Tìm nguyên x
$\frac{2}{2.4}+\frac{2}{4.6}+…+\frac{2}{x(x+2)}=\frac{4}{9}$
Cảm ơn ạ!
Tìm nguyên x $\frac{2}{2.4}+\frac{2}{4.6}+…+\frac{2}{x(x+2)}=\frac{4}{9}$ Cảm ơn ạ!
By Anna
By Anna
Tìm nguyên x
$\frac{2}{2.4}+\frac{2}{4.6}+…+\frac{2}{x(x+2)}=\frac{4}{9}$
Cảm ơn ạ!
`2/(2.4) + 2/(4.6) + …+ 2/(x.(x+2)) = 4/9`
`=> 1/2 – 1/4 + 1/4 – 1/6 + ….+1/x – 1/(x+2) = 4/9`
`=> 1/2 – 1/(x+2) = 4/9`
`=> 1/(x+2) = 1/2 – 4/9`
`=> 1/(x+2) = 1/18`
`=> x+2 = 1.18`
`=> x+2 = 1.18`
`=> x+2 = 18`
`=> x = 16`
(Chúc bạn học tốt)
Đáp án + Giải thích các bước giải:
`(2)/(2.4)+(2)/(4.6)+…+(2)/(x(x+2))=(4)/(9)`
`=>(1)/(2)-(1)/(4)+(1)/(4)-(1)/(6)+…+(1)/(x)-(1)/(x+2)=(4)/(9)`
`=>(1)/(2)-(1)/(x+2)=(4)/(9)`
`=>(1)/(x+2)=(1)/(2)-(4)/(9)`
`=>(1)/(x+2)=(1)/(18)`
`=>x+2=18`
`=>x=16`