Giúp mình vs. Mình cảm ơn trc ạ
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2 tháng 9 2021
a, Vì ME là tiếp tuyến đường tròn O và M là tiếp điểm
=> \(MO\perp MF\) ( t/c tiếp tuyến ) hay ^OME = 900
Vậy tam giác EMO là tam giác vuông tại M
b, mình sửa đề là OE = 60 cm nhé
Theo định lí Pytago cho tam giác EMO vuông tại M
\(ME=\sqrt{EO^2-OM^2}=48\)cm
c, sửa ON vuông OE tại N
đến đây thì mình chả hiểu đề kiểu gì, chịu, bạn chép đề kiểu gì ấy, sai tào lao sao á, xem lại nhé
2 tháng 9 2021
a: Xét ΔMEO có \(\widehat{OME}=90^0\)
nên ΔMEO vuông tại M
TN
2
9 tháng 8 2021
1, What would he like to have for breakfast?
He would like to have a sandwich
2,Who would you like to go fishing with?
I would like to go fishing with my father
3,What would her children like to do in the summer?
They would like to go swimming
4,When would Mrs Tam like to go shopping?
She would like to go shopping at weekends
5,Where would Hung and Tung like to study
They would like to study in the library
DT
Đỗ Thanh Hải
CTVVIP
9 tháng 8 2021
1 What would he like for breakfast?
He'd like a sandwich
2 Who would you like to go fishing with?
I would like to go with my father
3 What would her children like to do in summer?
They would like to swim inpool
4 When would Mrs Tam like to go shopping?
She would like to go shopping on the weekends
5 Where would Tung and Hung like to study ?
THey would like to study in the school library
LT
3
Y
4 tháng 8 2016
1-x-2x^2
= 1-x-2x.2x
= 1 - ( x + 2x.2x)
= 1 - 5x
Để 1-x-2x^2 mang giá trị lớn nhất thì x phài là số âm.
OP
4 tháng 8 2016
\(A=1-x-2x^2\)
\(=-2\left(x^2+2\times x\times\frac{1}{4}+\left(\frac{1}{4}\right)^2-\left(\frac{1}{4}\right)^2-\frac{1}{2}\right)\)
\(=-2\left[\left(x+\frac{1}{4}\right)^2-\frac{9}{16}\right]\)
\(\left(x+\frac{1}{4}\right)^2\ge0\)
\(\left(x+\frac{1}{4}\right)^2-\frac{9}{16}\ge-\frac{9}{16}\)
\(-2\left[\left(x+\frac{1}{4}\right)^2-\frac{9}{16}\right]\le\frac{9}{8}\)
Vậy Max A = \(\frac{9}{8}\) khi x = \(-\frac{1}{4}\)
YH
2
MH
7 tháng 5 2022
\(\dfrac{x\left(x-8\right)+3\left(x+6\right)}{\left(x+6\right)\left(x-8\right)}=\dfrac{-12x+33}{\left(x+6\right)\left(x-8\right)}\left(đk:x\ne-6;8\right)\)
\(x^2-8x+3x+18=-12x+33\)
\(x^2-5x+18+12x-33=0\)
\(x^2+7x+15=0\)
\(\text{∆}=7^2-4.15=-11< 0\)
⇒ pt vô nghiệm
8 tháng 5 2022
đk : x khác -6 ; 8
\(x^2-8x+3x+18=-12x+33\Leftrightarrow x^2+7x-25=0\)
\(\Leftrightarrow x=\dfrac{-7\pm\sqrt{149}}{2}\)
giúp mình với ạ, mình đang cần gấp, cảm ơn trc ạ
DT
Đỗ Thanh Hải
CTVVIP
17 tháng 4 2021
V Transformation
1 In spite being quite poor, the villagers live a happy and healthy way
2 Despite studying very hard, he still didn't pass the exam
3 Although Sylvia had no interest in folklore, she still enjoyed the story
4 Despite having much experience in machinery, he didn't succeed in repairing this machine
5 Though it was dark, they continued to work
6 In spite of having health problems, he is always smiling
7 Despite the difficult exam, Kieu Anh got good marks
8 THe man about whom I told you works in the hospital
9 Do you know the girl to whom Tom is talking?
10 The tree which stankds near the gate of my house has lovely flowers
DT
Đỗ Thanh Hải
CTVVIP
17 tháng 4 2021
11 The book which I was reading yesterday was a lovely story
12 Do you know the new student whose name I can't remember?
13 I will never forget the day when I first met her
14 The country where I was born is beautiful
15 Because of the polluted water, it was unsafe to drink
16 Because he work hard and methodically, John succeeded in his exam
17 I suggest getting together and talking about our presentation before we do it in class
18 When did you built this stilt house?
19 If you don't hurry up, you will be late for school
20 You must turn off the TV before 11p.m
14 tháng 7 2021
c) Ta có: \(\sqrt{\sqrt{x}+3}=3\)
\(\Leftrightarrow\sqrt{x}+3=9\)
\(\Leftrightarrow\sqrt{x}=6\)
hay x=36
Ta có: \(\sqrt{x-2\sqrt{x-1}}=2\)
\(\Leftrightarrow x-2\sqrt{x-1}-4=0\)
\(\Leftrightarrow x-1-2\cdot\sqrt{x-1}\cdot1+1=4\)
\(\Leftrightarrow\left(\sqrt{x-1}-1\right)^2=4\)
\(\Leftrightarrow\sqrt{x-1}-1=2\)
\(\Leftrightarrow\sqrt{x-1}=3\)
\(\Leftrightarrow x-1=9\)
hay x=10
23 tháng 2 2022
a: Xét ΔBAD có BA=BD
nên ΔBAD cân tại B
hay \(\widehat{BAD}=\widehat{BDA}\)
b: \(\widehat{HAD}+\widehat{BDA}=90^0\)
\(\widehat{CAD}+\widehat{BAD}=90^0\)
mà \(\widehat{BAD}=\widehat{BDA}\)
nên \(\widehat{HAD}=\widehat{CAD}\)
hay AD là tia phân giác của góc HAC
c: Xét ΔADH vuông tại H và ΔADK vuông tại K có
AD chung
\(\widehat{HAD}=\widehat{KAD}\)
Do đó:ΔADH=ΔADK
Suy ra: AH=AK
a. 1.2 + 32 + 16.4 = 98đvC
b. 31.2 + 16.3 = 110đvC
c. 23 + 16 + 1 = 40đvC
d. 39 + 14 + 16.3 = 101đvC