Vì \(\Delta AMB=\Delta AMC\) (cmt)

\(\Rightarrow\) góc BAM = góc CAM

Xét \(\Delta AHM\) và \(\Delta AKM\) có:

góc AHM = góc AKM = \(90^0\)(vì \(MH\perp AB;MK\perp AC\)) (gt)

AM chung

góc HAM = góc KAM (vì góc BAM = góc CAM)

\(\Rightarrow\Delta HAM=\Delta KAM\) (cạnh huyền - góc nhọn)

\(\Rightarrow MH=MK\) (2 cạnh tương ứng)

Lưu ý: từ "góc" trong bài là phải kí hiệu lên nhé; ở trong này mik ko biết kí hiệu mũ nên mik viết từ "góc" rồi ak. Mong bạn chú ý ak chứ viết vào vở có chữ "góc" là ko có điểm đâu nhé!

1. I enjoy swimming in a deep pool

2. I miss taking the last bus of the day

3. I don't feel like talking to him after what he did to me

4. It's no use taking that solution.

5. They spend time playing video games every day

Câu 1: C

Câu 2: A

Câu 3: A

Câu 4: C

Câu 5: C

\(x^2-y^2-2x+2y\)

\(\Leftrightarrow\left(x^2-y^2\right)-2.\left(x-y\right)\)

\(\Leftrightarrow\left(x-y\right).\left(x+y\right)-2.\left(x-y\right)\)

\(\Leftrightarrow\left(x-y\right).\left(x+y-2\right)\)

1. She did not cheat on the test, for it was the wrong thing to do.

2. I really need to go to work, but I am too sick to drive.

3. He ran out of money, so he had to stop playing poker.

4. They got there early, and the got really good seats.

5. Everyone was busy, so I went to the movie alone

\(\left|2y-5\right|=0\)

\(2y-5=0\)

\(2y=5\)

\(y=\dfrac{5}{2}\)

Xét \(\Delta ABD\) vuông tại D (vì \(BD\perp AC\)) có:

\(BD^2+AD^2=AB^2\) (định lí Pytago)

\(15^2+AD^2=17^2\)

\(225+AD^2=289\)

\(AD^2=289-225\)

\(AD^2=64\)

\(AD=8cm\)

Ta có AD+DC=AC

\(\Rightarrow8+DC=17\)

\(\Rightarrow DC=9\)cm

Xét \(\Delta BCD\) vuông tại D có

\(BD^2+DC^2=BC^2\)(định lí Pytago)

\(15^2+9^2=BC^2\)

\(BC^2=225+81\)

\(BC^2=306\)

\(BC=\sqrt{306}cm\)

nhầm  môn rồi ak

nhầm môn rồi bn