`3x^2 -8x+4`

`=3x^2-2x-6x+4`

`=x(3x-2) - 2(3x-2)`

`=(3x-2)(x-2)`

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`4x^2-4x-3`

`=4x^2 +2x-6x-3`

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

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

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`x^2-6x+5`

`=x^2-x-5x+5`

`=x(x-1)-5(x-1)`

`=(x-1)(x-5)`

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`x^4 +2x^2-3`

`=x^4+x^3+3x^2+3x-x^3-x^2-3x-3`

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

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

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

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

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`x^4+64`

`=x^4+16x^2+64-16x^2`

`=(x^2)^2 +16x^2 +8^2 -(4x)^2`

`=(x^2 +8)^2 -(4x)^2`

`= (x^2+8-4x)(x^2 +8+4x)`

`(3/4)^5 * x=(3/4)^7`

`=> x= (3/4)^7 : (3/4)^5`

`=> x= (3/4)^(7-5)`

`=>x=(3/4)^2`

`=>x= 9/16`

\(\dfrac{3-3x}{x^2-9}\cdot\dfrac{x-3}{x-1}\\ =\dfrac{3\left(1-x\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)\left(x-1\right)}\\ =\dfrac{-3\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)\left(x-1\right)}\\ =-\dfrac{3}{x+3}\\ \dfrac{6x+4}{x^2-4}\cdot\dfrac{x^2-2x}{3x+2}\\ =\dfrac{2\left(3x+2\right)x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)\left(3x+2\right)}\\ =\dfrac{2x}{x+2}\)

tớ làm r nhe cậu^^

\(\dfrac{2x+4}{x^3-1}-\dfrac{2}{x-1}+\dfrac{x+2}{x^2+x+1}\\ =\dfrac{2x+4}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{2}{x-1}+\dfrac{x+2}{x^2+x+1}\\ =\dfrac{2x+4}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{2\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{\left(x+2\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\\ =\dfrac{2x+4-2x^2-2x-2+x^2-x+2x-2}{\left(x-1\right)\left(x^2+x+1\right)}\\ =\dfrac{-x^2+x}{\left(x-1\right)\left(x^2+x+1\right)}\\ =\dfrac{-x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)

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

Khoảng cách : `4`

Số số hạng là : \(\dfrac{201-1}{4}+1=51\)

Tổng là : \(\dfrac{\left(201+1\right)\cdot51}{2}=5151\)

`a, 2/(x+1)` hay `2/(x-1)` cậu nhỉ?

`b,`

\(\dfrac{x-1}{x^2-5x+6}-\dfrac{x-3}{x-2}+\dfrac{x-2}{x-3}\\ =\dfrac{x-1}{\left(x-2\right)\left(x-3\right)}-\dfrac{x-3}{x-2}+\dfrac{x-2}{x-3}\\ =\dfrac{x-1}{\left(x-2\right)\left(x-3\right)}-\dfrac{\left(x-3\right)^2}{\left(x-2\right)\left(x-3\right)}+\dfrac{\left(x-2\right)^2}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{x-1-\left(x^2-6x+9\right)+x^2-4x+4}{\left(x-2\right)\left(x-3\right)}\\ =\dfrac{x-1-x^2+6x-9+x^2-4x+4}{\left(x-2\right)\left(x-3\right)}\\ =\dfrac{3x-6}{\left(x-2\right)\left(x-3\right)}\)

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

\(\sqrt{x^2+x}=x\\ \Leftrightarrow\sqrt{\left(x^2+x\right)^2}=x^2\\ \Leftrightarrow x^2+x-x^2=0\\ \Leftrightarrow x=0\)

Vậy \(S=\left\{0\right\}\)

Ui, ko dám nhận cj, cj hơn tuổi e mà^^

`3^9 * 9^2`

`=3^9 * (3^2)^2`

`= 3^9 * 3^4`

`=3^(9+4)`

`= 3^(13)`