giúp mình mình còn 1 phút nữa đi học

giúp mình đi huhuhu

b) [3x - 6] . 3⁷ = 3⁹

3x - 6 = 3⁹ : 3⁷

3x - 6 = 3²

3x = 3² - 6

3x = 9 - 6

3x = 3

x = 1

c) 2^x-2 + 13 . 2² = 17 . 2²

2^x-2 + 13 . 2² = 68

2^x-2 + 13 = 68 : 2²

2^x-2 + 13 = 17

2^x-2 = 17 - 13

2^x-2 = 4

=> 2^x-2 = 2²

=> x = 4

\(2^x.2^{x+4+}=256\\ 2^{x+x+4}=256\\ 2^{2x+4}=256\\ 2^{2x+4}=2^8\\ \Rightarrow 2x+4=8\\ \Rightarrow2x=8-4\\ \Rightarrow2x=4\\ \Rightarrow x=\dfrac{4}{2}=2\)

=>2^(x+x+4)=2^8

=>2x+4=8

=>2x=4

=>x=2

Đặt bểu thức là A

Ta có :

\(A=2^0.2^1.2^2.....2^8\)

\(\Rightarrow A=2^{0+1+2+3+...+8}\)

\(\Rightarrow A=2^{36}\)

1x2x4x8x16x...x256=2⁰x2¹x2²x2³x...x2⁸

                                =2^{0+1+2+3+4+...+8}

                                =2³⁶

   vậy 1x2x4x8x..x256=2³⁶

giúp mình nha

a.\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{2015}\right)=\frac{1}{2}.\frac{2}{3}...\frac{2014}{2015}=\frac{1.2.3...2014}{2.3...2015}=\frac{1}{2015}\)

b.\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{256}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{128}-\frac{1}{256}=1-\frac{1}{256}=\frac{255}{256}\)

c.\(\frac{5}{2}+\frac{5}{4}+\frac{5}{8}+...+\frac{5}{256}=5\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{256}\right)=5.\frac{255}{256}=\frac{1275}{256}\)

d.14,35+(13,7-13,6).1=14,35+0,1.1=14,35+0,1=14,45

a.1/2015

(x + 32) - 17.2=42

(x + 32) - 34 = 42

x + 32 = 42 + 34

x + 32 = 76

      x = 76 - 32

      x = 44

Đặt x/1=y/2=k

=>x=k; y=2k

x^4*y^4=256

=>k^4*16k^4=256

=>16k^8=256

=>k^8=16

=>k^4=4

=>k^2=2

TH1: \(k=\sqrt{2}\)

=>\(x=\sqrt{2};y=2\sqrt{2}\)

TH2: \(k=-\sqrt{2}\)

=>\(x=-\sqrt{2};y=-2\sqrt{2}\)