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\(b,\left(2\chi-7\right)^{4-1}=4^{2\times5}\)\(a,3\times2^{\chi-7}=17\)

a) \(3.2^x-7=17\)

\(3\cdot2^x=24\)

\(2^x=8=2^3\)

=> x = 3

b) \(\left(2x-7\right)^4-1=4^2\cdot5\)

\(\left(2x-7\right)^4-1=80\)

\(\left(2x-7\right)^4=81=\left(\pm3\right)^4\)

+) 2x - 7 = 3

2x = 10

x = 5

+) 2x - 7 = -3

2x = 4

x = 2

Vậy,...........

a. 3 . 2^x - 7 = 17

3 . 2^x          = 17 + 7

2^x               = 24 : 3

2^x               = 8

2^x               = 2³

x                 = 3

b. (2x - 7)⁴ - 1 = 4² . 5

(2x - 7)⁴ - 1     = 80

(2x - 7)⁴          = 81

(2x - 7)⁴          = 3⁴

2x - 7              = 3

2x                   = 3 + 7

x                     = 10 : 2

x                     = 5

\(\frac{x+1}{3}=\frac{9}{2}\)

\(\left(x+1\right).2=9.3\)

\(\left(x+1\right).2=27\)

\(x+1=27:2\)

\(x+1=13,5\)

\(x=13,5-1=12,5\)

vậy x = 12.5

\(\frac{x+1}{3}=\frac{9}{2}\)

\(\Leftrightarrow2\left(x+1\right)=3\times9\)

\(\Leftrightarrow2\left(x+1\right)=27\)

\(\Leftrightarrow x+1=\frac{27}{2}\)

\(\Leftrightarrow x=\frac{25}{2}\)

\(\frac{x+1}{3}\)= \(\frac{9}{2}\)

:> 2( x +1) = 9 . 3

:> 2x + 2 = 27

:>       2x = 27 - 2

:>       2x = 25

:>        x = \(\frac{25}{2}\)

(8x−3)(3x+2)−(4x+7)(x+4)=(2x+1)(5x−1)(8x−3)(3x+2)−(4x+7)(x+4)=(2x+1)(5x−1)

 20x2−16x−34=10x2+3x−120x2−16x−34=10x2+3x−1

 10x2−19x−33=010x2−19x−33=0

 (10x+11)(x−3)=0

chỉ bt lm con b thoy

..army,,,,,,,,,,

a) \(\left(2x+3\right)\left(x-4\right)+\left(x-5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)

\(\Leftrightarrow3x^2-12x-2=3x^2-17x+20\)

\(\Leftrightarrow3x^2-12x=3x^2-17x+20+2\)

\(\Leftrightarrow3x^2-12x=3x^2-17x+22\left(3x^2-17x\right)\)

\(\Leftrightarrow5x=22\)

\(\Rightarrow x=\frac{22}{5}\)

b) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)

\(\Leftrightarrow20x^2-16x-34=10x^2+3x+1\)

\(\Leftrightarrow20x^2-16x-33=10x^2+3x\)

\(\Leftrightarrow20x^2-16x-33=10x^2+3x-3x\)

\(\Leftrightarrow20x^2-16x-33=10x^2\)

\(\Leftrightarrow20x^2-16x-33=10x^2-10x^2\)

\(\Leftrightarrow20x^2-16x-33=0\)

\(\Rightarrow\orbr{\begin{cases}x=3\\x=-\frac{11}{10}\end{cases}}\)

a, ( 2x + 3 )(x - 4) + ( x - 5 )( x - 2) =( 3x - 5 ) ( x - 4 )
<-> 2x² - 8x + 3x - 12 + x² - 2x - 5x + 10 = 3x² - 12x - 5x + 20
<-> 3x² - 10x - 2 = 3x² - 15x + 20
<-> 3x² - 3x² - 10x + 15x = 20 + 2
<-> 5x = 22
<-> x = 22/5

\(\left(x-1\right)^x=10^{11}\)

\(\Leftrightarrow\left(x-1\right)^x=\left(11-1\right)^{11}\)

\(\Rightarrow x=11\)

(x  -  1)^x = 10¹¹

=> x - 1 = 10

=> x  = 11

vậy_

     \(\left(X-1\right)^x=10^{11}\)

      \(=>X=11\)  

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Bạn bị sai đề nhé

\(\frac{3}{7.10}+\frac{3}{10.13}+....+\frac{3}{100.103}\)

\(=\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+....+\frac{1}{100}-\frac{1}{103}\)

\(=\frac{1}{7}-\frac{1}{103}\)

\(=\frac{96}{721}\)

\(\frac{2}{7.10}+\frac{2}{10.13}+...+\frac{2}{100.103}\)

\(=\frac{2}{3}\left(\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{100}-\frac{1}{103}\right)\)

\(=\frac{2}{3}\left(\frac{1}{7}-\frac{1}{103}\right)\)

\(=\frac{2}{3}.\frac{96}{721}\)

\(=\frac{64}{721}\)