11 D

12 B

13 A

14 B

15 A

16 A

17 A

18 B

19 C

20 B

21 B

22 C

23 C

24 C

25 B

26 heated

27 solar

28 pollution

if you open the window,you will enjoy the air

if you are not healthy,you can live longer

if you don't work harder ,you will get higher position

if you are quick ,you can meet Cao Thanh Dung

if Mary is short ,she can't play basketball

b: Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

Ta có: \(\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}\)

\(=\dfrac{\left(bk+dk\right)^2}{\left(b+d\right)^2}\)

\(=k^2\)(1)

Ta có: \(\dfrac{a^2-c^2}{b^2-d^2}\)

\(=\dfrac{\left(bk\right)^2-\left(dk\right)^2}{b^2-d^2}\)

\(=k^2\left(2\right)\)

Từ (1) và (2) suy ra \(\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}=\dfrac{a^2-c^2}{b^2-d^2}\)

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

Ta có: \(\dfrac{4a+3c}{4b+3d}=\dfrac{4bk+3dk}{4b+3d}=k\)

\(\dfrac{4a-3c}{4b-3d}=\dfrac{4bk-3dk}{4b-3d}=k\)

Do đó: \(\dfrac{4a+3c}{4b+3d}=\dfrac{4a-3c}{4b-3d}\)​

14/9

Lời giải:

a. 

$A=(-2x^3+6x^3)+(6x^3y-25x^3y)+(7y+8y)-20$

$=4x^3-19x^3y+15y-20$

b.

$B=(5x^3-14x^3)+(9xy^4-13xy^4)-3y^2+(17-23)$

$=-9x^3-4xy^4-3y^2-6$

b. 

Bậc của $A$ gắn với $-19x^3y$, là  $3+1=4$

Bậc của $B$ gắn với $-4xy^4$, là $1+4=5$

ab, \(A=-2x^3-19x^3y+15y-20\)

-> bậc 4 

\(B=-9x^3-4xy^4-3y^2-5\)

-> bậc 5 

1. Products

2. disappointingly

3. consuming

4. polluting

5. respiratory

6. seriously

7. harmful

8. Conservationists

9. deforestation

10. necessary

11. celebrations

12. ineffectively

13. innovations

14. respiratory

15. respiratory

b: Ta có: 2x=3y=5z

\(\Leftrightarrow\dfrac{x}{\dfrac{1}{2}}=\dfrac{y}{\dfrac{1}{3}}=\dfrac{z}{\dfrac{1}{5}}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{\dfrac{1}{2}}=\dfrac{y}{\dfrac{1}{3}}=\dfrac{z}{\dfrac{1}{5}}=\dfrac{x-2y+z}{\dfrac{1}{2}-\dfrac{2}{3}+\dfrac{1}{5}}=\dfrac{14}{\dfrac{1}{30}}=420\)

Do đó: x=210; y=140; z=84

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y-z}{2+3-4}=-\dfrac{20}{1}=-20\)

⇒\(\left\{{}\begin{matrix}x=-20.2=-40\\y=-20.3=-60\\z=-20.4=-80\end{matrix}\right.\)