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27 tháng 10 2021

a) \(x=x_0+v_0t+\dfrac{1}{2}at^2=10+5t+\dfrac{1}{2}\cdot1\cdot t^2=10+5t+0,5t^2\)

b) Sau 20s:

   Vận tốc vật: \(v=v_0+at=5+1\cdot20=25\)m/s

   Vật cách gốc tọa độ: \(x=10+5\cdot20+0,5\cdot20^2=310\left(m\right)\)

Ta có: \(\sqrt{4x^2-4x+9}=3\)

\(\Leftrightarrow4x^2-4x=0\)

\(\Leftrightarrow4x\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

14 tháng 8 2021

\(\sqrt{x^2-x+16}=4\)

\(\Rightarrow x^2-x+16=16\\ \Rightarrow x^2-x=0\\ \Rightarrow x\left(x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

Ta có: \(\sqrt{x^2-x+16}=4\)

\(\Leftrightarrow x^2-x=0\)

\(\Leftrightarrow x\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

6 tháng 11 2021

24 C

25 A

26 C

27 C

28 D

29 D

30 C

6 tháng 11 2021

Thank bn❤️

13 tháng 4 2022

58.They have learned English since 2010. 
59. It's time for you to go to school now.
60.She has bought that house since 5 years ago.
61.The car is too expensive for him to buy.
62. Who was this novel written by?

13 tháng 4 2022

60&62 còn có thể có cách làm khác:

60.She has bought that house for 5 years.

62. Who was the author of this novel?

 

63. Lan is always forgetting her homework.

64. Tom is the worst Vietnamese speaker in his Vietnamese club.

65. Is the Temple of Literature surrounded by four busy streets?

66. Despite the rough sea, they still went to the island.

67. How far is it from your house to your school?

68, My brother used to go to school late when he was small.

69. Although it rained heavily, they still went out.

70. Some trees have been planted along the roads to the school by the children.

e: \(=3x^6-x^3+4\)

12 tháng 3 2022

a, \(40x-20+45x-30=48x-36\Leftrightarrow37x=14\Leftrightarrow x=\dfrac{14}{37}\)

b, đk : x khác -3 ; 3 

\(5x+15+4x-12=x-5\Leftrightarrow8x=-38\Leftrightarrow x=-\dfrac{19}{4}\)(tm) 

c, \(\left[{}\begin{matrix}2x+3=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)