a) 3A=1.2.3 + 2.3.3 + 3.4.3 +... + n.(n+1).3

=1.2.(3-0) + 2.3.(4-1) + ... + n.(n+1).[(n+2)-(n-1)]

=[1.2.3+ 2.3.4 + ...+ (n-1).n.(n+1)+ n.(n+1)(n+2)] - [0.1.2+ 1.2.3 +...+(n-1).n.(n+1)] 

=n.(n+1).(n+2) 

=>S=[n.(n+1).(n+2)] /3

b)

Nhân 4 vào hai vế ta được:

4A = 4.[1.2.3 + 2.3.4 + 3.4.5 + … + (n – 1).n.(n + 1)]

4A = 1.2.3.4 + 2.3.4.4 + 3.4.5.4 + … + (n – 1).n.(n + 1).4

4A = 1.2.3.4 + 2.3.4.(5 – 1) + 3.4.5.(6 – 2) + … + (n – 1).n.(n + 1).[(n + 2) – (n – 2)]

4A = 1.2.3.4 + 2.3.4.5 – 1.2.3.4 + 3.4.5.6 – 2.3.4.5 + … + (n – 1).n(n + 1).(n + 2) – (n – 2).(n – 1).n.(n + 1)

4A = (n – 1).n(n + 1).(n + 2)

A = (n – 1).n(n + 1).(n + 2) : 4.

3A=1.2.3 + 2.3.3 + 3.4.3 +... + n.(n+1).3

=1.2.(3-0) + 2.3.(4-1) + ... + n.(n+1).[(n+2)-(n-1)]

=[1.2.3+ 2.3.4 + ...+ (n-1).n.(n+1)+ n.(n+1)(n+2)] - [0.1.2+ 1.2.3 +...+(n-1).n.(n+1)] 

=n.(n+1).(n+2) 

=>S=[n.(n+1).(n+2)] /3

\(\left(4x-5\right)\left(2x+30\right)-4\left(x+2\right)\left(2x-1\right)+\left(10x+7\right)\)

\(=8x^2+110x-150-8x^2-12x+8+10x+7\)

\(=108x-135\)

$(4x-5)(2x+30)-4(x+2)(2x-1)+(10x+7)\\=4x(2x+30)-5(2x+30)-4[x(2x-1)+2(2x-1)]+10x+7\\=8x^2+120x-10x-150-4[2x^2-x+4x-2]+10x+7\\=8x^2+120x-143-4[2x^2+3x-2]\\=8x^2+120x-143-8x^2-12x+8\\=108x-135$

`@` `\text {Ans}`

`\downarrow`

`1)`

\(2x+\dfrac{1}{2}=\dfrac{5}{3}\)

`\Rightarrow`\(2x=\dfrac{5}{3}-\dfrac{1}{2}\)

`\Rightarrow`\(2x=\dfrac{7}{6}\)

`\Rightarrow`\(x=\dfrac{7}{6}\div2\)

`\Rightarrow`\(x=\dfrac{7}{12}\)

Vậy, `x = 7/12`

`2)`

\(\dfrac{1}{7}+\dfrac{4}{5}x=\dfrac{5}{3}\)

`\Rightarrow`\(\dfrac{4}{5}x=\dfrac{5}{3}-\dfrac{1}{7}\)

`\Rightarrow`\(\dfrac{4}{5}x=\dfrac{32}{21}\)

`\Rightarrow`\(x=\dfrac{32}{21}\div\dfrac{4}{5}\)

`\Rightarrow`\(x=\dfrac{40}{21}\)

Vậy, `x = 40/21`

`3)`

\(\dfrac{3}{5}-\dfrac{3}{5}x=\dfrac{1}{7}\)

`\Rightarrow`\(\dfrac{3}{5}x=\dfrac{3}{5}-\dfrac{1}{7}\)

`\Rightarrow`\(\dfrac{3}{5}x=\dfrac{16}{35}\)

`\Rightarrow`\(x=\dfrac{16}{35}\div\dfrac{3}{5}\)

`\Rightarrow`\(x=\dfrac{16}{21}\)

Vậy, `x = 16/21`

`4)`

\(\dfrac{5}{6}-3x=\dfrac{3}{4}\)

`\Rightarrow`\(3x=\dfrac{5}{6}-\dfrac{3}{4}\)

`\Rightarrow`\(3x=\dfrac{1}{12}\)

`\Rightarrow`\(x=\dfrac{1}{12}\div3\)

`\Rightarrow`\(x=\dfrac{1}{36}\)

Vậy, `x  = 1/36`

`5)`

\(\dfrac{5}{3}-\dfrac{1}{2}x=\dfrac{3}{7}\)

`\Rightarrow`\(\dfrac{1}{2}x=\dfrac{5}{3}-\dfrac{3}{7}\)

`\Rightarrow`\(\dfrac{1}{2}x=\dfrac{26}{21}\)

`\Rightarrow`\(x=\dfrac{26}{21}\div\dfrac{1}{2}\)

`\Rightarrow`\(x=\dfrac{52}{21}\)

Vậy, `x = 52/21`

`6)`

\(5x+\dfrac{1}{2}=\dfrac{2}{3}\)

`\Rightarrow`\(5x=\dfrac{2}{3}-\dfrac{1}{2}\)

`\Rightarrow`\(5x=\dfrac{1}{6}\)

`\Rightarrow`\(x=\dfrac{1}{6}\div5\)

`\Rightarrow`\(x=\dfrac{1}{30}\)

Vậy, `x = 1/30.`

c: \(=\dfrac{x^3+2x+2x^2+2x+x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(=\dfrac{x^3+3x^2+3x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{x^2+2x+1}{x^2-x+1}\)