1/2.4 . 1/4.6. …. . 1/(x(x+2))= 7/16 Giúp mik vs mik đg cần gấp 09/11/2021 Bởi Alaia 1/2.4 . 1/4.6. …. . 1/(x(x+2))= 7/16 Giúp mik vs mik đg cần gấp
Giải thích các bước giải: Có: $\frac{1}{2.4} + \frac{1}{4.6} + … + \frac{1}{x(x+2)}=\frac{7}{16}$ $=>\frac{2}{2.4} + \frac{2}{4.6} + … +\frac{2}{x.(x+2)}=\frac{7}{8}$ $=>\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+…+\frac{1}{x}-\frac{1}{x+2}=\frac{7}{8}$ $=>\frac{1}{2}-\frac{1}{x+2}=\frac{7}{8}$ $=>\frac{3}{3x+6}=\frac{3}{-8}$ $=>3x+6=-8$ $=>3x=-14$ $=>x=\frac{-14}{3}$ Bình luận
Sửa đề: $⇔$$\dfrac{1}{2.4}.\dfrac{1}{4.6}….. \dfrac{1}{(x.(x+2)}=\dfrac{7}{16}$ $⇔$$\dfrac{2}{2.4}.\dfrac{2}{4.6}….. \dfrac{2}{(x.(x+2)}=\dfrac{7}{16}.2$ $⇔$$\dfrac{1}{2}-\dfrac{1}{4} + \dfrac{1}{4} – \dfrac{1}{6} + …… + \dfrac{1}{x} – \dfrac{1}{x+2}=\dfrac{7}{8}$ $⇔$$\dfrac{1}{2} – \dfrac{1}{x+2}=\dfrac{7}{8}$ $⇔$$\dfrac{1}{x+2} = \dfrac{1}{2} – \dfrac{7}{8}$ $⇔$$\dfrac{1}{x+2} = \dfrac{-3}{8}$ $⇔$$\dfrac{3}{3x+6} = \dfrac{3}{-8}$ $⇔$$3x+6=-8$ $⇔$$3x=-14$ $⇔$$x = \dfrac{-14}{3}$ Vậy $x = \dfrac{-14}{3}$ Bình luận
Giải thích các bước giải:
Có:
$\frac{1}{2.4} + \frac{1}{4.6} + … + \frac{1}{x(x+2)}=\frac{7}{16}$
$=>\frac{2}{2.4} + \frac{2}{4.6} + … +\frac{2}{x.(x+2)}=\frac{7}{8}$
$=>\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+…+\frac{1}{x}-\frac{1}{x+2}=\frac{7}{8}$
$=>\frac{1}{2}-\frac{1}{x+2}=\frac{7}{8}$
$=>\frac{3}{3x+6}=\frac{3}{-8}$
$=>3x+6=-8$
$=>3x=-14$
$=>x=\frac{-14}{3}$
Sửa đề:
$⇔$$\dfrac{1}{2.4}.\dfrac{1}{4.6}….. \dfrac{1}{(x.(x+2)}=\dfrac{7}{16}$
$⇔$$\dfrac{2}{2.4}.\dfrac{2}{4.6}….. \dfrac{2}{(x.(x+2)}=\dfrac{7}{16}.2$
$⇔$$\dfrac{1}{2}-\dfrac{1}{4} + \dfrac{1}{4} – \dfrac{1}{6} + …… + \dfrac{1}{x} – \dfrac{1}{x+2}=\dfrac{7}{8}$
$⇔$$\dfrac{1}{2} – \dfrac{1}{x+2}=\dfrac{7}{8}$
$⇔$$\dfrac{1}{x+2} = \dfrac{1}{2} – \dfrac{7}{8}$
$⇔$$\dfrac{1}{x+2} = \dfrac{-3}{8}$
$⇔$$\dfrac{3}{3x+6} = \dfrac{3}{-8}$
$⇔$$3x+6=-8$
$⇔$$3x=-14$
$⇔$$x = \dfrac{-14}{3}$
Vậy $x = \dfrac{-14}{3}$