1. to, with

2. in

3. on, for

4. about

5. off

6. from

7. for

8. at

9. after

10. for

Xét ΔABC có DE//BC

nên \(\dfrac{AD}{DB}=\dfrac{AE}{EC}\)

=>\(\dfrac{14}{EC}=\dfrac{10}{20}=\dfrac{1}{2}\)

=>\(EC=14\cdot2=28\left(m\right)\)

Bn ơi cho mik hỏi chút. Vì sao DE//BC?

\(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{z}{5}=\dfrac{6x}{9}=\dfrac{6y}{8}=\dfrac{6z}{30}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\dfrac{6x}{9}=\dfrac{6y}{8}=\dfrac{6z}{30}=\dfrac{6\left(x+y+z\right)}{9+8+30}=\dfrac{6.20}{47}=\dfrac{120}{47}\\ \Rightarrow x=\dfrac{120}{47}.3:2=\dfrac{180}{47}\\ \Rightarrow y=\dfrac{120}{47}.4:3=\dfrac{160}{47}\\ \Rightarrow z=\dfrac{120}{47}.5=\dfrac{600}{47}\)

Bạn xem xem cách này có đúng không nha!

úi, mik cảm ơn bạn nha!!! =))

A = 2010 × 2011 - 1945 / 2010 × 2010 + 65

A = 2010 × 2010 + ( 2010 - 1945) / 2010 × 2010 + 65

A = 2010 × 2010 + 65 / 2010 × 2010 + 65

A = 1

B = 2001 × 2001 + 111 / 2001 × 2002 - 1890

B = 2001 × 2001 + 111 / 2001 × 2001 + ( 2001 - 1890)

B = 2001 × 2001 + 111 / 2001 × 2001 + 111

B = 1

=> A = B

b=1

a=1

a=b

:}

Câu 15: 

f(2)=2

f(-1)=-1

f(0)=-2

vt rõ đề đi

Nguyễn Huy Thắng onl nhiều ghê

\(\dfrac{2001.2002+1981+2003.21}{2002.2003-2001.2002}\)

\(=\dfrac{2001.2002+1981+\left(2002+1\right).21}{2002.\left(2003-2001\right)}\)

\(=\dfrac{2001.2002+1981+21+2002.21}{2002.2}\)

\(=\dfrac{2001.2002+2002+2002.21}{2002.2}\)

\(=\dfrac{2002\left(2001+1+21\right)}{2002.2}=\dfrac{2023}{2}\)