2:

a: \(A=\dfrac{x_1+x_2}{x_1x_2}=\dfrac{-6}{3}=-2\)

b: \(B=\dfrac{\left(x_1+x_2\right)^2-3x_1x_2}{1-x_1x_2}=\dfrac{36-3\cdot3}{1-3}=\dfrac{36-9}{-2}=-\dfrac{27}{2}\)

c: \(C=\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}\)

\(=\sqrt{\left(-6\right)^2-4\cdot3}=2\sqrt{6}\)

d: \(D=\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)-3x_1x_2\)

\(=\left(-6\right)^3-3\cdot3\cdot\left(-6\right)-3\cdot3\)

=261

a) Xét ∆ADE và ∆ABC có:

AD = AB (do A là trung điểm BD)

∠DAE = ∠BAC (đối đỉnh)

AE = AC (do A là trung điểm CE)

⇒ ∆ADE = ∆ABC (c-g-c)

b) Do ∆ADE = ∆ABC (cmt)

⇒ ∠ADE = ∠ABC (hai góc tương ứng)

Mà ∠ADE và ∠ABC so le trong

⇒ DE // BC

3 bài nào vậy bạn!?

có thấy bài đâu 

V

1 an

2 some

3 some

4 some

5 some

6 a

7 some

8 some

VI

1 some

2 any

3 any

4 some

5 some

6 any

IX

1 as

2 as

3 as

4 from

5 as

1. an

2. some

3. some

4. some

5. a

6. some

7. some

VI,

1. some

2. any

3. any

4. some

5. some

6. any

VII,

1. as

2. as

3. as

4. from

5. as

V)

1. an

2. some 

3. some

4. some

5. some

6. a

7. some

8. some

VI)

1. some

2. any

3. any

4. some

5. some

6. any

IX)

1. as

2. as

3. as

4. from

5. as

8: \(x\sqrt{x}-1=\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\)

9: \(x\sqrt{x}+1=\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)\)

10: \(x\sqrt{x}-y\sqrt{y}=\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)\)

11: \(x\sqrt{x}+y\sqrt{y}=\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)\)

các bn ơi...!!

à mình nhầm!!! đó là sách lớp 6 nha mn. ko phải lớp 5 đâu ạ. mong mn giúp !!! 

mn ai biết thì giúp với ạ