A(x) ở đâu

 tìm A(x) biết A(x)=M(x)-N(x) ko thấy à 

=> \(\left(a-2\right)\left(a+1\right)\left(a+3\right)=0\)

=> TH1 : a-2=0 => a=2

TH2: a+1=0 => a=-1

TH3: a+3=0 => a=-3 

Vậy a={-3;-1;2}

Còn cách phân tích là : 

=> \(a^3+a^2+a^2+a-6a-6=0\)

=> \(a^2\left(a+1\right)+a\left(a+1\right)-6\left(a+1\right)=0\)

=> \(\left(a^2+a-6\right)\left(a+1\right)=0\)

=> \(\left(a^2-2a+3a-6\right)\left(a+1\right)=0\)

=> \(\left(a+1\right)\left(a-2\right)\left(a+3\right)=0\)

1/4 ???

a = 48/21

Ta có: \(\frac{1}{1+2\times a}=\frac{3}{21}\)

=>\(\frac{1}{2a+1}=\frac17\)

=>2a+1=7

=>2a=6

=>a=3

A=3+3^2+3^3+..........+3^99+3^100

3A=3^2+3^3+...............+3^100+3^101

=> 3A-A= (3^2+3^3+......+3^100+3^101) - (3+3^2+3^3+........+3^99+3^100)

=> 2A= 3^101 - 3

=>2A+3=3^101

=>3^n=3^101

=> n=101

chán không muốn dùng x² nữa ^.^ !

Câu a:

A = \(\overline{ab}\) + \(\overline{ba}\)

A = 10a + b + 10b + a

A = (10a + a) + (b + 10b)

A = 11a + 11b

A = 11(a + b) ⋮ 11 (đpcm)

Câu c:

A = 1 + 3 + 3^2 + 3^3 + ... + 3^2009 + 3^2010

3A = 3 + 3^2 + 3^3 + ... + 3^2010 + 3^2011

3A - A = (3 + 3^2 + 3^3 + ... + 3^2010 + 3^2011) - (1 + 3 + 3^2 + 3^3 + ... + 3^2009 + 3^2010)

2A = 3 + 3^2 + 3^3 + ... + 3^2010 + 3^2011 - 1 - 3 - 3^2-..-3^2010

2A = (3 - 3) + (3^2-3^2) + ...+(3^2010-3^2010) + (3^2011 - 1)

2A = 0 + 0 + ..+ 0 + 3^2011 - 1

2A = 3^2011 - 1

2A + 1 = 3^2011 - 1 + 1

2A + 1 = 3^2011 - (1 - 1)

2A + 1 = 3^2011 - 0

2A + 1 = 3^2011

\(a,f\left(\dfrac{3}{2}\right)=2\cdot\dfrac{3}{2}=3\\ b,f\left(a\right)+f\left(-a\right)=2a-2a=0\)