Câu 1:

`a)`

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

`= 2x.x^2 + 2x . ( - 3x) + 2x.4`

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

`b)`

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

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

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

`= 2x^2 + x - 6`

`c)`

`(3x^3 + 9x^2 - 2x) : 3x`

`= (3x^3 : 3x) + (9x^2 : 3x) - ( 2x : 3x)`

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

Câu 2:

`a)`

`{3x + 4}/{3xy} + {6x - 4}/{3xy}`

`= {3x + 4 + 6x - 4}/{3xy}`

`= {9x}/{3xy}`

`= 3/y`

`b)`

`x/{x - 1} + 3/{x + 1} - {6x - 4}/{x^2 - 1}`

`= x/{x - 1} + 3/{x + 1} - {6x - 4}/{(x - 1)(x + 1)}`

`= {x(x + 1)}/{(x - 1)(x + 1)} + {3(x - 1)}/{(x - 1)(x + 1)} - {6x - 4}/{(x - 1)(x + 1)}`

`= {x^2 + x}/{(x - 1)(x + 1)} + {3x - 3}/{(x - 1)(x + 1)} - {6x - 4}/{(x - 1)(x + 1)}`

`= {x^2 + x + 3x - 3 - (6x - 4)}/{(x - 1)(x + 1)}`

`= {x^2 + 4x - 3 - 6x + 4}/{(x - 1)(x + 1)}`

`= {x^2 - 2x + 1}/{(x - 1)(x + 1)}`

`= {(x - 1)^2}/{(x - 1)(x + 1)}`

`= {x - 1}/{x + 1}`

`#SunHee`

Đáp án+Giải thích các bước giải:

Câu 1: Thực hiện phép tính

${a) 2x(x^2 -3x+4)=2x^3-6x^2+8x}$

${b) (x+2)(2x-3)=2x^2-3x+4x-6=2x^2+x-6}$

${c) (3x^3+9x^2-2x): 3x=x^2 +3x-}$ $\frac{2}{3}$

Câu 2: Rút gọn biểu thức

$\frac{3x+4}{3xy}$ ${+}$ $\frac{6x-4}{3xy}$ ${=}$ $\frac{3x+4+6x-4}{3xy}$ ${=}$ $\frac{9x}{3xy}$ ${=}$ $\frac{3}{y}$

${b) }$ $\frac{x}{x-1}$ ${+}$ $\frac{3}{x+1}$ ${-}$ $\frac{6x-4}{x^2-1}$ ${=}$ $\frac{x}{x-1}$ ${+}$ $\frac{3}{x+1}$ ${-}$ $\frac{6x-4}{(x-1)(x+1)}$${=}$ $\frac{x(x+1)}{(x-1)(x+1)}$ ${+}$ $\frac{3(x-1)}{(x-1)(x+1)}$ ${-}$ $\frac{6x-4}{(x-1)(x+1)}$${=}$ $\frac{x(x+1)+3(x-1)-6x+4}{(x-1)(x+1)}$ ${=}$ $\frac{x^2+x+3x-3-6x+4}{(x-1)(x+1)}$ ${=}$$\frac{x^2-2x+1}{(x-1)(x+1)}$ ${=}$ $\frac{(x-1)^2}{(x-1)(x+1)}$ ${=}$ $\frac{x-1}{x+1}$