\(a,x\left(y-2\right)=8\\ \Rightarrow x;\left(y-2\right)\inƯ\left(8\right)=\left\{-8;-4;-2;-1;1;2;4;8\right\}\)

Vậy \(\left(x;y\right)=\left(-8;1\right),\left(-4;0\right),\left(-2;-2\right),\left(-1;-6\right),\left(2;6\right),\left(4;4\right),\left(8;3\right)\)

\(b,\left(x-1\right)\left(y-2\right)=9\\ \Rightarrow\left(x-1\right),\left(y-2\right)\inƯ\left(9\right)=\left\{-9;-3;-1;1;3;9\right\}\)

Vậy \(\left(x;y\right)=\left(-8;1\right),\left(-2;-1\right),\left(0;-7\right),\left(2;11\right),\left(4;5\right),\left(10;3\right)\)

\(c,\left(x+1\right)\left(y-2\right)=15\\ \Rightarrow\left(x+1\right),\left(y-2\right)\inƯ\left(15\right)=\left\{-15;-1;1;15\right\}\)

Vậy \(\left(x;y\right)=\left(-16;1\right),\left(-2;-13\right),\left(0;17\right),\left(14;3\right)\)

2^x . 4 = 128

2^x = 128 : 4

2^x = 32

=> 2⁵ = 32 => x = 5

2^x . 4 = 128

=> 2^x = 128 : 4

=> 2^x = 32 = 2⁵

=> x = 5

2^x . 4 = 128

2^x = 128 : 4

2^x = 32

2^x = 2⁵

=> x = 5

a)2^Xx4=128

   2^X    =128:4

   2^X    =32

   2^X    =2^5

=> X=5

b)X^15=X

=>X=1.Vì 1 mũ bao nhiêu vẫn = 1

c)16^x<128

X=1 vì 16^2= 256 mà lại lớn hơn 128,16^3 thì còn lớn hơn

=>X=1

đúng thì k nha :)

\(128^4=\left(2^7\right)^4=2^{28}=\left(2^5\right)^{5,6}=32^{5,6}\)

theo đề ta có

\(32^x< 128^4=32^{5,6}\Rightarrow x\in\left\{1;2;3;4;5\right\}\)

a, 128 – 3(x+4) = 23

b,  12 x - 4 3 . 8 3 = 4 . 8 4

c, [(4x+28).3+55]:5 = 35

d, 720:[41 – (2x – 5)] =  2 3 . 5