Hộ e nốt ah
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a. Mg + 2HCl ---> MgCl2 + H2↑
b. Fe2O3 + 6HCl ---> 2FeCl3 + 3H2O
c. 2Al + 6HCl ---> 2AlCl3 + 3H2↑
d. 2C2H6 + 7O2 ---to---> 4CO2↑ + 6H2O
e. BaCl2 + 2AgNO3 ---> Ba(NO3)2 + 2AgCl↓
f. Al2(SO4)3 + 3Ba(OH)2 ---> 2Al(OH)3 + 3BaSO4↓
Trong đó:
to: nhiệt độ
↑: bay hơi
↓: kết tủa
(Câu f sai nên mik sửa từ Al2(SO4)3 thành Al(OH)3)
III
1 appreciated
2 worried
3 tense
4 confident
5 delighted
6 frustrated
7 calm
8 relaxed
9 self-disciplined
10 depressed
V
1 had resolved
2 has taken
3 Did you say
4 not to take
5 wanted
6 is being repaired
7 taking
8 to think
9 had worked
10 was washing - dropped
11 will find
12 faces
13 turned
14 to give
15 will empathise
Đặt \(A=\dfrac{1}{\sqrt[3]{a+7b}}+\dfrac{1}{\sqrt[3]{b+7c}}+\dfrac{1}{\sqrt[3]{c+7a}}\)
\(A=\dfrac{\sqrt[3]{64}}{\sqrt[3]{8.8\left(a+7b\right)}}+\dfrac{\sqrt[3]{64}}{\sqrt[3]{8.8\left(b+7c\right)}}+\dfrac{\sqrt[3]{64}}{\sqrt[3]{8.8\left(c+7a\right)}}\)
\(\ge\dfrac{4}{\dfrac{8+8+a+7b}{3}}+\dfrac{4}{\dfrac{8+8+b+7c}{3}}+\dfrac{4}{\dfrac{8+8+c+7a}{3}}\ge\dfrac{\left(2+2+2\right)^2}{\dfrac{8+8+a+7b+8+8+b+7c+8+8+c+7a}{3}}\)
\(=\dfrac{36.3}{8\left(a+b+c\right)+48}=\dfrac{3}{2}\)
Vậy \(A_{min}=\dfrac{3}{2}\Leftrightarrow a=b=c=1\)
4, were
5, was
6, were
7, was
8, were
9, was
10, were
11, were
12, was
13, was
14, was
4.was
5.was
6.were
7.was
8.were
9.was
10.were
11.were
12.was
13.was
14.was
Câu 4:
\(\dfrac{25}{26}-\dfrac{15}{26}=\dfrac{5}{13}\)
\(\dfrac{46}{39}-\dfrac{11}{13}=\dfrac{1}{3}\)
\(\dfrac{37}{45}-\dfrac{5}{9}=\dfrac{4}{15}\)
\(1-\dfrac{1}{3}=\dfrac{2}{3}\)
Câu 5:
Cạnh ngắn là:
\(2-\dfrac{1}{3}=\dfrac{5}{3}\left(m\right)\)
Chu vi hình bình hành là:
\(\left(2+\dfrac{5}{3}\right)\times2=\dfrac{22}{3}\left(m\right)\)
Nửa chu vi hình bình hành là:
\(\dfrac{22}{3}:2=\dfrac{11}{3}\left(m\right)\)
Chọn D
Phương trình \(\Delta\) có dạng:
\(y=m\left(x+1\right)-2\Leftrightarrow y=mx+m-2\)
Phương trình hoành độ giao điểm (P) và \(\Delta\):
\(-\dfrac{1}{2}x^2=mx+m-2\Leftrightarrow x^2+2mx+2m-4=0\) (1)
\(\Delta'=m^2-2m+4=\left(m-1\right)^2+3>0\) ; \(\forall m\)
\(\Rightarrow\) (1) luôn có 2 nghiệm pb với mọi m hay (P) luôn cắt \(\Delta\) tại 2 điểm pb
b.
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_A+x_B=-2m\\x_Ax_B=2m-4\end{matrix}\right.\)
Đặt \(A=x_A^2x_B+x_Ax_B^2=x_Ax_B\left(x_A+x_B\right)\)
\(A=-2m\left(2m-4\right)=-4m^2+8m=-4\left(m-1\right)^2+4\le4\)
\(A_{max}=4\) khi \(m=1\)
V
1 has won
2 haven't seen
3 have taken
4 has visited
5 have read
41. A
42. B
V.
1. has won
2. haven't studied
3. have taken
4. has visited
5. have read