`Answer:`

\(\left(x+\frac{1}{3}\right)+\left(x+\frac{1}{9}\right)+\left(x+\frac{1}{27}\right)+...+\left(x+\frac{1}{729}\right)=\frac{4209}{729}\)

\(\Leftrightarrow\left(x+\frac{1}{3}\right)+\left(x+\frac{1}{3^2}\right)+\left(x+\frac{1}{3^3}\right)+...+\left(x+\frac{1}{3^6}\right)=\frac{4209}{729}\)

\(\Leftrightarrow6x+\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^6}\right)=\frac{4209}{729}\text{(*)}\)

Đặt \(N=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^6}\)

\(\Leftrightarrow3N=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^5}\)

\(\Leftrightarrow3N-N=\left(1+\frac{1}{3}+\frac{1}{3^2}+..+\frac{1}{3^5}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^6}\right)\)

\(\Leftrightarrow2N=1-\frac{1}{3^6}\)

\(\Leftrightarrow2N=\frac{728}{729}\)

\(\Leftrightarrow N=\frac{364}{729}\)

\(\text{(*)}\Leftrightarrow6x+\frac{364}{729}=\frac{4209}{729}\)

\(\Leftrightarrow6x=\frac{3845}{729}\)

\(\Leftrightarrow x=\frac{3845}{4374}\)

x ở đâu em

(2x+1)³=729
=>2x+1=\(\sqrt[3]{729}\)
=>2x+1=9
=>2x=8
=>x=4

tick đúng nha

3^{x}.3^{x+1}.3^{x+2}>=729`

`=>3^{x+x+1+x+2}>=3^6`

`=>3^{3x+3}>=3^6`

`=>3x+3>=6`

`=>3x>=3=>x>=1`

Ta có: \(A=\left[6.\left(\frac{-1}{3}\right)^2-\left(-\frac{1}{3}\right)+1\right]:\left(\frac{-1}{3}-1\right)\)

\(\Rightarrow A=\left[6.\frac{1}{9}+\frac{1}{3}+1\right]:\left(\frac{-1}{3}-\frac{3}{3}\right)\)

\(\Rightarrow A=\left[\frac{2}{3}+\frac{1}{3}+1\right]:\frac{-4}{3}\)

\(\Rightarrow A=\left[1+1\right].\frac{-3}{4}=2.\frac{-3}{4}=\frac{-3}{2}\)

Mà \(B=\left(729-1^3\right)\left(729-2^3\right)\left(729-3^3\right)...\left(729-125^3\right)\)

\(=\left(729-1^3\right)\left(729-2^3\right)...\left(729-9^3\right)...\left(729-125^3\right)\)

\(=\left(729-1^3\right)\left(729-2^3\right)...0...\left(729-125^3\right)=0\)

Vì \(\frac{-3}{2}< 0\)nên A < B

x nguyên dương => x>0

x<7 vì nếu x = 7 =>19*7^2>729

x=1 => 28*y^2 = 729-19*1^2=710 =>y^2=355/14 => y không nguyên dương (loại)

thử dần ta chả thấy trường hợp nào đúng 

=> không thể tìm x,y thỏa mãn

trong đó có  ___  để ghi phân số      ^   là số mũ

\(x\) \(\times\) \(\dfrac{1}{4}\) = 6 : 1 : 2

\(x\) \(\times\) \(\dfrac{1}{4}\) = 6:2

\(x\) \(\times\) \(\dfrac{1}{4}\) =  3

\(x\)         = 3 : \(\dfrac{1}{4}\)

\(x\)        = 12

Mình nhầm đoạn 6:1/2.nhờ bạn giải lại hộ mình với

\(x\)\(\times\) \(\dfrac{1}{4}\) = 6 : \(\dfrac{1}{2}\)

\(x\) \(\times\) \(\dfrac{1}{4}\) = 6 \(\times\) \(\dfrac{2}{1}\)

\(x\) \(\times\) \(\dfrac{1}{4}\) = 12

\(x\)       =  12 : \(\dfrac{1}{4}\)

\(x\)      = 48

Tính nhanh:

                         A =         \(\dfrac{2}{3}\) + \(\dfrac{2}{9}\) + \(\dfrac{2}{27}\) + \(\dfrac{2}{81}\) + \(\dfrac{2}{243}\) + \(\dfrac{2}{729}\)

                  A \(\times\) 3 =  2  +  \(\dfrac{2}{3}\) + \(\dfrac{2}{9}\) + \(\dfrac{2}{27}\) + \(\dfrac{2}{81}\) + \(\dfrac{2}{243}\) 

            A \(\times\) 3 - A = 2 - \(\dfrac{2}{729}\)

           A \(\times\) ( 3 - 1) = 2 \(\times\) ( 1 - \(\dfrac{1}{729}\))

             A \(\times\) 2       = 2 \(\times\) \(\dfrac{728}{729}\)

            A = 2 \(\times\) \(\dfrac{728}{729}\) : 2

           A = \(\dfrac{728}{729}\)

c, tính:

 \(\dfrac{4}{5}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{6}\)

= \(\dfrac{4}{5}\) + \(\dfrac{1}{3}\)

= \(\dfrac{17}{15}\)

\(\left(\frac{3^x}{3}\right)^2=\frac{1}{729}\Rightarrow3^{x-1}=\frac{1}{27}\Rightarrow x-1=-3\Rightarrow x=-2\)

\(x=-2\)

x = -2 nhé bn

tích mk nha các bn

>_< học tốt