Tìm x
( x+1/3)+(x+1/9)+(x+1/27)+(x+1/81)=56/81
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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`
Bài 1:
$(y+\frac{1}{3})+(y+\frac{1}{9})+(y+\frac{1}{27})+(y+\frac{1}{81})=\frac{56}{81}$
$(y+y+y+y)+(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81})=\frac{56}{81}$
$4\times y+\frac{40}{81}=\frac{56}{81}$
$4\times y=\frac{56}{81}-\frac{40}{81}=\frac{16}{81}$
$y=\frac{16}{81}:4=\frac{4}{81}$
Bài 2:
$18: \frac{x\times 0,4+0,32}{x}+5=14$
$18: \frac{x\times 0,4+0,32}{x}=14-5=9$
$\frac{x\times 0,4+0,32}{x}=18:9=2$
$x\times 0,4+0,32=2\times x$
$2\times x-x\times 0,4=0,32$
$x\times (2-0,4)=0,32$
$x\times 1,6=0,32$
$x=0,32:1,6=0,2$
\(x+1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}=2\)
\(\Leftrightarrow x+\dfrac{121}{81}=2\)
hay \(x=\dfrac{41}{81}\)
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 \(x=\frac{7}{5}-1=\frac{2}{5}\)