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\(\dfrac{3x-y}{x+y}=\dfrac{3}{4}\\ \Rightarrow4\cdot\left(3x-y\right)=3\cdot\left(x+y\right)\\ \Leftrightarrow12x-4y=3x+3y\\ \Leftrightarrow12x-3x=3y+4y\\ \Leftrightarrow9x=7y\\ \Rightarrow\dfrac{x}{7}=\dfrac{y}{9}\\ \Rightarrow\dfrac{x}{y}=\dfrac{7}{9}\)
Vậy \(\dfrac{x}{y}=\dfrac{7}{9}\)
Ta có: \(\dfrac{3x-y}{x+y}=\dfrac{3}{4}\)
\(\Rightarrow\left(3x-y\right)4=\left(x+y\right)3\)
\(\Rightarrow12x-4y=3x+3y\)
\(\Rightarrow12x-3x=4y+3y\)
\(\Rightarrow9x=7y\)
\(\Rightarrow\dfrac{x}{y}=\dfrac{7}{9}\)
Vậy \(\dfrac{x}{y}=\dfrac{7}{9}.\)
1.
\(\left(\dfrac{-1}{8}+\dfrac{-5}{6}\right)\cdot\dfrac{6}{23}\\ =-\dfrac{23}{24}\cdot\dfrac{6}{23}\\ =-\dfrac{6}{24}=-\dfrac{1}{4}\)
2. Xem lại đề nha!
4.
\(x+0,75=-1\dfrac{1}{4}\\ x+\dfrac{3}{4}=-\dfrac{3}{4}\\ x=-\dfrac{3}{4}-\dfrac{3}{4}\\ x=-\dfrac{3}{4}+\left(-\dfrac{3}{4}\right)=-\dfrac{6}{4}=-\dfrac{3}{2}\)
5.
\(\dfrac{x}{28}=-\dfrac{4}{7}\\ \Leftrightarrow7x=-4.28\\ \Rightarrow7x=-112\\ \Rightarrow x=-112:7=-16\)
6.
\(\dfrac{3x-y}{x+y}=\dfrac{3}{4}\\ \Leftrightarrow\left(3x-y\right).4=3\left(x+y\right)\\ \Rightarrow12x-4y=3x+3y\\ \Rightarrow12x-3x=4y+3y\\ \Rightarrow9x=7y\\ \Leftrightarrow\dfrac{x}{7}=\dfrac{y}{9}\Leftrightarrow\dfrac{x}{y}=\dfrac{7}{9}\)
Vậy giá trị của tỉ số \(\dfrac{x}{y}=\dfrac{7}{9}\).
a: k=y/x=-10,8/3,6=-3
=>y=-3x
b: Khi x=-3 thì y=9
Khi x=24 thì y=-72
Khi x=-2/3 thì y=2
Khi x=7/6 thì y=-3*7/6=-7/2
Khi x=-1/15 thì y=-3*(-1/15)=1/5
c: y=-3x
=>x=-1/3y
Khi y=4 thì x=-4/3
Khi y=12 thì y=-1/3*12=-4
Khi y=-26 thì x=-1/3*(-26)=26/3
Khi y=4/3 thì x=-1/3*4/3=-4/9
Khi y=-26/15 thì y=1/3*26/15=26/45
\(\dfrac{3x-y}{x+y}=\dfrac{3}{4}\)
\(\Rightarrow4\left(3x-y\right)=3\left(x+y\right)\)
\(\Rightarrow12x-4y=3x+3y\)
\(\Rightarrow12x-4y-3y=3x\)
\(\Rightarrow12x-7y=3x\)
\(\Rightarrow12x-3x=7y\)
\(\Rightarrow9x=7y\)
\(\Rightarrow\dfrac{x}{7}=\dfrac{y}{9}\)
\(\Rightarrow\dfrac{x}{y}=\dfrac{7}{9}\)
Lời giải:
\(\frac{5x+3}{2\frac{1}{7}}=\frac{\frac{7}{15}}{5x+3}\)
\(\Rightarrow (5x+3)(5x+3)=2\frac{1}{7}.\frac{7}{15}=1\)
\(\Leftrightarrow (5x+3)^2=1=1^2=(-1)^2\)
\(\Rightarrow \left[\begin{matrix} 5x+3=1\\ 5x+3=-1\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{-2}{5}\\ x=-\frac{4}{5}\end{matrix}\right.\)
Ta có:
\(\dfrac{3x-y}{x+y}=\dfrac{3}{4}\\ \Rightarrow12x-4y=3x+3y\\ \Rightarrow9x=7y\\ \Rightarrow\dfrac{x}{y}=\dfrac{7}{9}\)
Vậy.........
\(\dfrac{3x-y}{x+y}=\dfrac{3}{4}\)
\(\Leftrightarrow4\left(3x-y\right)=3\left(x+y\right)\)
\(\Leftrightarrow12x-4y=3x+3y\)
\(\Leftrightarrow12x=3x+7y\)
\(\Leftrightarrow9x=7y\)
\(\Leftrightarrow\dfrac{x}{7}=\dfrac{y}{9}\Leftrightarrow\dfrac{x}{y}=\dfrac{7}{9}\)
\(\dfrac{3x-y}{x+y}=\dfrac{3}{4}\)
=> 4(3x-y)=3(x+y)
=> 12x-4y=3x+3y
=> 12x-3x=4y+3y
=> 9x=7y
=> \(\dfrac{x}{y}=\dfrac{7}{9}\)
1 a) \(\dfrac{\left(-2\right)}{5}\)= \(\dfrac{-6}{15}\); \(\dfrac{15}{-6}\)= \(\dfrac{5}{-2}\); \(\dfrac{-6}{-2}\)= \(\dfrac{15}{5}\); \(\dfrac{-2}{-6}\)= \(\dfrac{5}{15}\)
7/x = -3/15
⇒ 7⨯15 = x ⨯ (-3)
⇒ 105 = x ⨯ (-3)
⇒ x =105/-3
⇒ x =-35
\(\dfrac{7}{x}=\dfrac{-3}{15}\)
\(\Rightarrow x\cdot\left(-3\right)=7\cdot15\)
\(\Rightarrow-3x=105\)
\(\Rightarrow x=105:\left(-3\right)\)
\(\Rightarrow x=-35\)