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a) (x+1)/3 + (x+1)/9 + (x+1)/27 + (x+1)/81 = 56/81
<=> (27x+27)/81 + (9x+9)/81 + (3x+3)/81 + (x+1)/81 = 56/81 (quy đồng)
<=> 27x + 9x + 3x + x + 27 + 9 + 3 + 1 = 56 (khử mẫu)
<=> 40x = 56- 40 = 16
<=> x = 16/40 = 2/5
~ hok tốt ~
\(3S=241+81+27+9+...+\dfrac{1}{9}+\dfrac{1}{27}\)
\(2S=3S-S=241-\dfrac{1}{81}=\dfrac{241x81-1}{81}\)
\(\Rightarrow S=\dfrac{241x81-1}{2x81}\)
4 x y + (1/3 + 1/9 + 1/27 + 1/81) = 56/81
4 x y + (27/81 + 9/81 + 3/81 + 1/81) = 56/81
4 x y + 40/81 = 56/81
4 x y = 56/81 - 40/81 = 16/81
y = 4/81
\(\left(y+\frac{1}{3}\right) +\left(y+\frac{1}{9}\right)+ \left(y+\frac{1}{27}\right)+\left(y+\frac{1}{81}\right)=\frac{56}{81}\)
\(y\cdot4+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\right)=\frac{56}{81}\)
\(y\cdot4+\frac{40}{81}=\frac{56}{81}\)
\(y\cdot4=\frac{56}{81}-\frac{40}{81}\)
\(y\cdot4=\frac{16}{81}\)
\(y=\frac{16}{81}:4\)
\(y=\frac{4}{81}\)
\(G=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\\ G=\dfrac{81}{243}+\dfrac{27}{243}+\dfrac{9}{243}+\dfrac{3}{243}+\dfrac{1}{243}\\ G=\dfrac{121}{243}\)
Đặt \(V=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+...+\dfrac{1}{729}+\dfrac{1}{2187}\)
\(\Rightarrow3V=3.\left(\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+...+\dfrac{1}{729}+\dfrac{1}{2187}\right)\)
\(\Rightarrow3V=1+\left(\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+...+\dfrac{1}{729}\right)\)
\(\Rightarrow3V=1+V-\dfrac{1}{2187}\)
\(\Rightarrow2V=1-\dfrac{1}{2187}\)
\(\Rightarrow V=\dfrac{1093}{2187}\).
A = 1/3 + 1/9 + 1/27 + 1/81 +...+1/729 + 1/2187
3A = 1 + 1/3 + 1/9 + 1/27 + 1/81 +...+1/729
=>2A = 1 - 1/2187
=> A = ....
`@` `\text {Ans}`
`\downarrow`
\(1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}?\)
Đặt \(A=1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\)
`3A=`\(3\times\left(1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\right)\)
`3A =`\(3+\dfrac{3}{3}+\dfrac{3}{9}+\dfrac{3}{27}+\dfrac{3}{81}+\dfrac{3}{243}+\dfrac{3}{729}\)
`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 =`\(3-\dfrac{1}{729}\)
`2A=`\(\dfrac{2186}{729}\)
\(A=\dfrac{2186}{729}\div2=\dfrac{1093}{729}\)
ta có :
\(\frac{x+1}{3}+\frac{x+1}{9}+\frac{x+1}{27}+\frac{x+1}{81}=\left(x+1\right)\times\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\right)\)
\(=\left(x+1\right)\times\frac{40}{81}=\frac{56}{81}\text{ nên }x+1=\frac{56}{40}=\frac{7}{5}\)
vậy \(x=\frac{7}{5}-1=\frac{2}{5}\)
a, 1+1/3+1/9+1/27+1/81
=81/81+27/81+9/81+3/81+1/81
=81+27+9+3+1/81
=121/81