\(\dfrac{3}{x+2}=\dfrac{5}{x-3}\left(x\ne-2;x\ne3\right)\)

suy ra: \(3\left(x-3\right)=5\left(x+2\right)\\ < =>3x-9=5x+10\\ < =>3x-5x=10+9\\ < =>-2x=19\\ < =>x=-\dfrac{19}{2}\left(tm\right)\)

\(\dfrac{3}{x+2}=\dfrac{5}{x-3}\)ĐKXĐ \(\left\{{}\begin{matrix}x+2\ne0\\x-3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-2\\x\ne3\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{3\left(x-3\right)}{\left(x+2\right)\left(x-3\right)}=\dfrac{5\left(x+2\right)}{\left(x+2\right)\left(x-3\right)}\)

`<=> 3(x-3) =5 (x+2)`

`<=> 3x-9 = 5x+10`

`<=>3x -5x=10+9`

`<=> -2x=19`

`<=>x=-19/2`

\(\dfrac{3}{x+2}=\dfrac{5}{x-3}\text{ĐKXĐ}:x\ne-2;3\)

\(\Leftrightarrow\dfrac{3\left(x-3\right)}{\left(x+2\right)\left(x-3\right)}=\dfrac{5\left(x+2\right)}{\left(x+2\right)\left(x-3\right)}MTC:\left(x+2\right)\left(x-3\right)\)

\(\Rightarrow3x-9=5x+10\)

\(\Leftrightarrow3x-9-5x-10=0\)

\(\Leftrightarrow-2x-19=0\)

\(\Leftrightarrow-2x=19\)

\(\Leftrightarrow x=\dfrac{-19}{2}\left(\text{nhận}\right)\)

\(\text{Vậy phương trình có tập nghiệm là }S=\left\{\dfrac{-19}{2}\right\}\)

a)=8/15 + 10/15 = 18/15 = 6/5

b)=24/56 + 28/56 = 52/56 = 13/14

a) 6 phan 5

b) 13 phan 14

= \(\frac{10}{18}\) = \(\frac{2}{5}\).

= \(\frac{7}{15}\) ( minh khong rut gon duoc vi no khong chia het co so nao ca )

Bn viết lại đề bài rõ hơn được ko?

6\85/7

\(\frac{2}{11}:\frac{1}{3}.\frac{3}{2}\)

\(=\frac{2}{11}.3.\frac{3}{2}\)

\(=\frac{2.3.3}{11.2}\)

\(=\frac{9}{11}\)

Dấu . là dấu nhân nha!

Mk trả lời lúc 14:12

\(\frac{2}{11}:\frac{1}{3}.\frac{3}{2}\)

\(=\frac{6}{11}.\frac{3}{2}\)

\(=\frac{9}{11}\)

9/11 nha bn

\(x=\dfrac{5}{3}-\dfrac{2}{7}=\dfrac{29}{21}\)

X + 2/7 = 5/3

X          = 5/3 - 2/7

X          = 29/21

X=5/3-2/7

X=29/21