Ai giải giúp em vs 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)
\(1,\\ a,=6x^4y^4-x^3y^3+\dfrac{1}{2}x^4y^2\\ b,=4x^3+5x^2-8x^2-10x+12x+15\\ =4x^3-3x^2+2x+15\\ 2,\\ a,=7\left(x^2-6x+9\right)=7\left(x-3\right)^2\\ b,=\left(x-y\right)^2-36=\left(x-y-6\right)\left(x-y+6\right)\\ 3,\\ \Leftrightarrow x\left(x^2-0,36\right)=0\\ \Leftrightarrow x\left(x-0,6\right)\left(x+0,6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=0,6\\x=-0,6\end{matrix}\right.\)
![](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)
II
1 C
2A
3A
4B
5A
III
1 A
2 B
3 C
4 B
5 C
6 B
7 A
8 C
9 D
10 C
11 A
12 B
13 C
14 A
15 B
16 C
17 C
18 B
19 A
20 B
21 C
22 A
23 D
24 D
25 D
26 A
27 A
28 C
29 B
30 B
31 A
32 D
33 B
34 B
35 C
36 C
37 B
38 D
39 A
40 B
41 C
42 A
![](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)
Ta có: \(3x=4y\Rightarrow\dfrac{x}{4}=\dfrac{y}{3}\Rightarrow\dfrac{x}{20}=\dfrac{y}{15}\)
\(2y=5z\Rightarrow\dfrac{y}{5}=\dfrac{z}{2}\Rightarrow\dfrac{y}{15}=\dfrac{z}{6}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{20}=\dfrac{y}{15}=\dfrac{z}{6}=\dfrac{x+z}{20+6}=\dfrac{52}{26}=2\)
\(\Rightarrow\left\{{}\begin{matrix}x=20.2=40\\y=15.2=30\\z=6.2=12\end{matrix}\right.\)
4 D
5 A
6 A
7 A
8 D
8 D
10 B