\(S=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}\)

\(3S=3+1+\frac{1}{3}+...+\frac{1}{3^6}\)

\(3S-S=\left(3+1+\frac{1}{3}+...+\frac{1}{3^6}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\right)\)

\(2S=3-\frac{1}{3^7}\)

\(S=\frac{3-\frac{1}{3^7}}{2}\)

S= 1+ \(\frac{1}{3}\)+ \(\frac{1}{9}\)+...+ \(\frac{1}{729}\)+ \(\frac{1}{2187}\).

=> S= 1+ \(\frac{1}{3}\)+ \(\frac{1}{3^2}\)+...+ \(\frac{1}{3^6}\)+ \(\frac{1}{3^7}\).

=>3S= 3+ 1+ \(\frac{1}{3}\)+...+ \(\frac{1}{3^5}\)+ \(\frac{1}{3^6}\).

=> 3S- S=( 3+ 1+ \(\frac{1}{3}\)+...+ \(\frac{1}{3^5}\)+ \(\frac{1}{3^6}\))-( 1+ \(\frac{1}{3}\)+ \(\frac{1}{3^2}\)+...+ \(\frac{1}{3^6}\)+ \(\frac{1}{3^7}\)).

=> 2S= 3- \(\frac{1}{3^7}\).

=> 2S= 3- \(\frac{1}{2187}\).

=> 2S= \(\frac{6560}{2187}\).

=> S= \(\frac{6560}{2187}\): 2.

=> S= \(\frac{3280}{2187}\).

Vậy S= \(\frac{3280}{2187}\).

x = 1 hoặc x=-1/3

k nha

\((3x-1)+3^2=5^2\)

\(\Rightarrow(3x-1)^4+9=25\)

\(\Rightarrow(3x-1)^4=25-9\)

\(\Rightarrow(3x-1)^4=16\)

\(\Rightarrow(3x-1)=(-2)^4=2^4\)

\(\Rightarrow\hept{\begin{cases}3x-1=2\\3x-1=-2\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}3x=2+1\\3x=-2+1\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}3x=3\\3x=-1\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=3\div3\\x=-1\div3\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=1\\x=-\frac{1}{3}\end{cases}}\)

-3701 k mik nha

\([16\times1+(153-x)]=3870\)

\(\Rightarrow16+(153-x)=3870\)

\(\Rightarrow153-x=3870-16\)

\(\Rightarrow153-x=3854\)

\(\Rightarrow x=153-3854\)

\(\Rightarrow x=-3701\)

đè sai hả kg ra kq nếu đề sai k tui đưa đề ms lm cho

\(2^3+(3x+1)^3=24\)

\(\Rightarrow8+(3x+1)^3=24\)

\(\Rightarrow(3x+1)^3=16\)

\(\Rightarrow x\in\varnothing\)

\(x-\frac{3}{4}=1-\frac{5}{6}\)

\(x-\frac{3}{4}=\frac{1}{6}\)

\(x=\frac{1}{6}+\frac{3}{4}\)

\(x=\frac{11}{12}\)

\(x-\frac{3}{4}=1-\frac{5}{6}\)

\(x-\frac{3}{4}=\frac{1}{6}\)

\(x=\frac{1}{6}+\frac{3}{4}\)

\(x=\frac{11}{12}\)

\(2\cdot x+1292=2\)

\(2\cdot x=2-1292\)

\(2\cdot x=-1290\)

\(x=-1290:2\)

\(x=-645\)

135 - 3 ( x + 1 ) = 30

3 ( x + 1 ) = 105

x + 1 = 35

x = 34

\(1005^2\cdot1005^x=1005^7\)

\(1005^{2+x}=1005^7\)

=> x + 2 = 7

=> x = 5

Vậy,........

Tính dọc ta đc

 abcd

+

   abc

bddbc

-=> d+c=c => chịu 

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Bạn phải tự suy nghĩ chứ 

:)    ( ko bt )