Bài 3:

theo đề bài ta có:

\(\left\{{}\begin{matrix}2a-3b=0\\5b-7c=0\\3a-7b+5c=30\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=42\\b=28\\c=20\end{matrix}\right.\)

Bài 4: 

Đặt \(\dfrac{x}{4}=\dfrac{y}{5}=\dfrac{z}{6}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=4k\\y=5k\\z=6k\end{matrix}\right.\)

Ta có: \(x^2-2y^2+z^2=18\)

\(\Leftrightarrow16k^2-50k^2+36k^2=18\)

\(\Leftrightarrow k^2=9\)

Trường hợp 1: k=3

\(\Leftrightarrow\left\{{}\begin{matrix}x=4k=4\cdot3=12\\y=5k=5\cdot3=15\\z=6k=6\cdot3=18\end{matrix}\right.\)

Trường hợp 2: k=-3

\(\Leftrightarrow\left\{{}\begin{matrix}x=4k=-3\cdot4=-12\\y=5k=-3\cdot5=-15\\z=6k=-3\cdot6=-18\end{matrix}\right.\)

16C

17A

18B

19D

20C

21C

22A 

16 C

17 A

18 B

19 D

20 C

21 C

22 A 

24 Yes,they do

Còn lại đúng hết rồi nhé

\(\dfrac{-188}{2017}< 0;\dfrac{1}{3678}>0\\ \Rightarrow\dfrac{-188}{2017}< \dfrac{1}{3678}\)

 Dấu <

HT

vì\(\dfrac{-188}{2017}\)bé hơn 0 nên \(\dfrac{-188}{2017}\)<\(\dfrac{1}{3678}\)

Para 1 - b

Para 2 - a

Para 3 - c

T - F - T - T - NG

1 B

2 A

3 D

4 D

5 A

1 C

2 D

3 A

4 A

5 C

6 B

7 A

8 C

9 B

12. C

13. D

14. A

15. A

16. C

17. B

18. D

19. C

20. B

tk mik nha

1.C

2.D

3.A

4.A

5.C

6.B

7.A

8.C