Bài 2: 

a: Xét ΔABD và ΔEBD có 

BA=BE

\(\widehat{ABD}=\widehat{EBD}\)

BD chung

Do đó: ΔABD=ΔEBD

Suy ra: AD=ED

b: Ta có: ΔABD=ΔEBD

nên \(\widehat{BAD}=\widehat{BED}\)

mà \(\widehat{BAD}=90^0\)

nên \(\widehat{BED}=90^0\)

hay DE\(\perp\)BC

bn cần giúp bài nào?

a: xy//BC

=>\(\hat{A_1}=\hat{ABC}\) (hai góc so le trong)

xy//BC

=>\(\hat{A_2}=\hat{ACB}\) (hai góc so le trong)

b: \(\hat{A_1}+\hat{A_2}+\hat{BAC}=180^0\)

=>\(\hat{ABC}+\hat{ACB}+\hat{BAC}=180^0\)

1 younger

2 taller

3 the smartest

4 more exciting

5 less

6 worst

7 cleverer

8 more expensive

9 more modern

10 more skillful

11 more peaceful

12 friendly

13 more valuable

14 more suitable

15 louder

16 better than

17 more slowly

18 higher

19 farther

20 more carefully

21 more often

22 the fastest

23 worse

24 harder

Ai giúp mik vs

⇒ Đáp án: C. Nghề trồng rau

Mìnk nghĩ lè C.nghề trồng rau

a) \(x^2+2x+1=\left(x+1\right)^2\)

\(x^2-2x+1=\left(x-1\right)^2\)

\(x^2+4x+4=\left(x+2\right)^2\)

\(x^2-4x+4=\left(x-2\right)^2\)

\(x^2+6x+9=\left(x+3\right)^2\)

\(x^2-6x+9=\left(x-3\right)^2\)

\(x^2-10x+25=\left(x-5\right)^2\)

\(x^2+10x+25=\left(x+5\right)^2\)

b) \(16x^2-8x+1=\left(4x-1\right)^2\)

c) \(4x^2+12xy+9y^2=\left(2x+3y\right)^2\)

d) \(x^2+x+\dfrac{1}{4}=\left(x+\dfrac{1}{2}\right)^2\)

e) \(x^2-x+\dfrac{1}{4}=\left(x-\dfrac{1}{2}\right)^2\)

f) \(9x^2+30x+25=\left(3x+5\right)^2\)

a=\(\left(x+1\right)^2\)

=\(\left(x-1\right)^2\)

=\(\left(x+2\right)^2\)

=\(\left(x-2\right)^2\)

=\(\left(x+3\right)^2\)

=\(\left(x-3\right)^2\)

=\(\left(x-5\right)^2\)

=\(\left(x+5\right)^2\)

b, =\(\left( 4 𝑥 − 1\right)^2\)

c,=\(\left(2x-3y\right)^2\)

\(\left(x^2-x-2\right)\sqrt{x-1}=0\left(đk:x\ge1\right)\)

\(\Leftrightarrow\left(x-2\right)\left(x+1\right)\sqrt{x-1}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\end{matrix}\right.\) (do x+1>0)

Ý B.

B

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