kết quả bằng :36*x^2

\(A=-x^2+2xy-4y^2+x-10y-8\)

=> \(-4A=4x^2-8xy+16y^2-4x+40y+32\)

                \(=\left(4x^2-8xy+4y^2\right)-\left(4x-4y\right)+1+12y^2+36y+31\)

                  \(=\left(2x-2y\right)^2-2\left(2x-2y\right)+1+3\left(4y^2+2.2y.3+9\right)+4\)

                   \(=\left(2x-2y+1\right)^2+3\left(2y+3\right)^2+4\ge4\)

=> \(A\le4:-4=-1\)

"=" xảy ra <=> \(\hept{\begin{cases}2x-2y+1=0\\2y+3=0\end{cases}\Leftrightarrow}\hept{\begin{cases}y=-\frac{3}{2}\\x=2\end{cases}}\)

Vậy max A=-1 <=> x=2 y=-3/2

Câu b em làm tương tự nhé!

dài thế sao làm đc

Trường bn gắt quá

câu a ) a*x^19+1 

câu b ) 

đa thức chia có bậc 2 nên đa thức dư có bậc không quá 1. vậy đa thức dư có bậc nhất dạng ax+b

Ta có: 

Cho x=1 rồi x=-1 ta được: 

Vậy dư trong phép chia trên là 5x+1

1 Life in the country is not as good as life in the city

2 These room are cleaned every evening

3 Would you mind decorating the christmas tree

4 It's the first time I have talked to the headmaster

V

1 have ever met

2 worked

3 reading

4 are looking - seeing

5 was

VI

1A

2A

3D

4D

5B

IV/

1. is not as good as life in the city

2. are cleaned every evening

3. decorating the Christmas tree?

4. the first time I have (ever) talked to the headmaster

V/

1. have ever met

2. worked

3. reading

4. are looking - seeing

5. was

VI/

1a 2a 3d 4d 5b

a: \(\frac{-2x+3}{5xy}+\frac{3x-4}{5xy}\)

\(=\frac{-2x+3+3x-4}{5xy}=\frac{x-1}{5xy}\)

b: \(\frac{x\left(x-2\right)}{x-3}+\frac{3}{3-x}\)

\(=\frac{x^2-2x-3}{x-3}\)

\(=\frac{\left(x-3\right)\left(x+1\right)}{x-3}=x+1\)

c: \(\frac{2x+7}{3\left(x+2\right)}-\frac{x+5}{3\left(x+2\right)}\)

\(=\frac{2x+7-x-5}{3\left(x+2\right)}\)

\(=\frac{x+2}{3\left(x+2\right)}=\frac13\)

d: \(\frac{5}{2x+6}+\frac{3}{x^2-9}\)

\(=\frac{5}{2\left(x+3\right)}+\frac{3}{\left(x-3\right)\left(x+3\right)}\)

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

e: \(\frac{3x}{x+2}+\frac{x+3}{x^2-4}\)

\(=\frac{3x\left(x-2\right)+x+3}{\left(x-2\right)\left(x+2\right)}=\frac{3x^2-6x+x+3}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{3x^2-5x+3}{\left(x-2\right)\left(x+2\right)}\)

f: \(\frac{3x+2}{\left(x+2\right)^2}-\frac{1}{x+2}\)

\(=\frac{3x+2-\left(x+2\right)}{\left(x+2\right)^2}=\frac{2x}{\left(x+2\right)^2}\)

B -> IS GOING
D -> DIDN'T THEY
B -> COULD GO
A -> HASN'T WORN
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