sao đang toán lại chèn cả hóa vào v?????

a) Áp dung TC của dãy tỉ số bằng nhau ta có :

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{-2y}{-6}=\dfrac{3z}{15}=\dfrac{x-2y+3z}{2-3+15}=\dfrac{x-2y+3z}{14}=\dfrac{38}{14}=\dfrac{19}{7}\)

Vậy\(\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{19}{7}\Leftrightarrow x=\dfrac{38}{7}\\\dfrac{y}{3}=\dfrac{19}{7}\Leftrightarrow y=\dfrac{57}{7}\\\dfrac{z}{5}=\dfrac{19}{7}\Leftrightarrow x=\dfrac{95}{7}\end{matrix}\right.\)

Kết luận:..

câu a sao lại là \(\dfrac{16}{15}\)

a)\(\dfrac{-3}{15}+\dfrac{-5}{15}\)

b)\(\dfrac{4}{3}-\dfrac{4}{5}\)

c)\(-1+\dfrac{7}{15}\)

d)\(\dfrac{-1}{15}-\dfrac{7}{15}\)

tôi xin!!!!

/a/ là gì vậy? giá trị tuyệt đối?

b)

\(\dfrac{2}{5}x+\dfrac{-3}{7}=\dfrac{-1}{5}x+\dfrac{8}{14}\\ \Rightarrow\dfrac{2}{5}x-\dfrac{-1}{5}x=\dfrac{8}{14}-\dfrac{-3}{7}\\ \)

\(\Rightarrow\dfrac{2}{5}x+\dfrac{1}{5}x=\dfrac{8}{14}+\dfrac{3.2}{7.2}\\ \Rightarrow\dfrac{3}{5}x=\dfrac{8}{14}+\dfrac{6}{14}\\ \Rightarrow\dfrac{3}{5}x=\dfrac{14}{14}\\ \Rightarrow\dfrac{3}{5}x=1\\ \Rightarrow x=1:\dfrac{3}{5}\\ \Rightarrow x=\dfrac{5}{3}\)

Vậy x=\(\dfrac{5}{3}\)

a)\(x:3\dfrac{1}{5}=1\dfrac{1}{2}\)

\(\Rightarrow x:\dfrac{16}{5}=\dfrac{3}{2}\\ \Rightarrow x=\dfrac{3}{2}.\dfrac{16}{5}\\ \Rightarrow x=\dfrac{24}{5}\)

Vậy x=\(\dfrac{24}{5}\)