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

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

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

Lời giải:

 $\frac{x}{y}=\frac{2}{3}\Rightarrow \frac{x}{2}=\frac{y}{3}$. Đặt $\frac{x}{2}=\frac{y}{3}=k$ thì:

$x=2k; y=3k$

Khi đó: $3x-2y=3.2k-3.2k=0$. Mẫu số không thể bằng $0$ nên $A$ không xác định. Bạn xem lại.

$B=\frac{2(2k)^2-2k.3k+3(3k)^2}{3(2k)^2+2.2k.3k+(3k)^2}=\frac{29k^2}{33k^2}=\frac{29}{33}$

Lời giải:
Đặt $\sqrt{y}=b(b\geq 0)\Rightarrow y=b^2$

$M=2x^2+5b^2-4xb-4x-8b+2036$

$=2(x^2+b^2-2xb)+3b^2-4x-8b+2036$

$=2(x-b)^2-4(x-b)+3b^2-12b+2036$

$=2(x-b)^2-4(x-b)+2+3(b^2-4b+4)+2022$

$=2[(x-b)^2-2(x-b)+1]+3(b-2)^2+2022$

$=2(x-b-1)^2+3(b-2)^2+2022\geq 2022$

Vậy $M_{\min}=2022$

Thay x = 1 ; y = -1 ta đc

\(\dfrac{1}{2}.1\left(-1\right)-\dfrac{3}{4}.1\left(-1\right)+1\left(-1\right)=-\dfrac{1}{2}+\dfrac{3}{4}-1=\dfrac{1}{4}-1=-\dfrac{3}{4}\)

a) \(-9x^2+12x-15=-\left(9x^2-12x+4\right)-11=-\left(3x-2\right)^2-11\le11< 0\)

b) \(-2x^2+4x-9=-2\left(x^2-2x+1\right)-7=-2\left(x-1\right)^2-7\le-7< 0\)

c) \(xy-x^2-y^2-1=-\dfrac{1}{2}\left(2x^2+2y^2-2xy+2\right)=-\dfrac{1}{2}\left[\left(x-y\right)^2+x^2+y^2+2\right]< 0\)

a)Ta có:\(x-y=2\Rightarrow\left(x-y\right)^2=4\Rightarrow\left(x^2+y^2\right)-2xy=4\Rightarrow4-2xy=4\Rightarrow2xy=0\Rightarrow xy=0\)

Khi đó ta có:\(x^5y=xy^5=xy\left(x^4-y^4\right)=0\)

A = x 2 – 5x + xy – 5y = ( x 2 + xy) – (5x + 5y) = x(x + y) – 5(x + y)

= (x – 5)(x + y)

Tại x = -5; y = -8 ta có

A = (-5 – 5)(-5 – 8) = (-10)(-13) = 130

Đáp án cần chọn là: A