14. C

15. C

16. C

17. B

cảm ơn nha 

X x 14 + X x 15 = 292929
Xx(14+15)=292929
X x 29=292929
X=292929:29
X=10101
Học tốt!!!!
 

X x 14 + X x 15 = 292929

= X . ( 14 + 15 ) = 292929

                    X = 292929 : 29

                     X = 10101

Dấu . là nhân

ko ghi lại đề

 x.(14+15)=292929

x=2929:29

x=10101

hc tốt

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

\(\dfrac{\sqrt{5}-1}{\sqrt{5}+1}+\dfrac{6}{1-\sqrt{5}}=\dfrac{\left(\sqrt{5}-1\right).\left(1-\sqrt{5}\right)+6.\left(\sqrt{5}+1\right)}{-4}=\dfrac{6-2\sqrt{5}-6\sqrt{5}-6}{4}=\dfrac{-8\sqrt{5}}{4}=-2\sqrt{5}\)

\(\dfrac{\sqrt{2}-\sqrt{3}}{2-\sqrt{6}}+\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{6}+2}=\dfrac{\left(\sqrt{2}-\sqrt{3}\right).\left(\sqrt{6}+2\right)+\left(\sqrt{3}-\sqrt{2}\right).\left(2-\sqrt{6}\right)}{-2}=\dfrac{2\left(\sqrt{12}-\sqrt{18}\right)}{-2}=\sqrt{18}-\sqrt{12}\)

\(\dfrac{-31+8\sqrt{x}-x}{x-8\sqrt{x}+15}-\dfrac{\sqrt{x}+5}{\sqrt{x}-3}-\dfrac{3\sqrt{x}-1}{5-\sqrt{x}}\)

\(=\dfrac{-31+8\sqrt{x}-x}{\left(\sqrt{x}-5\right)\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}+5}{\sqrt{x}-3}+\dfrac{3\sqrt{x}-1}{\sqrt{x}-5}\)

\(=\dfrac{-31+8\sqrt{x}-x-x+25+3x-9\sqrt{x}-\sqrt{x}+3}{\left(\sqrt{x}-5\right)\left(\sqrt{x}-3\right)}=\dfrac{x-2\sqrt{x}-3}{\left(\sqrt{x}-5\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-5}\)

14. A

15. D

16. C

17. D

19. B

13. C

14.C

15.A

Câu 14:

\(n_{H_2O}=\dfrac{3,6}{18}=0,2\left(mol\right)\Rightarrow n_H=0,2.2=0,4\left(mol\right)\)

Có: m_C + m_H = 5,2 (g) ⇒ m_C = 5,2 - 0,4.1 = 4,8 (g) 

\(\Rightarrow n_{CO_2}=n_C=\dfrac{4,8}{12}=0,4\left(mol\right)\)

A: C_nH_2n-2

⇒ n_A = n_CO2 - n_H2O = 0,4 - 0,2 = 0,2 (mol)

\(\Rightarrow n=\dfrac{n_{CO_2}}{n_A}=2\)

→ A là C₂H₂.

→ Đáp án: B

Câu 15:

X: C_nH_2n-2

\(\Rightarrow\dfrac{2n-2}{12n+2n-2}=0,11111\Rightarrow n=4\)

→ X là C₄H₆.

CTCT: \(CH\equiv C-CH_2-CH_3\)

\(CH_3-C\equiv C-CH_3\)

→ Đáp án: B

Câu 14:

Ta có: \(n_{C_2H_4}+n_A=\dfrac{3,36}{22,4}=0,15\left(1\right)\)

m _{bình tăng} = m_C2H4 = 1,4 (g) \(\Rightarrow n_{C_2H_4}=\dfrac{1,4}{28}=0,05\left(mol\right)\) (2)

Từ (1) và (2) \(\Rightarrow n_A=0,1\left(mol\right)\)

Gọi CTPT của A là C_nH_2n+2.

Có: \(n_{CO_2}=\dfrac{8,8}{44}=0,2\left(mol\right)\)

\(\Rightarrow n=\dfrac{n_{CO_2}}{n_A}=2\)

Vậy: A là C₂H₆.

→ Đáp án: A.

Câu 15:

Ta có: \(n_{CO_2}=n_{H_2O}=x\left(mol\right)\)

\(n_{O_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

BTNT O, có: \(2n_{CO_2}+n_{H_2O}=2n_{O_2}\)

\(\Rightarrow2x+x=2.0,6\Leftrightarrow x=0,4\left(mol\right)\)

BTNT C, có: \(n_{CaCO_3}=n_{CO_2}=0,4\left(mol\right)\)

\(\Rightarrow m_{CaCO_3}=0,4.100=40\left(g\right)\)

→ Đáp án: C

Câu 13:

1: 

a: \(2x^2+2x=2x\cdot x+2x\cdot1=2x\left(x+1\right)\)

b: \(9x^2-4y^2\)

\(=\left(3x\right)^2-\left(2y\right)^2\)

=(3x-2y)(3x+2y)

2:

\(\dfrac{xy+2x+1}{xy+x+y+1}+\dfrac{yz+2y+1}{yz+y+z+1}+\dfrac{zx+2z+1}{zx+z+x+1}\)

\(=\dfrac{xy+2x+1}{\left(y+1\right)\left(x+1\right)}+\dfrac{yz+2y+1}{\left(z+1\right)\left(y+1\right)}+\dfrac{z\left(x+2\right)+1}{\left(z+1\right)\left(x+1\right)}\)

\(=\dfrac{\left(xy+2x+1\right)\left(z+1\right)+\left(yz+2y+1\right)\left(x+1\right)+\left(xz+2z+1\right)\left(y+1\right)}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\)

\(=\dfrac{xyz+xy+2xz+2x+z+1+xyz+yz+2xy+2y+x+1+\left(xz+2z+1\right)\left(y+1\right)}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\)

\(=\dfrac{2xyz+3xy+2xz+3x+z+2+yz+2y+x+xyz+xz+2zy+2z+y+1}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\)

\(=\dfrac{3xyz+3xy+3xz+3yz+3x+3z+3y+3}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\)

\(=\dfrac{3\left(xyz+xy+xz+yz+x+z+y+1\right)}{\left(xy+x+y+1\right)\left(z+1\right)}\)

=3

Câu 14:

1:

f(0)=0+5=5

2:
Vì hệ số góc của y=ax+b là -1 nên a=-1

=>y=-x+b

Thay x=1 và y=2 vào y=-x+b, ta được:

b-1=2

=>b=3