a) \(2x=5y\Leftrightarrow\dfrac{x}{5}=\dfrac{y}{2}\)

Áp dụng tính chất của dãy tỉ số bằng nhau:

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

Khi đó: \(\left\{{}\begin{matrix}\dfrac{x}{5}=-3\\\dfrac{y}{2}=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3.5=-15\\y=-3.2=-6\end{matrix}\right.\)

\(\Rightarrow O=x^2-xy+2y=\left(-3\right)^2-\left(-15\right).\left(-6\right)+2.\left(-6\right)=9-90-12=-93\)

b)

Đặt \(\dfrac{x}{2}=\dfrac{y}{5}=k\Rightarrow\left\{{}\begin{matrix}x=2k\\y=5k\end{matrix}\right.\)

\(\Rightarrow2k.5k=90\\ \Leftrightarrow10k^2=90\\ \Leftrightarrow k^2=9\\ \Leftrightarrow\left[{}\begin{matrix}k=-3\\k=3\end{matrix}\right.\)

Nếu k = -3

\(\Rightarrow\left\{{}\begin{matrix}x=-3.2=-6\\y=-3.5=-15\end{matrix}\right.\)

\(\Rightarrow Q=-93\)

Nếu k = 3

\(\Rightarrow\left\{{}\begin{matrix}x=3.2=6\\y=3.5=15\end{matrix}\right.\)

\(\Rightarrow Q=6^2-6.15+2.15=-24\)

Bài 1:
a. $y=ax$

Thay $x=-9; y=27$ thì: $27=a(-9)$

$\Rightarrow a=-3$

Công thức liên hệ giữa $y$ và $x$: $y=-3x$

b.

Khi $x=-4$ thì $y=-3x=-3(-4)=12$

Khi $x=0,25$ thì $y=-3.0,25=-0,75$

Khi $x=\frac{1}{3}$ thì $y=-3.\frac{1}{3}=-1$

c. $y=-3x\rightarrow x=\frac{-y}{3}$

Khi $y=-2$ thì $x=\frac{-(-2)}{3}=\frac{2}{3}$

Khi $y=0$ thì $x=\frac{-(0)}{3}=0$

Khi $y=\frac{3}{5}$ thì $x=\frac{-3}{5.3}=\frac{-1}{5}$

Bài 2:

`!`

`-3/5 .(4x-12)=-5/3`

`=>4x-12=-5/3 : (-3/5)`

`=>4x-12=-5/3 xx (-5/3)`

`=>4x-12= 25/9`

`=>4x=25/9 +12`

`=>4x=133/9`

`=>x=133/9 : 4`

`=>x=133/9 xx 1/4`

`=>x=133/36`

Ta có : \(\dfrac{x-2012}{8}+\dfrac{x-2008}{6}+\dfrac{x-2005}{5}=10-\dfrac{x-2004}{4}\)

\(\Leftrightarrow\left(\dfrac{x-2012}{8}-1\right)+\left(\dfrac{x-2008}{6}-2\right)+\left(\dfrac{x-2005}{5}-3\right)+\left(\dfrac{x-2004}{4}-4\right)=0\)\(\Leftrightarrow\dfrac{x-2020}{8}+\dfrac{x-2020}{6}+\dfrac{x-2020}{5}+\dfrac{x-2020}{4}=0\)

\(\Leftrightarrow\left(x-2020\right).\left(\dfrac{1}{8}+\dfrac{1}{6}+\dfrac{1}{5}+\dfrac{1}{4}\right)=0\)

\(\Leftrightarrow x-2020=0\Leftrightarrow x=2020\)

Vậy x = 2020