1: A=-1/2*xy^3*4x^2y^2=-2x^3y^5

Bậc là 8

Phần biến là x^3;y^5

Hệ số là -2

2:

a: P(x)=3x+4x^4-2x^3+4x^2-x^4-6

=3x^4-2x^3+4x^2+3x-6

Q(x)=2x^4+4x^2-2x^3+x^4+3

=3x^4-2x^3+4x^2+3

b: A(x)=P(x)-Q(x)

=3x^4-2x^3+4x^2+3x-6-3x^4+2x^3-4x^2-3

=3x-9

A(x)=0

=>3x-9=0

=>x=3

I. 1b 2d 3a 4c

II. 1c 2b 3a 4b 5a

Đk:\(y^2-2x-5y+6\ge0\)

Pt (1)\(\Leftrightarrow\left(x^2-1\right)-\left(xy-y\right)+\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)-y\left(x-1\right)+\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2-y\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\y=x+2\end{matrix}\right.\)

TH1: Thay x=1 vào pt (2) ta đc: \(3\sqrt{y^2-5y+4}=y+9\)

\(\Leftrightarrow\left\{{}\begin{matrix}y+9\ge0\\9\left(x^2-5y+4\right)=y^2+18y+81\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y\ge-9\\8y^2-63y-45=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}y=\dfrac{63+3\sqrt{601}}{16}\\y=\dfrac{63-3\sqrt{601}}{16}\end{matrix}\right.\) (tm)

TH2: Thay y=x+2 vào pt (2) ta đc:

\(\left(x-1\right)^2+3\sqrt{\left(x+2\right)^2-2x-5\left(x+2\right)+6}=x+2+9\)

\(\Leftrightarrow x^2-3x-10+3\sqrt{x^2-3x}=0\)

Đặt \(t=\sqrt{x^2-3x}\left(t\ge0\right)\)

Pttt: \(t^2-10+3t=0\)\(\Leftrightarrow\left[{}\begin{matrix}t=2\left(tm\right)\\t=-5\left(ktm\right)\end{matrix}\right.\)

\(\Rightarrow2=\sqrt{x^2-3x}\)\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}y=6\\y=1\end{matrix}\right.\) (tm)

Vậy \(\left(x;y\right)=\text{​​}\left\{\left(1;\dfrac{63+3\sqrt{601}}{16}\right);\left(1;\dfrac{63-3\sqrt{601}}{16}\right),\left(4;6\right),\left(-1;1\right)\right\}\)

Xét pt đầu:

\(\left(x^2+x-2\right)-y\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)-y\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2-y\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\y=x+2\end{matrix}\right.\)

- Với \(x=1\) thay xuống pt dưới:

\(3\sqrt{y^2-5y+4}=y+9\) \(\left(y\ge-9\right)\)

\(\Leftrightarrow9\left(y^2-5y+4\right)=y^2+18y+81\)

\(\Leftrightarrow8y^2-63y-45=0\)

\(\Rightarrow y=\dfrac{63\pm3\sqrt{601}}{16}\) (thỏa mãn)

- Với \(y=x+2\) thay xuống pt dưới:

\(\left(x-1\right)^2+3\sqrt{x^2-3x}=x+11\) (ĐKXĐ: ....)

\(\Leftrightarrow x^2-3x+3\sqrt{x^2-3x}-10=0\)

Đặt \(\sqrt{x^2-3x}=t\ge0\)

\(\Rightarrow t^2+3t-10=0\Rightarrow\left[{}\begin{matrix}t=2\\t=-5\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x^2-3x}=2\Leftrightarrow x^2-3x-4=0\)

\(\Leftrightarrow...\)

a) nAl= 0,2(mol)

PTHH: 4Al + 3 O2 -to-> 2 Al2O3

nO2= 3/4 . 0,2= 0,15(mol)

=>V(O2,đktc)=0,15.22,4=3,36(l)

Vkk(đktc)=5.V(O2,đktc)=3,36.5=16,8(l)

b) nAl=0,2(mol)

nO2=0,4(mol)

Ta có: 0,2/4 < 0,4/3

=> Al hết, O2 dư, tính theo nAl.

- Sau phản ứng có O2(dư) và Al2O3

nAl2O3= nAl/2= 0,2/2=0,1(mol)

nO2(dư)= 0,4- 0,2. 3/4=0,25(mol)

2D

3A

4B

5A

6B

7D

8C

9A

10C

11B