(2x-16)^7 = 128
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) (x - 1)2 = 1.
<=> x - 1 = 1 hoặc x - 1 = -1.
<=> x = 2 hoặc x = 0.
b) 72x - 6 = 49.
<=> 72x - 6 = 72.
<=> 2x - 6 = 2.
<=> x = 4.
c) (2x - 16)7 = 128.
<=> (2x - 16)7 = 27.
<=> 2x - 16 = 2.
<=> x = 9.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\left(2x-16\right)^7=128\\ \Rightarrow\left(2x-16\right)^7=2^7\\ \Rightarrow2x-16=2\\ \Rightarrow2x=18\\ \Rightarrow x=9\\ b,x^3.x^2=2^8:2^3\\ \Rightarrow x^5=2^5\\ \Rightarrow x=5\\ c,3^{x-3}-3^2=2.3^2\\ \Rightarrow3^{x-3}-9=18\\ \Rightarrow3^{x-3}=27\\ \Rightarrow3^{x-3}=3^3\\ \Rightarrow x-3=3\\ \Rightarrow x=6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b) 2x + 15 = -27
2x = -27 - 15
2x = -42
x = -42 : 2
x = -21
c) -765 - (305 + x) = 100
305 + x = -765 - 100
305 + x = -865
x = -865 - 305
x = -1170
d) 2x : 4 = 16
2x = 16 x 4
2x = 64
2x = 26
⇒ x = 6
e) 128 - 3.(x + 4) = 23
3.(x + 4) = 128 - 23
3.(x + 4) = 105
x + 4 = 105 : 3
x + 4 = 35
x = 35 - 4
x = 31
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt A=7/2+7/4+...+7/256
=>2A=7+7/2+...+7/128
=>A=7-7/256=7*255/256=1785/256
![](https://rs.olm.vn/images/avt/0.png?1311)
Tìm x :
a. -x + 17 = -12
x = (-12) -17
x = -29
b. -30 + (32-x) = 10
32-x = 10-(-30)
32-x = 40
x = 32 - 40
x = -8
c. \(|2x-1|\) = 7
=> 2x - 1 = 7 hoặc => 2x - 1 = -7
=> x = 4 => x = -3
d. \(|x+3|\) - 5 = 7
\(|x+3|\) = 7 + 5
\(|x+3|\) = 12
=> x + 3 = 12 hoặc => x + 3 = -12
=> x = 12 - 3 => x = (-12) - 3
=> x = 9 => x = -15
e. 16 - ( 42 + x ) = 50
42 + x = 16 - 50
42 + x = -34
x = (-34) - 42
x = -76
g. 128 - 3 . ( x + 4 ) = 23
3 . ( x + 4 ) = 128 - 23
3 . ( x + 4 ) = 105
x + 4 = 105 : 3
x + 4 = 35
x = 35 - 4
x = 31
![](https://rs.olm.vn/images/avt/0.png?1311)
( 2x + 1 )3 = 96
( 2x + 1 )3 = 63
2x + 1 = 6
2x = 6 - 1
2x = 5
x = 5 : 2
x = 2,5
16x < 1284
x = 0;1;2;3;4;........
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{16^7\times5^4-8^8\times16\times5^2\times125}{16^7\times25^2-8^7\times128\times5^5}\)
\(=\frac{16^7\times5^4-8^8\times2^4\times5^2\times5^3}{16^7\times5^4-8^7\times2^7\times5^5}\)
\(=\frac{8}{2^3}\)
\(=\frac{8}{8}\)
\(=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Trần Anh Dũng
b) \(720:\left[41-\left(2x-5\right)\right]=2^3.5\)
\(720:\left[41-2x+5\right]=40\)
\(46-2x=720:40\)
\(46-2x=18\)
\(2x=28\)
\(x=14\)
vậy \(x=14\)
a) \(2^x+2-2^x=96\)
\(0^x=96-2\)
\(0^x=94\)
\(\Rightarrow x\in\varnothing\)
vậy \(x\in\varnothing\)
c) \(16^x< 128^4\)
\(2^{4x}< 2^{28}\)
\(\Rightarrow4x< 28\)
\(\Rightarrow x< 7\)
vậy \(x< 7\)
a, Ta có \(2^{x+2}-2^x=96\)
\(\Rightarrow2^x.2^2-2^x=96\)
\(\Rightarrow2^x.\left(2^2-1\right)=96\)
\(\Rightarrow2^x.3=96\)
\(\Rightarrow2^x=32\)
\(\Rightarrow2^x=2^5\)
\(\Rightarrow x=5\)
b, \(720:\left[51-\left(2x-5\right)\right]=2^3.5\)
\(\Rightarrow720:\left[51-\left(2x-5\right)\right]=40\)
\(\Rightarrow\left[51-\left(2x-5\right)\right]=18\)
\(\Rightarrow2x-5=33\)
\(\Rightarrow2x=38\)
\(\Rightarrow x=19\)
c,\(16^x< 128^4\)
\(\Rightarrow\left(2^4\right)^x< \left(2^7\right)^4\)
\(\Rightarrow\left(2^4\right)^x< \left(2^4\right)^7\)
\(\Rightarrow x< 7\)
\(\left(2x-16\right)^7=128\\ \Rightarrow\left(2x-16\right)^7=2^7\\ \Rightarrow2x-16=2\\ \Rightarrow2x=18\\ \Rightarrow x=9\)
\(\left(2x-16\right)^7=128\)
\(\left(2x-16\right)^7=2^7\)
\(2x-16=2\)
\(2x=2+16\)
\(x=9\)