Làm trc cho 2 câu cuối

c) \(a^2-b^2-4a+4b\)

\(=\left(a+b\right)\left(a-b\right)-4\left(a-b\right)\)

\(=\left(a-b\right)\left[\left(a+b\right)-4\right]\)

d) \(a^2+2ab+b^2-2a-2b+1\)

\(=\left(a+b\right)^2-2\left(a+b\right)+1\)

\(=\left(a+b\right)\left[\left(a+b\right)-2\right]+1\)

$Q=1+2+2^2+2^3+\cdots+2^{50}$

$Q=(1+2+2^2+\cdots+2^{50})$

$Q=2^{51}-1$

Do đó: $Q+1=2^{51}$

Mà: $Q+1=2^n$

Suy ra: $n=51$

$1+4+4^2+4^3+\cdots+4^n$

Đây là tổng cấp số nhân:

$1+4+4^2+\cdots+4^n=\dfrac{4^{n+1}-1}{4-1}$

$=\dfrac{4^{n+1}-1}{3}$

Vậy $1+4+4^2+\cdots+4^n=\dfrac{4^{n+1}-1}{3}$

$A=1+(-2)+3+(-4)+\cdots+19+(-20)$

$=(1-2)+(3-4)+\cdots+(19-20)$

$=(-1)+(-1)+\cdots+(-1)$

$=-10$

$B=1-2+3-4+\cdots+99-100$

$=(1-2)+(3-4)+\cdots+(99-100)$

$=-1\times50$

$=-50$

Ta có:

$A=1+2+2^2+2^3+\cdots+2^{100}$

$A=2^0+2^1+2^2+\cdots+2^{100}$

Nhân $A$ với $2$:

$2A=2^1+2^2+2^3+\cdots+2^{100}+2^{101}$

Lấy $2A-A$:

$2A-A=2^{101}-1$

$A=2^{101}-1$

Vì:

$2^{101}-1<2^{101}$

Vậy $A<2^{101}$.

1. \(450\%+3=\dfrac{9}{2}+3=\dfrac{15}{2}\)

2. \(\dfrac{1}{4}-2,75+2=0,25-2,75+2=-0,5\)

3. \(\dfrac{3}{10}+2,7=0,3+2,7=3\)

A) |x| = |-7|

|x| = 7

=>x=7  hoặc x=(-7)

Vậy x thuộc {7;-7}

B) |x+1|=2

=>x+1=2    hoặc x+1=(-2)

  x=2-1                x=(-2)-1

 x=1                    x=(-3)

Vậy x thuộc {1;-3}

C) |x+1|=3

=>x+1=3 hoặc x+1=(-3)

Vì x+1<0

nên x+1=(-3)

x=(-3)-1

x=(-4)

D) x +|-2| = 0

x+2=0

x=0-2

x=(-2)

E) 4.(3x – 4) – 2 = 18

4.(3x – 4) =18+2

4.(3x – 4) =20

3x-4=20 : 4

3x-4=5

3x=5+4

3x=9

x=9 : 3

x=3

a) \(\left|x\right|=\left|-7\right|\)

\(\Rightarrow\left|x\right|=7\)

\(\Rightarrow\orbr{\begin{cases}x=7\\x=-7\end{cases}}\)

 Vậy ...

b) \(\left|x+1\right|=2\)

\(\Rightarrow\orbr{\begin{cases}x+1=2\\x+1=-2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-3\end{cases}}}\)

Vậy ...

d) \(x+\left|-2\right|=0\)

\(\Rightarrow x+2=0\)

\(\Rightarrow x=-2\)

Vậy ...

e) \(4\left(3x-4\right)-2=18\)

\(\Rightarrow4\left(3x-4\right)=20\)

\(\Rightarrow3x-4=5\)

\(\Rightarrow3x=9\Leftrightarrow x=3\)

Vậy ...

bằng 1

Ta có:

$A=\dfrac{1+(1+2)+(1+2+3)+\cdots+(1+2+\cdots+98)}{1\times98+2\times97+3\times96+\cdots+98\times1}$

Tử số:

$1+(1+2)+(1+2+3)+\cdots+(1+2+\cdots+98)$

$=1+2+2^2+?$

Cách dễ nhất là viết:

$1+2+\cdots+k=\dfrac{k(k+1)}2$

Nên tử số:

$=\dfrac{1\cdot2}{2}+\dfrac{2\cdot3}{2}+\cdots+\dfrac{98\cdot99}{2}$

$=\dfrac{1}{2}(1\cdot2+2\cdot3+\cdots+98\cdot99)$

$=\dfrac{1}{2}\cdot\dfrac{98\cdot99\cdot100}{3}$

$=161700$

Mẫu số:

$1\cdot98+2\cdot97+3\cdot96+\cdots+98\cdot1$

$=1(99-1)+2(99-2)+\cdots+98(99-98)$

$=99(1+2+\cdots+98)-(1^2+2^2+\cdots+98^2)$

$=99\cdot\dfrac{98\cdot99}{2}-\dfrac{98\cdot99\cdot197}{6}$

$=161700$

Do đó:

$A=\dfrac{161700}{161700}=1$

Vậy $A=1$.

B=\(1+3^2+3^4+...+3^{100}\)

9B=\(3^2+3^4+...+3^{100}\)

9B-B=\(\left(3^2+3^4+...+3^{102}\right)-\left(1+3^2+3^4+...+3^{100}\right)\)

8B=\(3^{102}-1\)

B=\(\left(3^{102}-1\right):8\)

C=\(1+5^3+5^6+...+5^{99}\)

125C=\(5^3+5^6+5^9+...+5^{102}\)

125C-C=\(\left(5^3+5^6+5^9+...+5^{102}\right)-\left(1+5^3+5^6+...+5^{99}\right)\)

124C=\(5^{102}-1\)

C=\(\left(5^{102}-1\right):124\)

2A=1-1/2+1/2^2-...+1/2^98-1/2^99

=>3A=1-1/2^100

=>\(A=\dfrac{2^{100}-1}{3\cdot2^{100}}\)