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16 tháng 10 2021

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

20 tháng 11 2021

B

4 tháng 1 2022

\(\left(\dfrac{5x-x-12}{x\left(x+3\right)}\right):\dfrac{x-3}{x+3}=\dfrac{4\left(x-3\right)}{x\left(x+3\right)}.\dfrac{x+3}{x-3}=\dfrac{4}{x}\)

19 tháng 3 2022

ĐK: x\(\ne\){-3;0;3}.

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

13 tháng 7 2017

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

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

   = 80/9x3+1/3x2-28/3x+27

12 tháng 10 2021

\(a,=\dfrac{2y^4}{3x\left(2x-3y\right)}\\ b,=-\dfrac{2y\left(3x-1\right)^2}{3x^2}\\ c,=\dfrac{5\left(4x^2-9\right)}{\left(2x+3\right)^2}=\dfrac{5\left(2x-3\right)\left(2x+3\right)}{\left(2x+3\right)^2}=\dfrac{5\left(2x-3\right)}{2x+3}\\ d,=\dfrac{5x\left(x-2y\right)}{-2\left(x-2y\right)^3}=-\dfrac{5x}{2\left(x-2y\right)^2}\)

Bài 2:

a: \(=2x^4-x^3-10x^2-2x^3+x^2+10x=2x^3-3x^3-9x^2+10x\)

b: \(=\left(x^2-15x\right)\left(x^2-7x+3\right)\)

\(=x^4-7x^3+3x^2-15x^3+105x^2-45x\)

\(=x^4-22x^3+108x^2-45x\)

c: \(=12x^5-18x^4+30x^3-24x^2\)

d: \(=-3x^6+2.4x^5-1.2x^4+1.8x^2\)

10 tháng 1 2021

\(\dfrac{7\left(3x^2-1\right)}{1-3x^2}\)

\(\dfrac{-7\left(3x^2-1\right)}{3x^2-1}\)

= -7 

Ta có: \(\dfrac{7\left(3x^2-1\right)}{1-3x^2}\)

\(=\dfrac{-7\cdot\left(1-3x^2\right)}{1-3x^2}\)

=-7

24 tháng 7 2017

câu d

\(D=\dfrac{\left(1-x^2\right)}{x}\left(\dfrac{x^2}{x+3}-1\right)+\dfrac{3x^2-14x+3}{x^2+3x}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{\left(1-x^2\right)\left(x^2-x-3\right)+3x^2-14x+3}{x\left(x+3\right)}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{x^2-x-3-x^4+x^3-3x^2+3x^2-14x+3}{x\left(x+3\right)}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-x^4+x^3+x^2-15x}{x\left(x+3\right)}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-x\left(x^3-x^2-x+15\right)}{x\left(x+3\right)}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-\left(x^3-x^2-x+15\right)}{\left(x+3\right)}\end{matrix}\right.\)