Giúp Mình Bài 4 Với Ạ🥺
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
1 B
2 B
3 A
4 A
5 D
6 A
7 C
8 C
9 C
10 B
11 A
12 C
13 C
14 C
15 A
II
1 friendship
2 impression
3 religious
4 tropical
5 compulsorily
6 tropical
III
1 I could swim well
2 to be late for school
3 to take me to the zoo when I was a child
4 to live in Hue
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 7 :
\(a) Na_2O + H_2O \to 2NaOH\\ SO_2 + H_2O \to H_2SO_3\\ b) Na_2O + 2HCl \to 2NaCl + H_2O\\ Fe_2O_3 + 6HCl \to 2FeCl_3 + 3H_2O\\ c) 2NaOH + SO_2 \to Na_2SO_3 + H_2O\)
Bài 8 :
a) Trích mẫu thử
Cho mẫu thử vào nước :
- mẫu thử tan là $Na_2O$
$Na_2O + H_2O \to 2NaOH$
- mẫu thử không tan là CuO
b) Trích mẫu thử
Cho giấy quỳ tím ẩm vào :
- mẫu thử hóa đỏ là $P_2O_5$
$P_2O_5 + 3H_2O \to 2H_3PO_4$
- mẫu thử hóa xanh là $Na_2O$
$Na_2O + H_2O \to 2NaOH$
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
4b.
\(\dfrac{\pi}{2}< a< \pi\Rightarrow cosa< 0\Rightarrow cosa=-\sqrt{1-sin^2a}=-\dfrac{4}{5}\)
\(\Rightarrow tana=\dfrac{sina}{cosa}=-\dfrac{3}{4}\)
\(tan\left(a+\dfrac{\pi}{3}\right)=\dfrac{tana+tan\left(\dfrac{\pi}{3}\right)}{1-tana.tan\left(\dfrac{\pi}{3}\right)}=\dfrac{-\dfrac{3}{4}+\sqrt{3}}{1-\left(-\dfrac{3}{4}\right).\sqrt{3}}=...\)
c.
\(\dfrac{3\pi}{2}< a< 2\pi\Rightarrow cosa>0\Rightarrow cosa=\sqrt{1-sin^2a}=\dfrac{5}{13}\)
\(cos\left(\dfrac{\pi}{3}-a\right)=cos\left(\dfrac{\pi}{3}\right).cosa+sin\left(\dfrac{\pi}{3}\right).sina=\dfrac{1}{2}.\dfrac{5}{13}+\left(-\dfrac{12}{13}\right).\dfrac{\sqrt{3}}{2}=...\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a.
Đặt \(x+2y+1=a\)
\(\Rightarrow P=a^2+\left(a+4\right)^2=2a^2+8a+16=2\left(a+2\right)^2+8\ge8\)
\(P_{min}=8\) khi \(a=-2\) hay \(x+2y+3=0\)
b.
\(\sqrt{x}-1=a\ge0\Rightarrow\sqrt{x}=a+1\Rightarrow x=a^2+2a+1\)
\(Q=\dfrac{\left(a^2+2a+1\right)+\left(a+1\right)+1}{a}=\dfrac{a^2+3a+3}{a}=a+\dfrac{3}{a}+3\ge2\sqrt{\dfrac{3a}{a}}+3=3+2\sqrt{3}\)
\(Q_{min}=3+2\sqrt{3}\) khi \(a=\sqrt{3}\) hay \(x=4+2\sqrt{3}\)
a: \(\sqrt{0.49a^2}=-0.7a\)
b: \(\sqrt{25\left(7-a\right)^2}=5\left(a-7\right)=5a-35\)
c: \(\sqrt{a^4\left(a-2\right)^2}=a^2\left(a-2\right)=a^3-2a^2\)