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2:
a: BC=căn 15^2+20^2=25cm
AH=15*20/25=12cm
góc ADH=góc AEH=góc DAE=90 độ
=>ADHE là hình chữ nhật
=>DE=AH=12cm
b: ΔAHB vuông tại H có HD vuông góc AB
nên AD*AB=AH^2
ΔAHC vuông tại H có HE vuông góc AC
nên AE*AC=AH^2
=>AD*AB=AE*AC
c: góc IAC+góc AED
=góc ICA+góc AHD
=góc ACB+góc ABC=90 độ
=>AI vuông góc ED
4:
a: góc BDH=góc BEH=góc DBE=90 độ
=>BDHE là hình chữ nhật
b: BDHE là hình chữ nhật
=>góc BED=góc BHD=góc A
Xét ΔBED và ΔBAC có
góc BED=góc A
góc EBD chung
=>ΔBED đồng dạng với ΔBAC
=>BE/BA=BD/BC
=>BE*BC=BA*BD
c: góc MBC+góc BED
=góc C+góc BHD
=góc C+góc A=90 độ
=>BM vuông góc ED
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a: \(\sqrt{0.49a^2}=-0.7a\)
b: \(\sqrt{25\left(a-7\right)^2}=5a-35\)
c: \(\sqrt{a^4\left(a-2\right)^2}=a^2\cdot\left(a-2\right)\)
d: \(\dfrac{1}{a-3b}\cdot\sqrt{a^6\left(a-3b\right)^2}\)
\(=\dfrac{1}{a-3b}\cdot a^3\cdot\left(a-3b\right)=a^3\)
Bài 2:
a: \(2\left(x+y\right)\cdot\sqrt{\dfrac{1}{x^2+2xy+y^2}}\)
\(=2\left(x+y\right)\cdot\dfrac{1}{x+y}\)
=2
b: \(\dfrac{3x}{7y}\cdot\sqrt{\dfrac{49y^2}{9x^2}}\)
\(=\dfrac{3x}{7y}\cdot\dfrac{-7y}{3x}\)
=-1
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,\\ a,=\dfrac{\left(3+2\sqrt{3}\right)\sqrt{3}}{3}+\dfrac{\left(2+\sqrt{2}\right)\left(\sqrt{2}-1\right)}{1}\\ =\dfrac{3\sqrt{3}+6}{3}+\sqrt{2}=\sqrt{3}+1+\sqrt{2}\\ b,=\left(\dfrac{\sqrt{5}+\sqrt{2}}{3}-\dfrac{\sqrt{5}-\sqrt{2}}{3}+1\right)\cdot\dfrac{1}{\left(\sqrt{2}+1\right)^2}\\ =\dfrac{\sqrt{5}+\sqrt{2}-\sqrt{5}+\sqrt{2}+3}{3}\cdot\dfrac{1}{\left(\sqrt{2}+1\right)^2}\\ =\dfrac{2\sqrt{2}+3}{3\left(3+2\sqrt{2}\right)}=\dfrac{1}{3}\)
\(2,\\ A=2x+\sqrt{\left(x-3\right)^2}=2x+\left|x-3\right|\\ =2\left(-5\right)+\left|-5-3\right|=-10+8=-2\\ B=\dfrac{\sqrt{\left(2x+1\right)^2}}{\left(x-4\right)\left(x+4\right)}\left(x-4\right)^2=\dfrac{\left|2x+1\right|\left(x-4\right)}{x+4}\\ B=\dfrac{17\cdot4}{12}=\dfrac{17}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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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
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(\Delta ABC\)
\(AB=\sqrt{BC^2-AC^2}=\sqrt{32^2-27^2}=17,18\)
\(sinB=\dfrac{AC}{BC}=\dfrac{27}{32}\Rightarrow\widehat{B}=57\)
\(\widehat{C}=90-\widehat{B}=90-57=33\)
![](https://rs.olm.vn/images/avt/0.png?1311)
P5:
28. A
29. B
30. A
31. C
32. B
33. B
34. A
35. C
P8:
51. 23 Mount Road
52. Japanese
53. engineer
54. Keiko
55. Monday 15 April
phần 5
CÂU 28-35
đọc bài báo về đà điểu
Chọn từ tốt nhất (A, B hoặc C) cho mỗi khoảng trống (28-35).
Đối với các câu hỏi 28-35, đánh dấu A, B hoặc C trên phiếu trả lời