TL:
\(\left(\dfrac{2}{5}-x\right):\left(-1\dfrac{1}{3}\right)=\dfrac{5}{4}\)

\(\dfrac{2}{5}-x=\dfrac{5}{4}\times\left(-1\dfrac{1}{3}\right)\)
\(\dfrac{2}{5}-x=\dfrac{5}{4}\times\left(-\dfrac{4}{3}\right)\)

\(\dfrac{2}{5}-x=-\dfrac{5}{3}\)
\(x=\dfrac{2}{5}-\left(-\dfrac{5}{3}\right)=\dfrac{31}{15}\)
Vậy \(x=\dfrac{31}{15}\)
HT!
 

\(\left(\dfrac{2}{5}-x\right)\div\left(-1\dfrac{1}{3}\right)=\dfrac{5}{4}\)

\(\Rightarrow\)\(\left(\dfrac{2}{5}-x\right)\div\left(-\dfrac{4}{3}\right)=\dfrac{5}{4}\)

\(\Rightarrow\)\(\left(\dfrac{2}{5}-x\right)=\dfrac{5}{4}\times\left(-\dfrac{4}{3}\right)\)

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

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

\(\Rightarrow\)\(x=\dfrac{2}{5}+\dfrac{5}{3}\)

\(\Rightarrow\)\(x=\dfrac{31}{15}\)

\(\left(\dfrac{2}{5}-x\right):\left(-1\dfrac{1}{3}\right)=\dfrac{5}{4}\)

\(\Leftrightarrow\left(\dfrac{2}{5}-x\right):-\dfrac{4}{3}=\dfrac{5}{4}\)

\(\Leftrightarrow\dfrac{2}{5}-x=\dfrac{5}{4}.\dfrac{-4}{3}\)

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

\(\Leftrightarrow x=\dfrac{31}{15}\)

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

\(\dfrac{3}{2}+x=\dfrac{3}{7}-\left(-\dfrac{3}{8}\right)\)

\(\dfrac{3}{2}+x=\dfrac{45}{56}\)

\(x=\dfrac{45}{56}-\dfrac{3}{2}\)

\(x=\dfrac{45-84}{56}=\dfrac{-39}{56}\)

10/7+x=1+4/9

10/7+x=13/9

x=13/9-10/7

x=1/63

\(2,5-\dfrac{1}{7}x=-2\)

\(\Rightarrow\dfrac{5}{2}-\dfrac{1}{7}x=-2\)

\(\Rightarrow\dfrac{1}{7}x=\dfrac{5}{2}-\left(-2\right)\)

\(\Rightarrow\dfrac{1}{7}x=\dfrac{9}{2}\)

\(\Rightarrow x=\dfrac{63}{2}\)

2/5-x=5/4x (-2/3)

2/5-x= (-5/6)

x=2/5-(-5/6)

x=37/30

\(\dfrac{3}{8}+\left(x-1\dfrac{1}{7}\right)=0\)

\(\Rightarrow x-\dfrac{8}{7}=-\dfrac{3}{8}\)

\(\Rightarrow x=\left(-\dfrac{3}{8}\right)+\dfrac{8}{7}\)

\(\Rightarrow x=\dfrac{43}{56}\)

Có một khu du lịch giá vé là342nghìn đồng ông chủ giảm giá vé số lượng người tăng thêm là 14% doanh thu tăng lên 6% hỏi ông chủ đã giảm bao nhiêu tiền 

Đặt \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}\)\(=k\) \(\Rightarrow\) \(\left\{{}\begin{matrix}x=3k\\y=4k\\z=5k\end{matrix}\right.\) (1)

Thay (1) vào \(2x^2+2y^2-3z^2=-100\)

Có 2(3k)² +2(4k)² -3(5k)² =-100

    2.9k² +2.16k²  -3.25k²=-100

18k² +32k² -75k²=-100

-25k²=-100

=>k²= 4

=>\(\left\{{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\)

TH1 k=2 =>\(\left\{{}\begin{matrix}x=2\cdot3\\y=2\cdot4\\z=2\cdot5\end{matrix}\right.\) <=>\(\left\{{}\begin{matrix}x=6\\y=8\\z=10\end{matrix}\right.\)

TH2 k=-2 =>\(\left\{{}\begin{matrix}x=-2\cdot3\\y=-2\cdot4\\z=-2\cdot5\end{matrix}\right.\) <=>\(\left\{{}\begin{matrix}x=-6\\y=-8\\z=-10\end{matrix}\right.\)