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S= 1/1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729

S= 3 x ( 1/1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729 )

S = 3 +1 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729

S= 3 +1 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729 - 1 - 1/9 -1/27 - 1/81 - 1/243 - 1/729

S = 3 - 1/729 

S= 142/729

1 +  1/3 + 1/9 + 1/27 + 1/81

= 1 + (1/3 + 1/27) + (1/9 + 1/81)

= 1 + (9/27 + 1/27) + (9/81 + 1/81)

= 1 + 10/27 + 10/81

= 1 + 30/81 + 10/81

= 1 + 40/81

= 121/81

ảm ơn bạn nhìu !!!!

A=1/3+1/9+...+1/243

=>3A=1+1/3+...+1/81

=>2A=1-1/243=242/243

=>A=121/243

\(A=1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\)

\(3A=3+1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\)

\(3A-A=\left(3+1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\right)-\left(1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\right)\)

\(2A=3-\dfrac{1}{729}=\dfrac{2186}{729}\)

\(A=\dfrac{2186}{729}\div2=\dfrac{1093}{729}\)

A = \(1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\)

3A = \(3+1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\)

3A - A = ( \(3+1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\) ) - ( \(1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\) )

2A = 3 - \(\dfrac{1}{729}=\dfrac{728}{729}\)

A = \(\dfrac{728}{729}:2=\dfrac{364}{729}\)

1) \(3^x=\dfrac{9^8}{27^3\cdot81^2}\)

\(\Rightarrow3^x=\dfrac{\left(3^2\right)^8}{\left(3^3\right)^3\cdot\left(3^4\right)^2}\)

\(\Rightarrow3^x=\dfrac{3^{16}}{3^{15}}\)

\(\Rightarrow3^x=3\)

\(\Rightarrow x=1\)

2) \(\dfrac{2^{4-x}}{16^5}=32^6\)

\(\Rightarrow\dfrac{2^{4-x}}{\left(2^4\right)^5}=\left(2^5\right)^6\)

\(\Rightarrow\dfrac{2^{4-x}}{2^{20}}=2^{30}\)

\(\Rightarrow2^{4-x}=2^{20}\cdot2^{30}\)

\(\Rightarrow2^{4-x}=2^{50}\)

\(\Rightarrow4-x=50\)

\(\Rightarrow x=-46\)

3) \(\dfrac{2^{2x-3}}{4^{10}}=8^3\cdot16^5\)

\(\Rightarrow\dfrac{2^{2x-3}}{\left(2^2\right)^{10}}=\left(2^3\right)^3\cdot\left(2^4\right)^5\)

\(\Rightarrow\dfrac{2^{2x-3}}{2^{20}}=2^{29}\)

\(\Rightarrow2^{2x-3}=2^{49}\)

\(\Rightarrow2x-3=49\)

\(\Rightarrow2x=52\)

\(\Rightarrow x=26\)

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`(x+1/3)+(x+1/9)+(x+1/27)+(x+1/81)=56/81`

`x+x+x+x+1/3+1/9+1/27=56/81-1/81`

`4x+13/27=55/81`

`4x=55/81-13/27`

`4x=55/81-52/81`

`4x=16/81`

`x=4/108`

Vậy `x=4/108`

\(\dfrac{2}{3}\sqrt{27}-\dfrac{9}{2}\sqrt{\dfrac{16}{81}}+\sqrt{\left(1-\sqrt{3}\right)^2}\)

\(=\dfrac{2}{3}\sqrt{3}-\dfrac{9}{2}.\dfrac{4}{9}+\left(1-\sqrt{3}\right)\)

\(=\dfrac{2\sqrt{3}}{3}-2+1-\sqrt{3}\)

\(=-\dfrac{\sqrt{3}}{3}-1\)

\(=-\dfrac{3+\sqrt{3}}{3}\)

a: A=2^0+2^1+...+2^9

2A=2+2^2+...+2^10

=>A=2^10-1

b: B=1+3+3^2+...+3^6

=>3B=3+3^2+...+3^7

=>2B=3^7-1

=>\(B=\dfrac{3^7-1}{2}\)

1+1/3 + 1/9 + 1/27 + 1/81 + 1/243

=243/243+81/243+27/243 +3/243 +1/243

=\(\dfrac{243+81+27+3+1}{243}\)

=355/243

Lời giải:

$A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}$

$3A=3+1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}$

$3A-A=3-\frac{1}{243}$

$2A=\frac{728}{243}$

$A=\frac{364}{243}$