Chọn C

ĐKXĐ: \(x< 2\)

\(m\sqrt{2-x}=\dfrac{x^2-2mx+2}{\sqrt{2-x}}\Rightarrow m\left(2-x\right)=x^2-2mx+2\)

\(\Leftrightarrow x^2+2=m\left(x+2\right)\Rightarrow m=\dfrac{x^2+2}{x+2}\)

Xét hàm \(f\left(x\right)=\dfrac{x^2+2}{x+2}\) với \(0< x< 2\)

\(f'\left(x\right)=\dfrac{2x\left(x+2\right)-\left(x^2+2\right)}{\left(x+2\right)^2}=\dfrac{x^2+4x-2}{\left(x+2\right)^2}=0\Rightarrow x=-2+\sqrt{6}\)

\(f\left(0\right)=1;f\left(2\right)=\dfrac{3}{2};f\left(-2+\sqrt{6}\right)=-4+2\sqrt{6}\)

\(\Rightarrow-4+2\sqrt{6}\le m< \dfrac{3}{2}\)

\(\Rightarrow m=1\)

Có đúng 1 giá trị nguyên m thỏa mãn

C

đáp,á:70

bằng 70

làm rồi kko đc loke mứt công

bye

(125.37.32):4=(125:4).(37:4).(32:4)=5.9,25.8=9,25.(5.8)=9,25. 40=9,25.4.10=37.10=370

\(M_X=17.2=34\left(g/mol\right)\)

Quy đổi hh \(X\left\{{}\begin{matrix}CH_4\\C_2H_4\\C_3H_4\\C_4H_4\end{matrix}\right.\rightarrow C_xH_4\left(1< x< 4\right)\)

\(m_X=34.0,075=2,55\left(g\right)\)

Mà \(n_H=4n_X=0,3\left(mol\right)\Rightarrow n_C=\dfrac{2,55-0,3}{12}=0,1875\left(mol\right)\)

BTNT: \(\left\{{}\begin{matrix}n_{CO_2}=n_C=0,1875\left(mol\right)\\n_{H_2O}=\dfrac{1}{2}n_H=0,15\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow m=m_{CO_2}+m_{H_2O}=0,1875.44+0,15.18=10,95\left(g\right)\)

Chọn D

Câu 31: 

PT: \(CH_2=CH_2+H_2O\underrightarrow{t^o,xt}CH_3CH_2OH\)

→ Đáp án: A

Câu 32:

Ta có: \(n_{CO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

\(n_X=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

Gọi CTPT của X là C_nH_2n.

\(\Rightarrow n=\dfrac{n_{CO_2}}{n_X}=5\)

Vậy: X là C₅H_10.

→ Đáp án: D

\(=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{4x+1}-\left(2x+1\right)+2x+1-\sqrt[3]{6x+1}}{x^2}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{-\dfrac{4x^2}{\sqrt{4x+1}+2x+1}+\dfrac{x^2\left(8x+12\right)}{\left(2x+1\right)^2+\left(2x+1\right)\sqrt[3]{6x+1}+\sqrt[3]{\left(6x+1\right)^2}}}{x^2}\)

\(=\lim\limits_{x\rightarrow0}\left(-\dfrac{4}{\sqrt{4x+1}+2x+1}+\dfrac{8x+12}{\left(2x+1\right)^2+\left(2x+1\right)\sqrt[3]{6x+1}+\sqrt[3]{\left(6x+1\right)^2}}\right)\)

\(=\dfrac{-4}{1+1}+\dfrac{12}{1+1+1}=2\)

a) \(\dfrac{3}{35}+\dfrac{7}{20}+\dfrac{32}{35}+\dfrac{13}{20}\)

Áp dụng tính chất giao hoán, kết hợp của phép cộng, ta được:

\(\left(\dfrac{3}{35}+\dfrac{32}{35}\right)+\left(\dfrac{7}{20}+\dfrac{13}{20}\right)\)

\(=1+1\)

\(=2\)

b) \(\dfrac{18\cdot15\cdot56}{42\cdot27\cdot25}=\dfrac{2\cdot3\cdot3\cdot5\cdot3\cdot4\cdot2\cdot7}{2\cdot3\cdot2\cdot2\cdot3\cdot9\cdot5\cdot5}\)

\(=\dfrac{2\cdot2\cdot3\cdot3\cdot3\cdot4\cdot5\cdot7}{2\cdot2\cdot2\cdot3\cdot3\cdot5\cdot5\cdot9}=\dfrac{3\cdot2\cdot2\cdot7}{2\cdot5\cdot9}=\dfrac{42}{45}\)