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27 tháng 6 2023

1.    John collects money.

Money is collected by John.

2.    Anna opened the window.

The window was opened by Anna.

3.    We do our homework.

Our homework is done.

4.    I  asked you a question.

You were asked a question.

5.    He cut out the picture.

The picture was cut out.

6.    The sheep ate a lot.

A lot was eaten by the sheep.

7.    We do not clean our rooms.

Our rooms are not cleaned.

8.    William does not repair the car.

The car isn't repaired by William.

9.    Did Sue draw this circle?

Was this circle drawn by Sue?

10.  Did you feed the dog?

Was the dog fed?

11. He teaches English.

English is taught.

12. The child eats bananas every day.

Bananas are eaten by the child every day.

13. She writes a letter.

A letter is written by her.

14. The master punished the servant.

The servant  was punished by the master.

15. He wrote a book.

A book was written by him.

16. Who wrote this letter?

By whom was this letter written?

17. Somebody cooks meal every day.

Meal is cooked every day.

18. He wore a blue shirt.

A blue shirt was worn by him.

19. They are building a house.

A house is being built.

20. I  finished the job 2 hours ago.

The job was finished 2 hours ago.

21. I sent the report yesterday.

The report was sent  yesterday.

22. She bought a diamond necklace.

A diamond necklace was bought by her.

23. Somebody stole my purse.

My purse was stolen.

Bị động thì quá khứ đơn: S + was/were + PII + (by O)+.....

Bị động thì hiện tại đơn: S + am/is/are + PII + (by O)+....

Bị  động thì hiện tại tiếp diễn: S + am/is/are + being +  PII + (by O)+....

-Trạng từ chỉ nơi chốn/mục đích/ phương tiện/....... + by O + trạng từ chỉ thời gian.

-Trong câu bị động bắt buộc bỏ: by somebody, by someone, by no one, by nobody, by everybody, by everyone, by something, by everything, by nothing.

-Trong câu bị động có thể bỏ: by us, by me, by you, by her, by them, by it, by his.

#\(Errink\times Cream\)

#\(yGLinh\)

27 tháng 6 2023

\(n_{HCl}=3,2.0,5=1,6\left(mol\right)\)

PTHH :

\(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)

0,2       0,4 

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)

0,4         1,2

\(\%m_{Mg}=\dfrac{0,2.24}{15,6}.100\%\approx30,77\%\)

\(\%m_{Al}=\dfrac{0,4.27}{15,6}.100\%\approx69,23\%\)

27 tháng 6 2023

Ủa em cơ sở nào em biết được số mol Mg, Al thế?

27 tháng 6 2023

\(\dfrac{x-1}{x+2}+\dfrac{6x}{x^2-4}=\dfrac{x+1}{2-x}\left(dkxd:x\ne\pm2\right)\)

\(\Leftrightarrow\dfrac{x-1}{x+2}+\dfrac{6x}{\left(x-2\right)\left(x+2\right)}=-\dfrac{x+1}{x-2}\)

\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x-2\right)+6x+\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow x^2-2x-x+2+6x+x^2+2x+x+2=0\)

\(\Leftrightarrow2x^2+6x+4=0\)

\(\Leftrightarrow2x^2+2x+4x+4=0\)

\(\Leftrightarrow2x\left(x+1\right)+4\left(x+1\right)=0\)

\(\Leftrightarrow\left(2x+4\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+4=0\\x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\left(ktm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)

Vậy \(S=\left\{-1\right\}\)

31 tháng 10 2025

Bài 5:

a: Xét ΔOAB và ΔOCD có

\(\hat{OAB}=\hat{OCD}\) (hai góc so le trong, AB//CD)

\(\hat{AOB}=\hat{COD}\) (hai góc đối đỉnh)

Do đó: ΔOAB~ΔOCD

=>\(\frac{OA}{OC}=\frac{OB}{OD}=\frac{AB}{CD}\)

=>\(OA\cdot OD=OB\cdot OC\)

b: Ta có: \(\frac{OA}{OC}=\frac{AB}{CD}\)

=>\(\frac{OA}{OC}=\frac{5}{10}=\frac12\)

=>\(OC=2OA\)

OA+OC=AC

=>2OA+OA=9

=>3OA=9

=>OA=3(cm)

=>\(OC=2\cdot3=6\left(\operatorname{cm}\right)\)

Bài 6:

a: Xét ΔBKA vuông tại K và ΔBAC vuông tại A có

góc KBA chung

Do đó: ΔBKA~ΔBAC

=>\(\frac{BK}{BA}=\frac{BA}{BC}\)

=>\(BK\cdot BC=BA^2\)

b: ΔABC vuông tại A

=>\(AB^2+AC^2=BC^2\)

=>\(AB^2=5^2-3^2=25-9=16=4^2\)

=>AB=4(cm)

Ta có: \(BK\cdot BC=BA^2\)

=>\(BK=\frac{4^2}{5}=3,2\left(\operatorname{cm}\right)\)

BK+CK=BC

=>CK=5-3,2=1,8(cm)

ΔABC~ΔKBA

=>\(\frac{AC}{KA}=\frac{BC}{BA}\)

=>\(AK=\frac{AC\cdot AB}{BC}=\frac{3\cdot4}{5}=2,4\left(\operatorname{cm}\right)\)

c: Xét ΔABC có AD là phân giác

nên \(\frac{BD}{CD}=\frac{AB}{AC}\)

=>\(\frac{BD}{DC}=\frac43\)

=>\(DC=\frac34BD\)

BD+DC=BC

=>\(BD+\frac34BD=5\)

=>\(\frac74BD=5\)

=>\(BD=5:\frac74=\frac{20}{7}\) (cm)

27 tháng 6 2023

A B C D

\(\widehat{A}+\widehat{D}=70^o+110^o=180^o\) 

=> ABCD là tứ giác nội tiếp (tứ giác có tổng 2 góc đối =180 là tứ giác nt)

\(\widehat{ABD}=\widehat{ACD}\) (góc nt cùng chắn cung AD) (1)

\(\widehat{CBD}=\widehat{CAD}\) (góc nt cùng chắn cung CD) (2)

Tg ADC cân tại D \(\Rightarrow\widehat{ACD}=\widehat{CAD}\) (3)

Từ (1) (2) (3) \(\Rightarrow\widehat{ABD}=\widehat{CBD}\)

27 tháng 6 2023

\(a,\left(x-2\right)\left(3x-1\right)=x\left(2-x\right)\)

\(\Leftrightarrow\left(x-2\right)\left(3x-1\right)+x\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(3x-1+x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\4x-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{4}\end{matrix}\right.\)

Vậy \(S=\left\{\dfrac{1}{4};2\right\}\)

\(b,\left|2x+3\right|=4x+1\)

\(TH_1:x\ge-\dfrac{3}{2}\)

\(2x+3=4x+1\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\left(tm\right)\)

\(TH_2:x< -\dfrac{3}{2}\)

\(-2x-3=4x+1\\ \Leftrightarrow-6x=4\Leftrightarrow x=-\dfrac{2}{3}\left(ktm\right)\)

Vậy \(S=\left\{1\right\}\)

\(c,\dfrac{x+1}{3}+1=3-\dfrac{5x}{10}\\ \Leftrightarrow\dfrac{10\left(x+1\right)+30-90+15x}{30}=0\\ \Leftrightarrow10x+10-60+15x=0\\ \Leftrightarrow25x=50\\ \Leftrightarrow x=2\)

Vậy \(S=\left\{2\right\}\)

\(d,\dfrac{1}{x+2}+\dfrac{3}{2-x}=\dfrac{2x-3}{x^2-4}\left(dk:x\ne\pm2\right)\)

\(\Leftrightarrow\dfrac{x-2-x-2-2x+3}{x^2-4}=0\)

\(\Leftrightarrow-2x=1\)

\(\Leftrightarrow x=-\dfrac{1}{2}\left(tmdk\right)\)

Vậy \(S=\left\{-\dfrac{1}{2}\right\}\)

27 tháng 6 2023

Câu d, Sửa từ dòng 2 :

\(\Leftrightarrow\dfrac{x-2-3x-6-2x+3}{x^2-4}=0\)

\(\Leftrightarrow-4x=5\)

\(\Leftrightarrow x=-\dfrac{5}{4}\left(tm\right)\)

Vậy ...

27 tháng 6 2023

\(a,\) P xác định \(\Leftrightarrow x^2-4\ne0\Leftrightarrow x\ne\pm2\)

\(b,P=\left(\dfrac{x}{x^2-4}+\dfrac{2}{2-x}+\dfrac{1}{x+2}\right):\dfrac{1}{x+2}\)

\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x+2}\right).\left(x+2\right)\)

\(=\dfrac{x-2\left(x+2\right)+x-2}{\left(x+2\right)\left(x-2\right)}.\left(x+2\right)\)

\(=\dfrac{x-2x-4+x-2}{x-2}\)

\(=\dfrac{-6}{x-2}\)