1) Vì x=25 thỏa mãn ĐKXĐ nên Thay x=25 vào biểu thức \(A=\dfrac{\sqrt{x}-2}{x+1}\), ta được:

\(A=\dfrac{\sqrt{25}-2}{25+1}=\dfrac{5-2}{25+1}=\dfrac{3}{26}\)

Vậy: Khi x=25 thì \(A=\dfrac{3}{26}\)

2) Ta có: \(B=\dfrac{\sqrt{x}-3}{\sqrt{x}+1}+\dfrac{2x+8\sqrt{x}-6}{x-\sqrt{x}-2}\)

\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}+\dfrac{2x+8\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x-5\sqrt{x}+6+2x+8\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{3x+3\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{3\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{3\sqrt{x}}{\sqrt{x}-2}\)

câu 3 chứ

1A

2B

3B

4C

5A

6B

7D

8A

9C

10B

11C

12C

13A

14A

15D

16B

17D

18B

19D

20B

1A

2 D

3B

4C

5A

6B

7D

8A

9C

10B

11C

12C

13B

14A

15D

16B

17D

18B

19D

20B

\(2x^2-2y^2+10x+10y\)

\(=2\left(x^2-y^2\right)+10\left(x+y\right)\)

\(=2\left(x-y\right)\left(x+y\right)+10\left(x+y\right)\)

\(=2\left(x+y\right)\left(x-y+5\right)\)

\(2x^2-2y^2+10x+10y=\left(2x^2-2y^2\right)+\left(10x+10y\right)=2\left(x^2-y^2\right)+10\left(x+y\right)=2\left(x-y\right)\left(x+y\right)+10\left(x+y\right)=\left(x+y\right)\left[2\left(x-y\right)+10\right]=\left(x+y\right)\left(2x-2y+10\right)=2\left(x+y\right)\left(x-y+5\right)\)

a: Tỉ số là 3/2

b: Tỉ số phần trăm là;

40/(30+40+20+20+5)=34,78%

a) n_CuCl2 = 0,15.2 = 0,3 (mol)

PTHH: 2Al + 3CuCl₂ --> 2AlCl₃ + 3Cu

____0,2<------0,3--------->0,2---->0,3

=> m = 0,3.64 - 0,2.27 = 13,8 (g)

b) \(C_{M\left(AlCl_3\right)}=\dfrac{0,2}{0,15}=1,3333M\)