b: Ta có: 2x=3y=5z

\(\Leftrightarrow\dfrac{x}{\dfrac{1}{2}}=\dfrac{y}{\dfrac{1}{3}}=\dfrac{z}{\dfrac{1}{5}}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{\dfrac{1}{2}}=\dfrac{y}{\dfrac{1}{3}}=\dfrac{z}{\dfrac{1}{5}}=\dfrac{x-2y+z}{\dfrac{1}{2}-\dfrac{2}{3}+\dfrac{1}{5}}=\dfrac{14}{\dfrac{1}{30}}=420\)

Do đó: x=210; y=140; z=84

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y-z}{2+3-4}=-\dfrac{20}{1}=-20\)

⇒\(\left\{{}\begin{matrix}x=-20.2=-40\\y=-20.3=-60\\z=-20.4=-80\end{matrix}\right.\)

3:

Xét ΔABO và ΔADO có

\(\widehat{AOB}=\widehat{AOD}\)

OA chung

\(\widehat{BAO}=\widehat{DAO}\)

Do đó: ΔABO=ΔADO

=>\(\widehat{ABO}=\widehat{ADO}\)(ĐPCM)

Bài 4:

Xét ΔBAM và ΔBDM có

\(\widehat{ABM}=\widehat{DBM}\)

BM chung

\(\widehat{AMB}=\widehat{DMB}\)

Do đó: ΔBAM=ΔBDM

=>\(\widehat{BAM}=\widehat{BDM}=90^0\)

=>DM\(\perp\)BC

5:

a: Xét ΔBAC có

\(\widehat{ABC}+\widehat{ACB}+\widehat{BAC}=180^0\)

=>\(\widehat{BAC}=180^0-70^0-30^0=80^0\)

b: AD là phân giác của góc BAC

=>\(\widehat{BAD}=\widehat{CAD}=\dfrac{80^0}{2}=40^0\)

Xét ΔADC có góc ADB là góc ngoài tại đỉnh D

nên \(\widehat{ADB}=\widehat{DAC}+\widehat{C}=40^0+30^0=70^0\)

ΔAHD vuông tại H

=>\(\widehat{HAD}+\widehat{HDA}=90^0\)

=>\(\widehat{HAD}=90^0-70^0=20^0\)

a, \(n_{N_2}=\dfrac{4,48}{22,4}=0,2mol\)

b, \(Số.phân.tử.N_2=0,2.6.10^{23}=1,2.10^{23}\) ( phân tử )

c, \(m_{N_2}=0,2.28=5,6g\)

\(\frac{-3}{7}\).\(^{\left(-3\right)^2}\)-\(\sqrt{\frac{4}{49}}\)

= \(\frac{-3}{7}.9-\sqrt{\frac{4}{49}}\)

=\(\frac{-27}{7}-\sqrt{\frac{4}{49}}\)

=\(\frac{-27}{7}-\frac{2}{7}\)

=\(\frac{-29}{7}\)

Chúc bạn học tốt

\(\left|-\frac{3}{7}\right|\cdot(-3)^2-\sqrt{\frac{4}{49}}\)

\(=\frac{3}{7}\cdot9-\frac{2}{7}\)

\(=\frac{27}{7}-\frac{2}{7}=\frac{25}{7}\)

11 D

12 B

13 A

14 B

15 A

16 A

17 A

18 B

19 C

20 B

21 B

22 C

23 C

24 C

25 B

26 heated

27 solar

28 pollution

if you open the window,you will enjoy the air

if you are not healthy,you can live longer

if you don't work harder ,you will get higher position

if you are quick ,you can meet Cao Thanh Dung

if Mary is short ,she can't play basketball

\(R1//R2\)

\(\Rightarrow\left\{{}\begin{matrix}Rtd=\dfrac{R1R2}{R1+R2}=8\Omega\\U2=U1=Ia1.R1=12V\Rightarrow Ia2=\dfrac{U2}{R2}=\dfrac{12}{40}=0,3A\\Ia=Ia1+Ia2=0,3+1,2=1,5A\\Uv=U2=12V\end{matrix}\right.\)

1. c

2. b

3. d

4. c

5. a

6. b

7. c

8. a

9. d

10. b