Đặt        \(A=1+\frac{1}{3}+\frac{1}{3}^2+\frac{1}{3}^3+....+\frac{1}{3}^{100}\)

                 \(\frac{1}{3}A=\frac{1}{3}+\frac{1}{3}^2+\frac{1}{3}^3+.....+\frac{1}{3}^{101}\)

          \(\frac{1}{3}A-A=\left(\frac{1}{3}+\frac{1}{3}^2+\frac{1}{3}^3+....+\frac{1}{3}^{101}\right)-\left(1+\frac{1}{3}+\frac{1}{3}^2+\frac{1}{3}^3+....+\frac{1}{3}^{100}\right)\)

          \(\frac{1}{3}A-A=\frac{1}{3}+\frac{1}{3}^2+\frac{1}{3}^3+....+\frac{1}{3}^{101}-1-\frac{1}{3}-\frac{1}{3}^2-\frac{1}{3}^3-....-\frac{1}{3}^{100}\)

          \(\frac{\left(-2\right)}{3}A=\frac{1}{3}-1\)

          \(\frac{\left(-2\right)}{3}A=\frac{\left(-2\right)}{3}\Rightarrow A=1\)

Vậy ......

ta có A=1+1/3+1/3^2+...+1/3^100

       A.3= 3+1+1/3+1/3^2+...+1/3^99

      A.3-A= (3+1+1/3+1/3^2+...+1/3^99)-(1+1/3+1/3^2+...+1/3^100)

      A.2=3-1/3^100

     A=(3-1/3^100):2

b) =(y^2-9)(y^2+9)-(y^4-4)

=y^4-81-y^4+4=-77

1) \(\left(x+1\right)^3-\left(x-4\right)\left(x+4\right)-x^3\)

\(=\left(x^3+3x^2+3x+1\right)-\left(x^2-16\right)-x^3\)

\(=x^3+3x^2+3x+1-x^2+16-x^3\)

\(=2x^2+3x+17\)

2) \(\left(x+2\right)^3-x\left(x+3\right)\left(x-3\right)-12x^2-8\)

\(=\left(x^3+6x^2+12x+8\right)-x\left(x^2-9\right)-12x^2-8\)

\(=x^3+6x^2+12x+8-x^3+9x-12x^2-8\)

\(=-6x^2+21x\)

`@` `\text {Ans}`

`\downarrow`

`1.`

\((x + 1) ^ 3 - (x - 4)(x + 4) - x ^ 3\)

`= x^3 + 3x^2 + 3x + 1 - [ x(x+4) - 4(x+4)] - x^3`

`= x^3 + 3x^2 + 3x + 1 - (x^2 + 4x - 4x - 16) - x^3`

`= x^3 + 3x^2 + 3x + 1 - (x^2 - 16) - x^3`

`= x^3 + 3x^2 + 3x + 1 - x^2 + 16 - x^3`

`= (x^3 - x^3) + (3x^2 - x^2) + 3x + (1+16)`

`= 2x^2 + 3x + 17`

`2.`

\((x + 2) ^ 3 - x(x + 3)(x - 3) - 12x ^ 2 - 8\)

`= x^3 + 6x^2 + 12x + 8 - [ (x^2 + 3x)(x-3)] - 12x^2 - 8`

`= x^3 + 6x^2 + 12x + 8 - (x^3 - 9x) - 12x^2 - 8`

`= x^3 + 6x^2 + 12x +8 - x^3 + 9x - 12x^2 - 8`

`= (x^3 - x^3) + (6x^2 - 12x^2) + (12x + 9x) + (8-8)`

`= -6x^2 + 21x `

(x - 3)³ - (x + 1)³ + 12x (x - 1)

= x³ - 3x² . 3 + 3x . 3² - 27 - (x³ + 3x² . 1 + 3x . 1² + 1³) + 12x . x + 12x . (-1)

= x³ - 9x² + 27x - 27 - x³ - 3x² - 3x - 1 + 12x² - 12x

= (x³ - x³) + (12x² - 9x² - 3x²) + (27x - 3x - 12x) - (27 + 1)

= 12x - 28

\(\left(x-3\right)^3-\left(x+1\right)^3+12x\left(x-1\right)\)

\(\Leftrightarrow\left(x^3-3x^23+3x3^2-3^3\right)-\left(x^3+3x^21+3x1^2+1^3\right)+12x^2-12x\)

\(\Leftrightarrow x^3-9x^2+27x-27-x^3-3x^2-3x-1+12x^2-12x\)

\(\Leftrightarrow12x-28=0\)

\(\Leftrightarrow12x=28\)

\(\Leftrightarrow x=\frac{7}{3}\)

Vậy S={\(\frac{7}{3}\)} là nghiệm pt

Nếu a+3 là dương

A=3a-3-2.(a+3)+9

A=3a-3-2a+6+9

A=a+12

Nếu a+3 là âm

A=3a-3-2.(-a-3)+9

A=3a-3-(-2).a-6+9

A=5.a+9-6-3

A=5.a

T..i..c..k nha

-2*trị tuyệt đối(a+3)+3*a+6

nếu Ia+3I là số dương thì Ia+3I=a+3 

=> A= 3a-3 - 2a+6 +9 

 A= 3a+2a -(-3+6+9)

= 5A - 12 

nếu Ia+3I là âm thì I a+3I = -a+3

=> A= 3a- 3 - -2a +6 +9

=> A= 3a-2a -( -3 +6+9)

 => A= a - 12 

a/ \(E=a^6+a^4+a^2b^2+b^4-b^6\)

\(E=\left[\left(a^2\right)^2+2a^2b^2+\left(b^2\right)^2\right]+\left(a^6-b^6\right)-a^2b^2\)

\(E=\left[\left(a^2+b^2\right)^2-\left(ab\right)^2\right]+\left(a^3-b^3\right)\left(a^3+b^3\right)\)

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

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

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

\(a^6+a^4+a^2b^2+b^4-b^6\)

\(a^2\left(a^4+a^2b^2+b^4\right)-b^2\left(a^4+a^2b^2+b^4\right)+\left(a^4+a^2b^2+b^4\right)\)

\(=\left(a^4+a^2b^2+b^4\right)\left(a^2-b^2+1\right)\)

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