a, 7\(\dfrac{3}{5}\) : \(x\) = 5\(\dfrac{4}{15}\) - 1\(\dfrac{1}{6}\)

     \(\dfrac{38}{5}\) : \(x\) = \(\dfrac{79}{15}\) - \(\dfrac{7}{6}\)

              \(x\) = \(\dfrac{41}{10}\)

             \(x\) = \(\dfrac{38}{5}\) : \(\dfrac{41}{10}\)

              \(x\) = \(\dfrac{76}{41}\)

b, \(x\) \(\times\) 2\(\dfrac{2}{3}\) = 3\(\dfrac{4}{8}\) + 6\(\dfrac{5}{12}\)

    \(x\) \(\times\) \(\dfrac{8}{3}\)  = \(\dfrac{7}{2}\) + \(\dfrac{77}{12}\)

     \(x\) \(\times\) \(\dfrac{8}{3}\) = \(\dfrac{119}{12}\)

     \(x\)          = \(\dfrac{119}{12}\)

     \(x\)          = \(\dfrac{119}{12}\): \(\dfrac{8}{3}\)

     \(x\)           = \(\dfrac{119}{32}\)

Trả lời :
0 x 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 = 0.

#hoctot

=0 

k mk

mk mún có

bằng 0

\(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+...+\dfrac{1}{x}=\dfrac{127}{256}\)

Đặt VT là A

\(\Rightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{2}{x}\)

\(2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{2}{x}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{x}\right)=\dfrac{127}{256}\)

\(\Leftrightarrow A=1-\dfrac{1}{x}=\dfrac{127}{256}\)

\(\Leftrightarrow\dfrac{1}{x}=\dfrac{129}{256}\)

\(\Rightarrow x=\dfrac{256}{129}\)

419 bn nha 

= 419^^

         HT

lớp 6 ai chơi được

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

Bạn ơi , 3x - 1 là số mũ phải không . Gỉa sử vậy , ta có :

\(\frac{256}{81}=\frac{4^4}{\left(-3\right)^4}=\frac{1}{\frac{\left(-3\right)^4}{4^4}}=\frac{1}{\left(\frac{-3}{4}\right)^4}=\left(\frac{-3}{4}\right)^{-4}=\left(\frac{-3}{4}\right)^{3x-1}\)=> 3x - 1 = -4 => 3x = -4 + 1 = -3  

=> x = -3 : 3 = -1 

\(\left(\frac{-3}{4}\right)^{3x-1}=\left(\frac{4}{3}\right)^4=\left(\frac{1}{\frac{3}{4}}\right)^4\left[\left(\frac{3}{4}\right)^{-1}\right]^4=\left(\frac{3}{4}\right)^{-4}=\left(-\frac{3}{4}\right)^{-4}\) 

(Lũy thừa số mũ chẵn thì \(a^x\)= \(\left(-a\right)^x\))

\(\Rightarrow\)3m - 1 = -4 \(\Rightarrow\)3m = -3 \(\Rightarrow\)m = -1