a)Hóa trị 4

b)Ba hóa trị 2

a) Mn hóa trị IV

b) Ba hóa trị II

Có thể giải ra lun được ko ạ 

a) Ta có: O hóa trị II, ta gọi hóa trị Mn cần tìm trong hợp chất MnO2 là x: \(Mn^xO_2^{II}\)

Theo quy tắc hóa trị, ta có:

1.x=2.II

=>x= (2.II)/1= IV

=> Hóa trị x của Mn cần tìm trong hợp chất MnO2 là IV.

a) Ta có: (PO4) hóa trị II, ta gọi hóa trị Mn cần tìm trong hợp chất  là y: \(Ba^y_3\left(PO_4\right)^{III}_2\)

Theo quy tắc hóa trị, ta có:

3.y=III.2

=>y=(III.2)/3=II

=> Hóa trị y của Ba cần tìm trong hợp chất Ba3(PO4)2 là II.

1 D

2 C

3 D

4 C

5 B

6 D

7 A

8 A

9 B

10 D

11 C

12 B

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#\(Vion.Serity\)

#\(yGLinh\)

1 telling

2 to come

3 having

4  talking

5 to speak

6 giving

7  carry

8 to study

9 waiting

10  to start

11 to help

12 going 

13 to bring

14to visit

15 going

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1 preparing

2 working - finishing

3 to give - smoking

4 talking - eating

5 arguing -  working

6  to think - making

7 to come - standing

8  solving

9to lock - going

10 to persuade - change

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1 watching - reading

2 playing - doing

3  to go

4 Did you see

5 to dream - were

6 showing - to send

7 going

8 doing

9  reading

10 to seeing

Bài 4

Ta có: \(\left(4+2x\right)\left(4-2x\right)+\left(2x-3\right)^2=2\)

\(\Leftrightarrow16-4x^2+4x^2-12x+9=2\)

\(\Leftrightarrow-12x=-23\)

hay \(x=\dfrac{23}{12}\)

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Bài 1:

\(54\left(\dfrac{km}{h}\right)=15\left(\dfrac{m}{s}\right);9\left(\dfrac{m}{s}\right)=32,4\left(\dfrac{km}{h}\right)\)

Baì 2:

\(t'=s':v'=5:\left(5.3,6\right)=\dfrac{5}{18}h\)

\(\Rightarrow v_{tb}=\dfrac{s'+s''}{t'+t''}=\dfrac{5+3,8}{\dfrac{5}{18}+\left(\dfrac{15}{60}\right)}\simeq16,67\left(\dfrac{km}{h}\right)\)

Câu 2: 

\(\Leftrightarrow\left(x+2\right)\left(10x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-\dfrac{3}{10}\end{matrix}\right.\)