\(a,2^{x+2}-2^x=96\)

\(=>2^x.2^2-2^x=96\)

\(=>2^x.\left(4-1\right)=96\)

\(=>2^x.3=96\)

\(=>2^x=96:3=32\)

\(=>2^x=2^5\)

\(=>x=5\)

\(b,720:\left[41.\left(2x-5\right)\right]=2^3.125:5^2\)

\(=>720:\left[41.\left(2x-5\right)\right]=8.125:25\)

\(=>720:\left[41.\left(2x-5\right)\right]=8.5=40\)

\(=>41.\left(2x-5\right)=720:40=18\)

\(=>2x-5=18:41=\frac{18}{41}\)

\(=>2x=\frac{18}{41}+5=\frac{223}{41}\)

\(=>x=\frac{223}{41}:2=\frac{223}{62}\)

\(c,\left(-2x+7\right)^{19}=\left(-2x+7\right)^{19}.\left(x+1\frac{1}{4}\right)^2\)

\(=>\left(-2x+7\right)^{19}:\left(-2x+7\right)^{19}=\left(x+\frac{5}{4}\right)^2\)

\(=>1=\left(x+\frac{5}{4}\right)^2\)

\(=>1^2=\left(x+\frac{5}{4}\right)^2\)

\(=>1=x+\frac{5}{4}\)

\(=>x=1-\frac{5}{4}=-\frac{1}{4}\)

Chúc bạn Hk tốt!!!!

Và giữ đúng lời hứa trên@@!!!!!

a) \(\left|-4x+1\frac{1}{3}\right|=x+2\frac{1}{7}\)

TH1: \(-4x+1\frac{1}{3}=x+2\frac{1}{7}\)

\(-4x-x=2\frac{1}{7}-1\frac{1}{3}\)

\(-5x=\frac{17}{21}\)

=> ...

TH2: \(-4x+1\frac{1}{3}=-x-2\frac{1}{7}\)

...

rùi bn tự lm típ nha!
b) 2^2x-1+4^x+2 = 264

=> 2^2x: 2 + (2²)^x+2=264

2^2x.1/2 + 2^2x+4=264

2^2x.1/2 + 2^2x.2⁴ = 264

2^2x.(1/2 + 2⁴) = 264

2^2x. 33/2 = 264

2^2x = 16

2^2x = 2⁴

=> 2x = 4

x = 2

a) 10-x-5=-5-7-11

=> 5 - x = -23

=> x = 28
b) |x| -3=0

=> |x| = 3

=> x = 3 hoặc x -3
c) ( 7-|x| ) .(2x-4)=0

=> 7 - |x| = 0 hoặc 2x - 4 = 0

=> |x| = 7 hoặc 2x = 4

=> x = 7 hoặc x = - 7 hoặc x = 2
c)2+3x=-15-19 

=> 2 + 3x = -34

=> 3x = 36

=> x = 12 
 

\(a,10-x-5=-5-7-11\)

\(10-x-5=-23\)

\(10-x=-18\)

\(x=28\)

\(b,|x|-3=0\)

\(|x|=3\)

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

1. 

2|x-6|+7x-2=|x-6|+7x

2|x-6| - |x-6|=7x-(7x-2)

     |x-6|       =    2

=>x-6 = +2

*x-6=2                                                                                                          *x-6 = -2

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

x      =8                                                                                                          x      =    4

2.

|x-5|-7(x+4)=5-7x

|x-5|-7x-28 =5-7x

|x-5|-28       =5-7x+7x

|x-5|-28       =      5

|x-5|            =   5+28

|x-5|            =      33

=>x-5          =    +33

*x-5=33                                                                      *x-5=-33

 x    =38                                                                      x   = -28

3.

3|x+4|-2(x-1)=7-2x

3|x+4|-2x+2 =7-2x

3|x+4|-2       =7-2x+2x

3|x+4|-2       =7

3|x+4|          =7+2

3|x+4|          =  9

  |x+4|          =9:3

  |x+4|          =  3

=>x+4          = +3

*x+4=3                                   *x+4=-3

 x     =-1                                   x    = -7

a) 1/7 - 3/5x = 3/5

3/5x= 1/7 - 3/5 

3/5x = -16/35

x= -16/35 : 3/5 = -16/21

b) 3/7 - 1/2x = 5/3

1/2x = 3/7 - 5/3 = -26/21

x= -26/21 : 1/2 = -52/21

 Thanh Nga làm nốt đc ko bn

Bài làm:

a) \(\left|\frac{1}{2}x-\frac{5}{2}\right|-1=-\frac{1}{2}\)

\(\Leftrightarrow\left|\frac{1}{2}x-\frac{5}{2}\right|=\frac{1}{2}\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x-\frac{5}{2}=\frac{1}{2}\\\frac{1}{2}x-\frac{5}{2}=-\frac{1}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x=3\\\frac{1}{2}x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=6\\x=4\end{cases}}\)

+ Nếu x = 6

\(\left|12-\frac{1}{3}y\right|=\frac{5}{6}\)

\(\Leftrightarrow\orbr{\begin{cases}12-\frac{1}{3}y=\frac{5}{6}\\12-\frac{1}{3}y=-\frac{5}{6}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{3}y=\frac{67}{6}\\\frac{1}{3}y=\frac{77}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}y=\frac{67}{2}\\y=\frac{77}{2}\end{cases}}\)

+ Nếu x = 4

\(\left|8-\frac{1}{3}y\right|=\frac{5}{6}\)

\(\Leftrightarrow\orbr{\begin{cases}8-\frac{1}{3}y=\frac{5}{6}\\8-\frac{1}{3}y=-\frac{5}{6}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{3}y=\frac{43}{6}\\\frac{1}{3}y=\frac{53}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}y=\frac{43}{2}\\y=\frac{53}{2}\end{cases}}\)

Vậy ta có 4 cặp số (x;y) thỏa mãn: \(\left(6;\frac{67}{2}\right);\left(6;\frac{77}{2}\right);\left(4;\frac{43}{2}\right);\left(4;\frac{53}{2}\right)\)

b) \(\frac{3}{2}x-\frac{1}{2}\left(x-\frac{2}{3}\right)=\frac{5}{3}\)

\(\Leftrightarrow\frac{3}{2}x-\frac{1}{2}x+\frac{1}{3}=\frac{5}{3}\)

\(\Leftrightarrow x=\frac{4}{3}\)

Thay vào ta được:

\(\frac{2.\frac{4}{3}+y}{\frac{4}{3}-2y}=\frac{5}{4}\)

\(\Leftrightarrow\frac{32}{3}+4y=\frac{20}{3}-10y\)

\(\Leftrightarrow14y=-4\)

\(\Rightarrow y=-\frac{2}{7}\)

Vậy ta có 1 cặp số (x;y) thỏa mãn: \(\left(\frac{4}{3};-\frac{2}{7}\right)\)

a)(x-5)^4=(x-5)^6

=>(x-5)^4-(x-5)^6=0

=>(x-5)^4.[1-(x-5)^6]=0

=>(x-5)^4=0 hoặc 1-(x-5)^6=0

=>(x-5)^4=0^4 hoặc (x-5)^6=1-0

=>x-5=0 hoặc (x-5)^6 =1

=> x=0+5 hoặc (x-5)^6=1^6  hoặc (x-5)^6=(-1)^6

=>x=5 hoặc x-5=1 hoặc x-5=-1

=>x=5 hoặc x=1+5 hoặc x=-1+5

=>x=5 hoặc x=6 hoặc x=4

 Vậy x=5 hoặc x=6 hoặc x=4

b)(2x+1)^3=125

=>(2x+1)^3=5^3

=>2x+1=5

=>2x=5-1

=>2x=4

=>x=4:2

=>x=2

Vậy x=2

\(\frac{7^{x+2}+7^{x+1}+7^x}{57}=\frac{5^{2x}+5^{2x+1}+5^{2x+3}}{131}\)

<=>\(\frac{7^x\left(7^2+7+1\right)}{57}=\frac{5^{2x}.\left(1+5+5^3\right)}{131}\)

<=>\(\frac{7^x.57}{57}=\frac{5^{2x}.131}{131}\)

<=>\(7^x=5^{2x}\)<=>\(7^x=10^x\)<=>x=0

Vậy x=0