Bài 4: 

a: Xét tứ giác OBAC có 

\(\widehat{OBA}+\widehat{OCA}=180^0\)

Do đó: OBAC là tứ giác nội tiếp

hay O,B,A,C cùng thuộc 1 đường tròn

Bài 5:

\(\sqrt{x+2021}-y^3=\sqrt{y+2021}-x^3\\ \Leftrightarrow\left(\sqrt{x+2021}-\sqrt{y+2021}\right)+\left(x^3-y^3\right)=0\\ \Leftrightarrow\dfrac{x-y}{\sqrt{x+2021}+\sqrt{y+2021}}+\left(x-y\right)\left(x^2+xy+y^2\right)=0\\ \Leftrightarrow\left(x-y\right)\left(\dfrac{1}{\sqrt{x+2021}+\sqrt{y+2021}}+x^2+xy+y^2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-y=0\\\dfrac{1}{\sqrt{x+2021}+\sqrt{y+2021}}+x^2+xy+y^2=0\left(1\right)\end{matrix}\right.\)

Dễ thấy \(\left(1\right)>0\) với mọi x,y

Do đó \(x-y=0\) hay \(x=y\)

\(\Leftrightarrow M=x^2+2x^2-2x^2+2x+2022=x^2+2x+1+2021\\ \Leftrightarrow M=\left(x+1\right)^2+2021\ge2021\)

Dấu \("="\Leftrightarrow x=y=-1\)

Bạn tham luận về gì thế?

mk tham luận về lao động và đạo đức, giúp mk với

Bạn tham khảo nhé (về lao động nè)

  Ai trong mỗi chúng ta cũng đều mong muốn được làm việc, đươc sinh hoạt trong một môi trường xanh- sạch- đẹp và an toàn .Trường học xanh, sạch, đẹp, an toàn đem lại cho ta không gian học tập, vui chơi hấp dẫn, giúp ta càng thêm yêu quý trường lớp, thầy cô, bạn bè .Vậy có ai trong mỗi chúng ta tự hỏi mình đã làm gì để bảo vệ môi trường xanh - sạch - đẹp an toàn chưa? Vâng, một câu hỏi không khó nhưng cũng không khiến chúng ta dễ trả lời phải không ạ! Và trong Đại hội hôm nay ,chúng ta hãy cùng tìm hiểu làm thế nào để trường lớp xanh –sạch- đẹp-an toàn nhé !

1 knowing the answer, they told us

2 the unique handicraft, we didn't like it

3 We looked into these machines so that we could find out the method 

We looked into these machines so as to find out the method 

4 Jack's poverty, everyone looks up to him

5 In order to choose the best chair, Linda looked through this catalouge 

6 is not as dirty as this one

7 I didn't play as successfully as Linda

8 A dog can't run as fast as a horse

A dog runs more slowly than a horse

9 No rivers in the world is as long as the Nile

10 No one in my family talks as politely as Linda

1 was arrested before the garden by the boy yesterday

2 his cats taken care of carefully by Jack everyday

3 Has the country been protected from Covid-19 successfully

4 is reported that the man is running out of money

is reported to be running out of money\

5 has been said that the thief got out of the prison

has been said to have got out of the prison

6 was thought that Mary had turned down the job

Mary was thought to have  turned down the job

\(1,\\ a,=\dfrac{\left(3+2\sqrt{3}\right)\sqrt{3}}{3}+\dfrac{\left(2+\sqrt{2}\right)\left(\sqrt{2}-1\right)}{1}\\ =\dfrac{3\sqrt{3}+6}{3}+\sqrt{2}=\sqrt{3}+1+\sqrt{2}\\ b,=\left(\dfrac{\sqrt{5}+\sqrt{2}}{3}-\dfrac{\sqrt{5}-\sqrt{2}}{3}+1\right)\cdot\dfrac{1}{\left(\sqrt{2}+1\right)^2}\\ =\dfrac{\sqrt{5}+\sqrt{2}-\sqrt{5}+\sqrt{2}+3}{3}\cdot\dfrac{1}{\left(\sqrt{2}+1\right)^2}\\ =\dfrac{2\sqrt{2}+3}{3\left(3+2\sqrt{2}\right)}=\dfrac{1}{3}\)

\(2,\\ A=2x+\sqrt{\left(x-3\right)^2}=2x+\left|x-3\right|\\ =2\left(-5\right)+\left|-5-3\right|=-10+8=-2\\ B=\dfrac{\sqrt{\left(2x+1\right)^2}}{\left(x-4\right)\left(x+4\right)}\left(x-4\right)^2=\dfrac{\left|2x+1\right|\left(x-4\right)}{x+4}\\ B=\dfrac{17\cdot4}{12}=\dfrac{17}{3}\)

\(3,\\ a,\dfrac{\left(1+\sqrt{x}\right)^2-4\sqrt{x}}{1-\sqrt{x}}\\ =\dfrac{\sqrt{x}-2\sqrt{x}+1}{1-\sqrt{x}}=\dfrac{\left(1-\sqrt{x}\right)^2}{1-\sqrt{x}}=1-\sqrt{x}=1-\sqrt{2}\)

\(b,\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2+4\sqrt{xy}}{1+\sqrt{xy}}\\ =\dfrac{x+2\sqrt{xy}+y}{1+\sqrt{xy}}=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^2}{1+\sqrt{xy}}\\ =\dfrac{\left(\sqrt{2}+\sqrt{3}\right)^2}{1+\sqrt{6}}=\dfrac{5+2\sqrt{6}}{1+\sqrt{6}}\\ =\dfrac{\left(5+2\sqrt{6}\right)\left(\sqrt{6}-1\right)}{5}\\ =\dfrac{3\sqrt{6}+7}{5}\)

a) Ta có: 

mOn=90omOn=90o

mà xOm+mOn+x′On=180oxOm+mOn+x′On=180o

⇒ xOm+90o+x′On=180oxOm+90o+x′On=180o

⇒ xOm+x′On=90oxOm+x′On=90o

⇒ (4x−10o)+(3x−5o)=90o(4x−10o)+(3x−5o)=90o

⇒ 4x−10o+3x−5o=90o4x−10o+3x−5o=90o

⇒ (4x+3x)+(−10o−5o)=90o(4x+3x)+(−10o−5o)=90o

⇒ 7x−15o=90o7x−15o=90o

⇒ 7x=105o7x=105o

⇒ x=15x=15

⇒ xOm=4.15o−10o=50oxOm=4.15o−10o=50o

    x′On=90o−50o=40ox′On=90o−50o=40o

b) Ta có: 

xOtxOt và nOx′nOx′ là 2 góc đối đỉnh

⇒ Ot là tia đối On (1)

mà tOy=90otOy=90o

⇒ Oy là tia đối Om (2)

Từ (1), (2) ⇒ mOnmOn và tOytOy là 2 góc đối đỉnh

1 My parents go shopping twice a week

2 Hoa's house has a balcony

3 My brother usually plays badminton with his friends

4 My favorite book is Tam and Cam. What is yours?

5 There are 10 pencil cases on the table

1, My parents go shopping twice a week.

2, Hoa's house has a balcony.

3, My brother usually plays badminton with his friends.

4, My favorite book is Tam and Cam. What's yours?

5, There are ten pencil cases on the table.

bài 1

\(\widehat{B}=90-\widehat{C}=90-30=60\)

\(sinC=\dfrac{AB}{BC}\Rightarrow BC=\dfrac{AB}{sinC}=\dfrac{30}{sin30}=60\)

áp dụng pytago vào \(\Delta ABC\)

\(AC=\sqrt{BC^2-AB^2}\)=\(\sqrt{60^2-30^2}\)=\(30\sqrt{3}\)=51,96

bài 2

\(\widehat{B}=90-\widehat{C}=90-30=60\)

\(sinC=\dfrac{AB}{BC}\Rightarrow AB=sinC.BC=sin30.5=2,5\)

áp 

áp dụng pytago vào \(\Delta ABC\)

\(AC=\sqrt{BC^2-AB^2}=\sqrt{5^2-2,5^2}\)=4,33

bài 3

\(\widehat{E}=90-\widehat{F}=90-47=43\)

\(sinF=\dfrac{ED}{EF}\Rightarrow EF=\dfrac{ED}{sinF}=\dfrac{9}{sin47}=12,31\)

áp dụng pytago vào \(\Delta DEF\)

\(DF=\sqrt{EF^2-ED^2}=\sqrt{12,31^2-9^2}\)=8,4

bài 4

áp dụng pytago vào \(\Delta ABC\)

\(AB=\sqrt{BC^2-AC^2}=\sqrt{32^2-27^2}=17,18\)

\(sinB=\dfrac{AC}{BC}=\dfrac{27}{32}\Rightarrow\widehat{B}=57\)

\(\widehat{C}=90-\widehat{B}=90-57=33\)