`(x+2)/3=(x-4)/5`
`<=> (x+2)*5=3(x-4)`
`<=> 5x+10=3x-12`
`<=> 5x-3x=-12-10`
`<=> 2x=-22`
`<=> x=-11`
(x+2)/3=(x-4)/5
<=> (x+2).5=(x-4).3
<=> 5x +10 =3x -12
<=> 5x-3x=-12-10
<=> 2x =-22
<=> x=-11
Vậy x=-11
`(x+2)/3=(x-4)/5`
`<=> (x+2)*5=3(x-4)`
`<=> 5x+10=3x-12`
`<=> 5x-3x=-12-10`
`<=> 2x=-22`
`<=> x=-11`
(x+2)/3=(x-4)/5
<=> (x+2).5=(x-4).3
<=> 5x +10 =3x -12
<=> 5x-3x=-12-10
<=> 2x =-22
<=> x=-11
Vậy x=-11
\(\dfrac{x+3}{-4}=\dfrac{1-x}{5}\)
Giúp mik với ạ
\(\dfrac{2}{3}x+\dfrac{1}{2}x=\dfrac{5}{2}:3\dfrac{3}{4}\) giúp với ạ
\(50\text{%}\dfrac{-3}{4}x^2=\dfrac{-5}{2}\)
Giúp mình với ạ
\(\left(\dfrac{2}{3}x+\dfrac{1}{2}\right).\left(-2x+3\right)=0\)
giúp mik với ạ
\(\left(x^2-1\dfrac{9}{16}\right).\left(x^3+\dfrac{1}{8}\right)=0\)
giúp mik với ạ
a, x =\(\dfrac{-2}{7}\) +\(\dfrac{9}{7}\)
b,\(\dfrac{x}{3}\) =\(\dfrac{2}{5}\) +\(\dfrac{-4}{3}\)
Làm giúp mik vs mình đang cần gấp!
\(a,\left(\dfrac{37}{9}+\dfrac{13}{4}\right)x\dfrac{9}{4}+\dfrac{11}{4}\) b,\(1+\left(\dfrac{9}{10}-\dfrac{-4}{5}\right):\dfrac{19}{6}\)
c,\(\dfrac{1}{4}-\dfrac{3}{2}+\dfrac{1}{2}x\dfrac{12}{5}\)
Giúp mik nha:>
\(\dfrac{3}{2}X-0,2=\dfrac{3}{5}\)
\(\dfrac{1}{3}+x=\dfrac{3}{4}\)
\(1\dfrac{1}{2}x-\dfrac{2}{5}=\dfrac{1}{4}\)
\(\dfrac{11}{8}-\dfrac{3}{8}.x=\dfrac{1}{8}\)
giúp với
\(\dfrac{1}{4}\) x \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) x \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) x \(\dfrac{1}{7}\)
Giúp mik với = <
Cảm ơn = >