B - A - D - C - C

a d b d b

\(d,-\dfrac{3}{26}.\left(-\dfrac{15}{19}\right)+\dfrac{2}{19}.\left(-\dfrac{3}{26}\right)\\=-\dfrac{3}{26}.\left(-\dfrac{15}{19}+\dfrac{2}{19}\right)\\ =-\dfrac{13}{19}.\left(-\dfrac{3}{26}\right)\\ =\dfrac{3}{38}. \)

d) \(\dfrac{-3}{26}\times\left(\dfrac{-15}{19}\right)+\dfrac{2}{19}\times\dfrac{-3}{26}\)

\(=\dfrac{-3}{26}\times\left[\left(\dfrac{-15}{19}\right)+\dfrac{2}{19}\right]\)

\(=\dfrac{-3}{26}\times\dfrac{-13}{19}\)

\(=\dfrac{3}{38}\)

`\sqrt{8+2x-x^2}+3x>6`

`đkxđ:-2<=x<=4`

`bpt<=>\sqrt{8+2x-x^2}>6-3x`

Đp bp 2 vế:`x<=2`

`<=>8+2x-x^2>9x^2-36x+36`

`<=>10x^3-38x+28<0`

`<=>5x^2-19x+14<0`

`<=>(x-1)(5x-14)<0`

`<=>1<x<14/5`

Kết hợp đkxđ:

`1<x<=2`

`=>S=(1,2]`

`=>ab=2`

`=>` chọn A

Ta có tổng hệ số của khai triển (a+b)^n là Cn⁰+C¹_n+....Cⁿ_n

theo đề ta có Cn⁰+C¹_n+....Cⁿ_n=4096=>(1+1)^n=4096=>2^n=4096=>n=12

vậy n=12

Xét C^k₁₂>C^k+1₁₂=>k<11/2=>k=1,2,3,4,5

=>C¹₁₂<C²₁₂<...<C⁵₁₂<C⁶₁₂>C⁷₁₂>.......>C¹²₁₂

vậy k max =6=> hệ số lớn nhất trong khai triển:C₁₂⁶=924=>A

Chọn A

Câu 27:

\(C_3H_4+AgNO_3+NH_3\rightarrow C_3H_3Ag+NH_4NO_3\\ C_2H_2+2AgNO_3+2NH_3\rightarrow Ag_2C_2+2NH_4NO_3\\ n_{C_3H_4}=\dfrac{1,8}{40}=0,045\left(mol\right);n_{C_2H_2}=\dfrac{0,52}{26}=0,02\left(mol\right)\\ n_{C_3H_3Ag}=n_{C_3H_4}=0,045\left(mol\right)\\ n_{Ag_2C_2}=n_{C_2H_2}=0,02\left(mol\right)\\ m_{kt}=m_{C_3H_3Ag}+m_{Ag_2C_2}=0,045.147+240.0,02=11,415\left(g\right)\\ Chọn.B\)

\(n_{Br_2}=\dfrac{32}{160}=0,2\left(mol\right)\\ Đặt:n_{C_3H_6}=a\left(mol\right);n_{C_3H_4}=b\left(mol\right)\\ PTHH:C_3H_6+Br_2\rightarrow C_3H_6Br_2\\ C_3H_4+2Br_2\rightarrow C_3H_4Br_4\\ \Rightarrow\left\{{}\begin{matrix}42a+40b=6,2\\a+2b=0,2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,05\end{matrix}\right.\\ \Rightarrow\%m_{C_3H_6}=\dfrac{0,1.42}{6,2}.100\approx67,74\%\\ Chọn.A\)

\(n_{Br_2}=\dfrac{0,8}{160}=0,005\left(mol\right)\\ C_2H_4\left(dư\right)+Br_2\rightarrow C_2H_4Br_2\\ n_{C_2H_4\left(dư\right)}=n_{Br_2}=0,005\left(mol\right)\\ n_{C_2H_4\left(đã.p.ứ.với.H_2\right)}=\dfrac{2,24}{22,4}-0,005=0,095\left(mol\right)\\ C_2H_4+H_2\rightarrow\left(Ni,t^o\right)C_2H_6\\ Vì:\dfrac{0,095}{1}< \dfrac{3,36:22,4}{1}\Rightarrow C_2H_4hết,H_2dư\\ H=\dfrac{0,095}{0,1}.100\%=95\%\\ Chọn.C\)

22 D => can

23 B => was known

24 D => well

25 B => in

26 B => happily