Biết: \(\frac{x+4}{45^2-5^2}+\frac{x-46}{45^2+5^2}-\frac{x-96}{50^2-20^2}=\frac{3x+6012}{50^2+20^2}+\frac{x}{2004}\)
Tính: \(S=\frac{|x|^2-1}{x+1}\)
K
Khách
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HN
1
25 tháng 3 2019
Bài 6: Make sentences with That-clauseusing the words in the box
1 I thought we would get in to trouble but now. I am relieved things going well
2 My parents are pleased to do fine at school
3 Lan is an excellent student.She is sure to win contest
4 They have planted to go to the beach but they were sad it was raining hard
5 She had an interview Y.T . She is now hopeful to get her job/
6 We had an accdient but we/ were elived that no one hurted/
7 Well done!I am delighted with you about the first prize
8 Will you join us?- I am afraid i can't DO lots homework
9 We are all happy in the summer vacation drawing near
10 Her parents are disappoited that she does not pass the entrane examination
25 tháng 3 2019
Bài 6: Make sentences with That-clauseusing the words in the box
1 I thought we would get in to trouble but now. I am relieved things going well
2 My parents are pleased to do fine at school
3 Lan is an excellent student.She is sure to win contest
4 They have planted to go to the beach but they were sad it was raining hard
5 She had an interview Y.T . She is now hopeful to get her job/
6 We had an accdient but we/ were elived that no one hurted/
7 Well done!I am delighted with you about the first prize
8 Will you join us?- I am afraid i can't DO lots homework
9 We are all happy in the summer vacation drawing near
10 Her parents are disappoited that she does not pass the entrane examination
LH
0
11 tháng 2 2019
A B C M P
a) Diện tích của tam giác ABC là:
\(S_{\Delta ABC}=\frac{1}{2}AB.AC=\frac{1}{2}.8.6=24\) (cm2)
b) Ta có: N là trung điểm của AB
M là trung điểm của BC
=> MN là đường trung bình của tam giác ABC
\(\Rightarrow MN//AC\)
Mà \(AB\perp AC\) (vì tam giác ABC vuông tại A)
Suy ra: \(MN\perp AB\)
c) Trong tứ giác AMBP:
Hai đường chéo PM và AB cắt nhau tại trung điểm mỗi đường (NP = NM ; NB = NA)
=> Tứ giác AMBP là hình bình hành
Mà \(MN\perp AB\) (cmt) cũng đồng nghĩa với \(MN\perp PM\) (vì P là điểm đối xứng với M qua AB)
=> AMBP là hình thoi (vì hình bình hành có hai đường chéo vuông góc là hình thoi)