a: Xét ΔAHK vuông tại H và ΔDHB vuông tại H có

HA=HD

HK=HB

Do đó: ΔAHK=ΔDHB

b: ΔAHK=ΔDHB

=>\(\widehat{HAK}=\widehat{HDB}\)

=>AK//DB

c: Xét ΔBAD có

BH là đường cao

BH là đường trung tuyến

Do đó: ΔBAD cân tại B

=>BA=BD

d: Xét ΔHAB vuông tại H và ΔHDK vuông tại H có

HA=HD

HB=HK

Do đó: ΔHAB=ΔHDK

=>\(\widehat{HAB}=\widehat{HDK}\)

=>AB//DK

ta có: IK\(\perp\)AC

AB\(\perp\)AC

Do đó: IK//AB

mà DK//AB

và IK,DK có điểm chung là K

nên I,K,D thẳng hàng

Đặt: \(\dfrac{x}{3}=\dfrac{y}{5}=k=>\left\{{}\begin{matrix}x=3k\\y=5k\end{matrix}\right.\)

Mà:

\(2x^2+y^2=43\\ =>2\cdot\left(3k\right)^2+\left(5k\right)^2=43\\ =>18k^2+25k^2=43\\ =>43k^2=43\\ =>k^2=1\\ =>k=\pm1\\ TH1:k=1=>\left\{{}\begin{matrix}x=3\cdot1=3\\y=5\cdot1=5\end{matrix}\right.\\ TH2:k=-1=>\left\{{}\begin{matrix}x=3\cdot\left(-1\right)=-3\\y=5\cdot\left(-1\right)=-5\end{matrix}\right.\)

Ex1

2 Have you heard the news yet? - Yes, I have already heard it

3 Have you eaten this fruit yet? - Yes, I've already eaten it

4 Has Jill written the report yet? - Yes, she has already written it

5 Has Ted repair his bike yet? - Yes, he has already repaired it

6 Has Messi signed the contract yet? - Yes, he has already signed it

7 Have Vera and Tom got engaged yet? - Yes, they have already done it

8 Has Paul bought a new car yet? - Yes, he has already bought it

9 Has Yen passed her driving test yet? - Yes, she has already passed it

10 Have you sent the invitations yet? - Yes, I have already sent them

Ex1

2 have met

3 has knwon

4 have read

5 haven't seen

6 has never eaten

7 Have you told

8 hasn't rained

9 have had

10 have just sold

\(\left|2\dfrac{1}{5}-x\right|+\left|x-\dfrac{1}{5}\right|+8\dfrac{1}{5}=1,2\\\Rightarrow\left|2\dfrac{1}{5}-x\right|+\left|x-\dfrac{1}{5}\right|=1,2-8\dfrac{1}{5}\\ \Rightarrow \left|2\dfrac{1}{5}-x\right|+\left|x-\dfrac{1}{5}\right|=-7\)

Nhận xét:

\(\left\{{}\begin{matrix}\left|2\dfrac{1}{5}-x\right|\ge0,\forall x\\\left|x-\dfrac{1}{5}\right|\ge0,\forall x\end{matrix}\right.\\ \Rightarrow\left|2\dfrac{1}{5}-x\right|+\left|x-\dfrac{1}{5}\right|\ge0,\forall x\)
Mà \(-7< 0\) nên:
Không tìm được giá trị \(x\) thỏa mãn đề bài
Vậy...

1 Mrs. Lien suggested organizing an English competition for their students
2 Bob denied taking Susie's calculator
3 David accused Henry of taking his pencil

E xem lại câu 2 với 3 bị lẫn lộn r

TH1: x<1

Phương trình sẽ trở thành:

\(1-x+2-x+3-x=-4x\)

=>-4x=-3x+6

=>-x=6

=>x=-6(nhận)

TH2: 1<=x<2

Phương trình sẽ trở thành:

\(x-1+2-x+3-x=-4x\)

=>-4x=-x+4

=>-3x=4

=>\(x=-\dfrac{4}{3}\left(loại\right)\)

TH3: 2<=x<3

Phương trình sẽ trở thành:

\(x-1+x-2+3-x=-4x\)

=>x=-4x

=>x=0(loại)

TH4: x>=3

Phương trình sẽ trở thành:

x-1+x-2+x-3=-4x

=>-4x=3x-6

=>-7x=-6

=>\(x=\dfrac{6}{7}\left(loại\right)\)

a: Oy nằm giữa Ox và Ot

=>\(\widehat{xOy}+\widehat{yOt}=\widehat{xOt}\)

=>\(\widehat{yOt}+30^0=50^0\)

=>\(\widehat{yOt}=20^0\)

b: Vì Ox là tia đối của tia Oz nên \(\widehat{xOz}=180^0\)

\(\left(2,5x-\dfrac{4}{7}\right):\dfrac{8}{21}=-1,5\)

=>\(\left(\dfrac{5}{2}x-\dfrac{4}{7}\right):\dfrac{8}{21}=-\dfrac{3}{2}\)

=>\(\dfrac{5}{2}x-\dfrac{4}{7}=\dfrac{-3}{2}\cdot\dfrac{8}{21}=\dfrac{-24}{42}=\dfrac{-4}{7}\)

=>\(\dfrac{5}{2}x=-\dfrac{4}{7}+\dfrac{4}{7}=0\)

=>x=0

\(-\dfrac{13}{38}=\dfrac{-13\cdot88}{38\cdot88}=\dfrac{-1144}{3344};\dfrac{29}{-88}=\dfrac{29\cdot\left(-38\right)}{\left(-88\right)\left(-38\right)}=\dfrac{-1102}{3344}\)

mà -1144<-1102

nên \(-\dfrac{13}{38}< \dfrac{29}{-88}\)

Ta có:

\(\dfrac{13}{38}>\dfrac{13}{39}=>\dfrac{-13}{38}< \dfrac{-13}{39}=-\dfrac{1}{3}\\\dfrac{29}{88} < \dfrac{29}{87}=>\dfrac{-29}{88}>-\dfrac{28}{87}=-\dfrac{1}{3}\)

\(=>\dfrac{-13}{38}< -\dfrac{1}{3}< \dfrac{-29}{88}\)