Bài 2: 

\(A=2+2^2+2^3+...+2^{60}\)

\(A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{59}+2^{60}\right)\)

\(A=2\cdot\left(1+2\right)+2^3\cdot\left(1+2\right)+...+2^{59}\cdot\left(1+2\right)\)

\(A=2\cdot3+2^3\cdot3+...+2^{59}\cdot3\)

\(A=3\cdot\left(2+2^3+...+2^{59}\right)\) ⋮ 3

Vậy: A ⋮ 3

_____________

\(A=2+2^2+...+2^{60}\)

\(A=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{58}+2^{59}+2^{60}\right)\)

\(A=2\cdot\left(1+2+4\right)+2^4\cdot\left(1+2+4\right)+...+2^{58}\cdot\left(1+2+4\right)\)

\(A=2\cdot7+2^4\cdot7+...+2^{58}\cdot7\)

\(A=7\cdot\left(2+2^4+....+2^{58}\right)\) ⋮ 7

Vậy: A ⋮ 7

___________________

\(A=2+2^2+...+2^{60}\)

\(A=\left(2+2^3\right)+\left(2^2+2^4\right)+...+\left(2^{58}+2^{60}\right)\)

\(A=2\cdot\left(1+4\right)+2^2\cdot\left(1+4\right)+...+2^{58}\cdot\left(1+4\right)\)

\(A=2\cdot5+2^2\cdot5+...+2^{58}\cdot5\)

\(A=5\cdot\left(2+2^2+...+2^{58}\right)\) ⋮ 5

Vậy: A ⋮ 5 

mn giúp e với

\(\left(x^3-8\right):\left(x^2+2x+4\right)\\ =\left(x-2\right)\left(x^2+2x+4\right):\left(x^2+2x+4\right)\\ =x-2\)

bài 2

a)

\(2xy^2-4y\\ =2y\left(xy-2\right)\)

b)

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

c)

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

d)

\(x^2+4x+3\\ =x^2+3x+x+3\\ =x\left(x+3\right)+\left(x+3\right)\\ =\left(x+3\right)\left(x+1\right)\)

4,\(x^2+4x+3=x^2+x+3x+3=x\left(x+1\right)+3\left(x+1\right)=\left(x+3\right)\left(x+1\right)\)

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

1,=\(2y\left(xy-2\right)\)

a: \(\sqrt{5a^2}=\left|a\sqrt{5}\right|=-a\sqrt{5}\left(a< =0\right)\)

c: A=\(\sqrt{72a^2b^4}=\sqrt{36a^2b^4\cdot2}=6\sqrt{2}\cdot b^2\cdot\left|a\right|\)

mà a<0

nên \(A=-6\sqrt{2}\cdot ab^2\)

d: \(\sqrt{24a^4b^8}=\sqrt{4a^4b^8\cdot6}=2a^2b^4\cdot\sqrt{6}\)

Bài 1:

a: \(\left|x-\dfrac{1}{2}\right|+\dfrac{1}{2}=x\)

=>\(\left|x-\dfrac{1}{2}\right|=x-\dfrac{1}{2}\)

=>\(x-\dfrac{1}{2}>=0\)

=>\(x>=\dfrac{1}{2}\)

b: \(\left|1-3x\right|+1=3x\)

=>\(\left|1-3x\right|=3x-1\)

=>\(1-3x< =0\)

=>3x-1>=0

=>3x>=1

=>\(x>=\dfrac{1}{3}\)

Bài 2:

a: \(C=\left|5-x\right|+x=\left|x-5\right|+x\)

TH1: x>=5

\(C=x-5+x=2x-5\)

TH2: x<5

C=5-x+x=5

b: D=|2x-1|-x

TH1: x>=1/2

\(D=2x-1-x=x-1\)

TH2: \(x< \dfrac{1}{2}\)

D=1-2x-x=1-3x

 1 Linh said that her uncle had worked in Hai Phong for 10 years

2 She told us that she was having dinner with Rachel the following day

3 The teacher said to the children that they must wash their hands

4 Susan said that  they hadn't left for London at 5 p.m the day before

5 Tom said to me that his brother would stay there the following Friday

6 He wanted to know  when the train left

7 Jack asked the old man how long he had worked in that company

8 My parents asked me what time your friend would arrive that weekend

9 They asked Mr. Quang if he had seen a UFO the evening before

10 The boy askes->asked Anna if he might use her phone

\(\text{#♪◈₰♣}\)

cho giá trị A1=5;B1=8.Hãy chọn kết quả của công thức=A1*2+B1*3
A.24
B.18
C.13
D.34

D.34

Ta có:

∠B₂ = ∠B₁ = 70⁰ (đối đỉnh)

⇒ ∠B₂ = ∠A₁ = 70⁰

Mà ∠B₂ và ∠A₁ là hai góc đồng vị

⇒ a // b

https://edward29.github.io/surprise/