Ơ đây là bài lớp mấy z

1.b

2d

3.a

4.c

5.c

6. b

7.a

8.d

9.d

10.d

cong ở 4 chỗ là sao? Mik khum hỉu cho lắm

\(\dfrac{3\sqrt{x}-9}{\sqrt{x}-5}-2\)

\(=\dfrac{3\sqrt{x}-9-2\left(\sqrt{x}-5\right)}{\sqrt{x}-5}\)

\(=\dfrac{3\sqrt{x}-9-2\sqrt{x}+10}{\sqrt{x}-5}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-5}\)

a, đk x >= 0 ; x khác 9 

\(P=\left(\dfrac{\sqrt{x}+1-\sqrt{x}+3}{x-9}\right)\left(\sqrt{x}-3\right)=\dfrac{4}{\sqrt{x}+3}\)

b, Ta có \(\dfrac{4-\sqrt{x}-3}{\sqrt{x}+3}=\dfrac{\sqrt{x}-1}{\sqrt{x}+3}\ge0\)

TH1 : \(\left\{{}\begin{matrix}\sqrt{x}-1>0\\\sqrt{x}+3>0\left(luondung\right)\end{matrix}\right.\Leftrightarrow x>1\)

Kết hợp đk vậy x > 1 ; x khác 9 

TH2 : \(\left\{{}\begin{matrix}\sqrt{x}-1< 0\\\sqrt{x}+3< 0\end{matrix}\right.\)( loại) 

mờ quá bạn ơi

1, How often do you have E?

2, When is Tien La Festival usually held every year in DHC?

MN là đường trung bình tam giác ABC \(\Rightarrow\overrightarrow{MN}=\dfrac{1}{2}\overrightarrow{BC}\)

NP là đường trung bình tam giác ABC \(\Rightarrow\overrightarrow{NP}=\dfrac{1}{2}\overrightarrow{AB}\Leftrightarrow\overrightarrow{AB}=2\overrightarrow{NP}\Leftrightarrow\overrightarrow{BA}=-2\overrightarrow{NP}\)

MP là đường trung bình \(\Rightarrow\overrightarrow{MP}=\dfrac{1}{2}\overrightarrow{AC}\)

Mà N là trung điểm AC \(\Rightarrow\overrightarrow{NA}=\dfrac{1}{2}\overrightarrow{CA}=-\dfrac{1}{2}\overrightarrow{AC}\)

\(\Rightarrow\overrightarrow{MP}=-\overrightarrow{NA}\)

3.

\(\left|\overrightarrow{AB}+\overrightarrow{BC}\right|=\left|\overrightarrow{BD}\right|=BD=\sqrt{AB^2+BC^2}=a\sqrt{2}\)

\(\left|\overrightarrow{u}\right|=\left|2\overrightarrow{AD}-\overrightarrow{BA}\right|\Rightarrow\left|\overrightarrow{u}\right|^2=\left(2\overrightarrow{AD}-\overrightarrow{BA}\right)^2=4AD^2+BA^2-4\overrightarrow{AD}.\overrightarrow{BA}\)

\(=4AD^2+BA^2=5a^2\Rightarrow\left|\overrightarrow{u}\right|=a\sqrt{5}\)

\(\left|\overrightarrow{CD}-2\overrightarrow{BO}\right|=\left|\overrightarrow{CD}-\overrightarrow{BD}\right|=\left|\overrightarrow{CD}+\overrightarrow{DB}\right|=\left|\overrightarrow{CB}\right|=a\)

Xet ΔOAB và ΔOCD có

góc OAB=góc OCD

góc AOB=góc COD

=>ΔOAB đồng dạng với ΔOCD

=>OA/OC=OB/OD

=>CE/DE=DF/CF

=>\(\dfrac{CF+FE}{DE}=\dfrac{DE+EF}{CF}\)

=>\(CF^2+FE\cdot CF=DE^2+DE\cdot FE\)

=>(CF-DE)(CF+DE)-FE(CF-DE)=0

=>(CF-DE)(CF+DE-FE)=0

=>CF=DE