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

Ai đó hãy giúp giải bà này đi 

Chọn A

\(\left(3x-4\right).2^3=64\)

\(\Rightarrow3x-4=64:8=8\)

\(\Rightarrow3x=4+8=12\)

\(\Rightarrow x=4\)

(3x-4).2^3 = 64

=> (3x-4).8 = 64

=) 3x-4 = 8

=> 3x = 12

=> x = 12:3

=> x = 4 (tmđk)

Vậy, x = 4

\(\left(3x-4\right).2^3=64\)

\(3x-4=\frac{64}{2^3}\)

\(3x-4=\frac{2^6}{2^3}\)

\(3x-4=2^{6-3}\)

\(3x-4=2^3\)

\(3x=8+4\)

\(3x=12\)

\(x=4\)

vậy \(x=4\)

b: Ta có: \(2^{x+3}+2^x=144\)

\(\Leftrightarrow2^x\cdot9=144\)

\(\Leftrightarrow2^x=16\)

hay x=4

a) (x ^ 54)^2 = x                                         

         x^108  = x

Để: x^108  = x 

=> x=0 hoặc x=1

b)   2^x+3 +2^x =144

     2^X . 2^3 + 2^x =144

      2^x.( 2^3+1) =144

      2^x. 9            =144

       2^x                =144:9

      2^x                = 16

=> 2^x                 = 2^4

-Vậy  x = 4

Lời giải :

1. \(\left(\frac{1}{2}a+b\right)^3+\left(\frac{1}{2}a-b\right)^3\)

\(=\frac{a^3}{8}+\frac{3a^2b}{4}+\frac{3ab^2}{2}+b^3+\frac{a^3}{8}-\frac{3a^2b}{4}+\frac{3ab^2}{2}-b^3\)

\(=\frac{a^3}{4}+3ab^2\)

Lời giải :

2. \(x^3-3x^2+3x-1=0\)

\(\Leftrightarrow\left(x-1\right)^3=0\)

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\)

Vậy...

3. \(\left(4x-1\right)^3-\left(4x-3\right)\left(16x^2+3\right)\)

\(=64x^3-48x^2+12x-1-64x^3+48x^2-12x+9\)

\(=8\)