\(3x^3+6x^2-3x\Rightarrow ChonD\)

D

ĐKXĐ: \(x\ne\left\{-\dfrac{1}{3};\dfrac{1}{3};0;-\dfrac{4}{3}\right\}\)

\(M=\left(\dfrac{3x\left(1+3x\right)}{\left(1-3x\right)\left(1+3x\right)}+\dfrac{2x\left(1-3x\right)}{\left(1-3x\right)\left(1+3x\right)}\right):\dfrac{2x\left(3x+5\right)}{\left(1-3x\right)^2}\)

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

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

1) D = 4(-xy³)²x = -4x³y⁶

=> bậc của D là: 3 + 6 = 9

2) C.D = (-3x³y⁶).(-4x³y⁶) = 12x⁶y¹²

`1)`

`D=4(-xy^3)^2x`

`D=4(x^2y^6)x`

`D=4x^3y^6`

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`2)C.D=(-3x^3y^6).(4x^3y^6)`

         `=(-3.4)(x^3 .x^3)(y^6 .y^6)`

         `=-12x^6y^12`

\(\left(\dfrac{3x}{1-3x}+\dfrac{2x}{3x+1}\right):\dfrac{6x^2+10x}{1-6x+9x^2}\)\(\left(ĐKXĐ:x\ne\pm\dfrac{1}{3}\right)\)

\(=\left[\dfrac{-3x\left(3x+1\right)+2x\left(3x-1\right)}{\left(3x-1\right)\left(3x+1\right)}\right]:\dfrac{2x\left(3x+5\right)}{\left(3x-1\right)^2}\)

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

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

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

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

\(=\dfrac{1-3x}{6x+2}\)

\(=\dfrac{3x-y+6x+4y}{3x+y}=\dfrac{9x+3y}{3x+y}=\dfrac{3\left(3x+y\right)}{3x+y}=3\)

lỗi ảnh rồi

a: \(\left(3x^2-6x\right):3x=x-2\)

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

ý D nhé