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24 tháng 7 2021

refer

1. like/dinner/house/would/have/tonight/you/to/my/at/?
Would you like to have dinner at my house tonight?

2.detective/parents/going/movie/this afternoon/we/see/with/are/a/our/to/.
We are going to see a detective movies with our parents this afternoon.

3.watching/prefers/to/brother/books/TV/my/reading.
My brother prefers watching TV and reading to books.

4.daughter / best/kinds/does /like/what/TV program/your /of/?
What kind of program does your daughter like best?

5. showing/local cinema/is/horror film / the/a/there/next week/at.
There is horror film showing at that local cinema next week.

6.durians /and/my/ like/much/does/I/very/father/so
I like durians very much so does my father.

7.went/she/stomachache/the doctor's/ an/Lan/because/awful to/ had.
Lan's went to the doctor's because she had an awful stomache.

8.have/the/farm/often/them/vegatables/from/dirt/on.
Vegetables from the farm of ten have dirt on time.

9.in/work/will/do/all/ the/machines/for/the/future/us.
In the future, machines will do all the work for us.

10.it/hours/about/two/Hoa Binh/to takes /by /get/to/couch.
It 's take us about to hours to get to Hoa Binh by couch.

24 tháng 7 2021

1 Would you like to have dinner at my house tonight?

2 We are going to see a detective movie with our parents this afternoon

3 My brother prefers watching TV to reading books

4 What kinds of TV program does your daughter like best?

5 There is a horror film showing at the local cinema next week

7 I like durians very much and so does my father

8 Vegetables often have dirt from the farm on them

9 Machines will do all the work for us in the future

10 It's takes about 2 hours to get to HoaBinh by coach

Bài 10:

a: \(\overrightarrow{AB}+\overrightarrow{BO}+\overrightarrow{OA}\)

\(=\overrightarrow{AO}+\overrightarrow{OA}=\overrightarrow{0}\)

b: \(\overrightarrow{OA}+\overrightarrow{BC}+\overrightarrow{DO}+\overrightarrow{CD}\)

\(=\overrightarrow{OA}+\overrightarrow{DO}+\overrightarrow{BD}\)

\(=\overrightarrow{OA}+\overrightarrow{BO}=\overrightarrow{BA}\)

26 tháng 10 2021

b: \(BC=\sqrt{89}\left(cm\right)\)

\(\sin\widehat{B}=\dfrac{5\sqrt{89}}{89}\)

\(\Leftrightarrow\widehat{B}\simeq32^0\)

\(\widehat{C}=58^0\)

Bài 8:

a: Ta có: \(M=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}\)

\(=\dfrac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)

b: Thay \(x=11-6\sqrt{2}\) vào M, ta được:

\(M=\dfrac{3-\sqrt{2}+1}{3-\sqrt{2}-3}=\dfrac{4-\sqrt{2}}{-\sqrt{2}}=-2\sqrt{2}+1\)

1 tháng 10 2021

Bài 8:

a) \(M=\dfrac{2\sqrt{x}-9-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{2\sqrt{x}-9-x+9+2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)

b) \(M=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=\dfrac{\sqrt{11-6\sqrt{2}}+1}{\sqrt{11-6\sqrt{2}}-3}=\dfrac{\sqrt{\left(3-\sqrt{2}\right)^2}+1}{\sqrt{\left(3-\sqrt{2}\right)^2}-3}=\dfrac{4-\sqrt{2}}{-\sqrt{2}}=1-2\sqrt{2}\)

c) \(M=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=3\)

\(\Leftrightarrow3\sqrt{x}-9=\sqrt{x}+1\Leftrightarrow2\sqrt{x}=10\Leftrightarrow\sqrt{x}=5\Leftrightarrow x=25\left(tm\right)\)

d) \(M=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}< 1\)

\(\Leftrightarrow\sqrt{x}+1< \sqrt{x}-3\Leftrightarrow1< -3\left(VLý\right)\)

Vậy \(S=\varnothing\)

e) \(M=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=1+\dfrac{4}{\sqrt{x}-3}\in Z\)

\(\Rightarrow\sqrt{x}-3\inƯ\left(4\right)=\left\{-4;-2;-1;1;2;4\right\}\)

Kết hợp đk:

\(\Rightarrow x\in\left\{1;16;25;49\right\}\)

26 tháng 10 2021

câu 5: 

x=3,6

y=6,4

câu 6: chụp lại đề

câu 7:

a)ĐKXĐ: \(x\ge0\)

\(3\sqrt{x}=\sqrt{12}\\ \Rightarrow9x=12\\ \Rightarrow x=\dfrac{4}{3}\)

b) ĐKXĐ: \(x\ge6\)

\(\sqrt{x-6}=3\\ \Rightarrow x-6=9\\ \Rightarrow x=15\)

26 tháng 10 2021

Câu 5: 

Áp dụng định lý Pi-ta-go ta có:

\(AB^2+AC^2=BC^2\\ \Rightarrow BC=\sqrt{6^2+8^2}\\ \Rightarrow BC=10\)

Áp dụng HTL ta có: \(x.BC=AB^2\Rightarrow x.10=6^2\Rightarrow x=3,6\)

Áp dụng HTL ta có: \(x.BC=AC^2\Rightarrow x.10=8^2\Rightarrow x=6,4\)

25 tháng 10 2021

Bài 8:

a) \(A=\dfrac{\sqrt{a}-1}{\sqrt{a}}.\dfrac{\sqrt{a}+1+\sqrt{a}-1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}=\dfrac{2\sqrt{a}}{\sqrt{a}\left(\sqrt{a}+1\right)}=\dfrac{2}{\sqrt{a}+1}\)

b) \(A=\dfrac{2}{\sqrt{a}+1}=\dfrac{2}{\sqrt{3-2\sqrt{2}}+1}=\dfrac{2}{\sqrt{\left(\sqrt{2}-1\right)^2}+1}=\dfrac{2}{\sqrt{2}-1+1}=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)

Bài 9:

\(pt\Leftrightarrow\sqrt{\left(3x+1\right)^2}=2\)\(\Leftrightarrow\left|3x+1\right|=2\)

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

24 tháng 11 2023

bài 9:

Kẻ OI vuông góc KH

=>OI là khoảng cách từ O đến KH

ΔOIM vuông tại I

=>OI<OM

Xét (O) có

OI,OM lần lượt là khoảng cách từ O đến KH,AB

KH,AB là các dây cung của (O)

OI<OM

Do đó: KH>AB

30 tháng 11 2021

Bài 10:

a: =254-254+135=135

17 tháng 5 2021

a)

$Mg + H_2SO_4 \to MgSO_4 + H_2$
$Mg + 2H_2SO_{4_{đặc}} \to MgSO_4 + SO_2 + 2H_2O$
$Cu + 2H_2SO_{4_{đặc}} \to CuSO_4 + SO_2 + 2H_2O$

Theo PTHH : 

n Mg = n H2 = 0,448/22,4 = 0,02(mol)

Mặt khác : n SO2 = 2,688/22,4 =0,12(mol)

4m gam X chứa 4n Cu và 0,02.4 = 0,08 mol Mg

Theo PTHH : n SO2 = n Cu + n Mg

=> n Cu = (0,12- 0 ,08)/4 = 0,01(mol)

Suy ra:  m = 0,01.64 + 0,02.24 = 1,12(gam)

b) $SO_2 + Br_2 + 2H_2O \to 2HBr + H_2SO_4$

n Br2 = n SO2 = 0,12(mol)

=> V dd Brom  = 0,12/0,02 = 6(lít)