1-4x)

=>9x=5-20x

=>29x=5

hay x=5/29

b: =>x/5=5/x

=>x²=25

=>x=5 hoặc x=-5

c: =>(x+2)³=1/343

=>x+2=1/7

hay x=-13/7