\(2,\Leftrightarrow\left\{{}\begin{matrix}20x+25y-10xy=0\\20x-30y+xy=0\end{matrix}\right.\Leftrightarrow55y-11xy=0\\ \Leftrightarrow11y\left(5-x\right)=0\Leftrightarrow\left[{}\begin{matrix}y=0\\x=5\end{matrix}\right.\)

Với \(y=0\Leftrightarrow4x+0=0\Leftrightarrow x=0\)

Với \(x=5\Leftrightarrow20+5y=10y\Leftrightarrow y=4\)

Vậy \(\left(x;y\right)=\left\{\left(0;0\right);\left(5;4\right)\right\}\)

a) \(\sqrt{x-8}\) xác định khi

\(x-8\ge0\Leftrightarrow x\ge8\)

b) \(\sqrt{3x+1}\) xác định khi

\(3x+1\ge0\Leftrightarrow3x\ge-1\Leftrightarrow x\le-\dfrac{1}{3}\)

c) \(\sqrt{x^2+1}\) 

Ta có: \(x^2\ge0\Rightarrow x^2+1\ge0\)

Vậy biểu thức được xác định với mọi x

d) \(\sqrt{\left(x-6\right)\left(x+3\right)}\)

Xác định khi

\(\left\{{}\begin{matrix}\left\{{}\begin{matrix}x-6\ge0\\x+3\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-6< 0\\x+3\ge0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x\ge6\\x\ge-3\end{matrix}\right.\\\left\{{}\begin{matrix}x< 6\\x< -3\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge6\\x< -3\end{matrix}\right.\)

e) \(\sqrt{\dfrac{-2}{x-5}}\) xác định khi

\(\left\{{}\begin{matrix}\dfrac{-2}{x-5}\ge0\\x-5\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le5\\x\ne5\end{matrix}\right.\)

\(\Leftrightarrow x< 5\)

f) \(\dfrac{4}{\sqrt{x+3}}\) xác định khi

\(x+3>0\)

\(\Leftrightarrow x>-3\)

g) \(\dfrac{6x-2}{\sqrt{x}-3}\)

Xác định khi:

\(\left\{{}\begin{matrix}\sqrt{x}\ge0\\\sqrt{x}-3\ne0\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)

h) \(\sqrt{x^2-16}=\sqrt{\left(x+4\right)\left(x-4\right)}\)

Xác định khi
\(\left\{{}\begin{matrix}\left\{{}\begin{matrix}x+4< 0\\x-4< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x+4\ge0\\x-4\ge0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x< -4\\x< 4\end{matrix}\right.\\\left\{{}\begin{matrix}x\ge-4\\x\ge4\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x< -4\\x\ge4\end{matrix}\right.\)

Bài 2: 

a: Xét tứ giác ABOC có

\(\widehat{ABO}+\widehat{ACO}=180^0\)

Do đó: ABOC là tứ giác nội tiếp

11 D

12 C

13 A

14 C

15 C

16 B

17 A

18 C

Câu 9:

a) Ta có: \(9x^2-16=0\)

\(\Leftrightarrow\left(3x-4\right)\left(3x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=-\dfrac{4}{3}\end{matrix}\right.\)

b) Ta có: \(4x^2=13\)

\(\Leftrightarrow x^2=\dfrac{13}{4}\)
\(\Leftrightarrow x\in\left\{\dfrac{\sqrt{13}}{2};-\dfrac{\sqrt{13}}{2}\right\}\)

c) Ta có: \(2x^2+9=0\)

\(\Leftrightarrow2x^2=-9\)(Vô lý)

d) Ta có: \(-x^2+324=0\)

\(\Leftrightarrow x^2=324\)

\(\Leftrightarrow\left[{}\begin{matrix}x=18\\x=-18\end{matrix}\right.\)

11 She asked me if I like English music

12 Two department stores have been built

13 The computer is being used at the moment

31 We should avoid playing computer games because it will hurt our eyes

32 Rescue workers sent water and food to victims yesterday

33 I like to travel to Europe but I don't have enough money

34 Our children will be taken care of by robots in 20 years' time

mik quên mất, làm hộ mik câu 1,11 và 12 nhé. cảm ơn mn rất nhiều

# ~Hok tốt~

11.

ngay thu nhat : 600m vai

ngay thu hai: 1050m vai

12.

a.4335m2

b.7225kg thoc

hoc tot

Tra loi:

1.A.23/27

hoc tot