tìm x
7+2.(x-3)=11
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`x . 3/7 = 2/3`
`=>x= 2/3 : 3/7`
`=>x= 2/3 . 7/3`
`=>x=14/9`
`-----------`
`x : 8/11 = 11/3`
`=>x= 11/3 . 8/11`
`=>x= 88/33`
`=>x=8/3`
`-----------`
`4/7 . x - 2/3 = 1/5`
`=> 4/7 . x = 1/5 +2/3`
`=>4/7 . x =3/15 + 10/15`
`=>4/7 . x =13/15`
`=>x= 13/15 : 4/7`
`=>x= 13/15 xx 7/4`
`=>x= 91/60`
Lời giải:
$x.\frac{3}{7}=\frac{2}{3}$
$x=\frac{2}{3}: \frac{3}{7}=\frac{14}{9}$
-----------
$x: \frac{8}{11}=\frac{11}{3}$
$x=\frac{11}{3}.\frac{8}{11}=\frac{8}{3}$
-----------
$\frac{4}{7}x-\frac{2}{3}=\frac{1}{5}$
$\frac{4}{7}x=\frac{2}{3}+\frac{1}{5}=\frac{13}{15}$
$x=\frac{13}{15}: \frac{4}{7}=\frac{91}{60}$
Bài 1:
a: \(x=\dfrac{2}{3}:\dfrac{3}{5}=\dfrac{2}{3}\cdot\dfrac{5}{3}=\dfrac{10}{9}\)
b: \(x=\dfrac{17}{8}:\dfrac{7}{17}=\dfrac{17}{8}\cdot\dfrac{17}{7}=\dfrac{289}{56}\)
c: \(x=-\dfrac{3}{4}:\dfrac{7}{12}=\dfrac{-3}{4}\cdot\dfrac{12}{7}=\dfrac{-63}{28}=-\dfrac{9}{4}\)
d: \(\Leftrightarrow x\cdot\dfrac{1}{6}=\dfrac{3}{8}-\dfrac{1}{4}=\dfrac{1}{4}\)
hay \(x=\dfrac{1}{4}:\dfrac{1}{6}=\dfrac{3}{2}\)
e: \(\Leftrightarrow\dfrac{1}{2}:x=-4-\dfrac{1}{3}=-\dfrac{17}{3}\)
hay \(x=-\dfrac{1}{2}:\dfrac{17}{3}=\dfrac{-3}{34}\)
a: x=9/2-3/7=57/14
b: =>x=7/5x5/7=1
c: =>x=11/3:3/8=11/3x8/3=88/9
a : x = \(\dfrac{9}{2}\) - \(\dfrac{3}{7}\) = \(\dfrac{57}{14}\)
b : = > x = \(\dfrac{7}{5}\) x\(\dfrac{5}{7}\) = 1
c : = > \(\dfrac{11}{3}\) : \(\dfrac{3}{8}\) = \(\dfrac{11}{3}\) x \(\dfrac{8}{3}\) = \(\dfrac{88}{9}\)
\(a,\frac{39}{7}:x=13\)
\(x=\frac{39}{7}:13=\frac{3}{7}\)
\(b,\frac{2}{3}\cdot x+\frac{1}{2}=\frac{1}{10}\)
\(\frac{2}{3}.x=\frac{1}{10}-\frac{1}{2}=-\frac{2}{5}\)
\(x=-\frac{2}{5}:\frac{2}{3}=-\frac{3}{5}\)
\(c,x:\frac{8}{11}=\frac{11}{3}\)
\(x=\frac{11}{3}\cdot\frac{8}{11}=\frac{8}{3}\)
\(d,\frac{2}{9}-\frac{7}{8}\cdot x=\frac{1}{3}\)
\(\frac{7}{8}.x=\frac{2}{9}-\frac{1}{3}=-\frac{1}{9}\)
\(x=-\frac{1}{9}:\frac{7}{8}=-\frac{8}{63}\)
cảm ơn Ngyễn Thị Ngọc Ánh nha bạn lm đúng mà mình k nhầm xl bạn nhìu
7+2.(x-3)=11
2.(x-3)=11-7
x-3=4:2
x=2+3
x=5
`7+2.(x-3)=11`
`2.(x-3)=11-7`
`2.(x-3)=4`
`x-3=4:2`
`x-3=2`
`x=2+3`
`x=5`
Vậy `x=5`
`#LeMichael`