Viết thế này có thể hiểu được ko bạn?

\(\left(3x^2-6x\right):3x-\frac{\left(1-3x\right)\left(1-3x\right)}{1-3x}\)

\(=\frac{3xx}{3x}-\frac{2\cdot3x}{3x}-\left(1-3x\right)\)

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

\(=4x-3\)

Ta có: \(\left(3x+2\right)\left(9x^2-6x+4\right)-9x\left(3x^2+1\right)\)

\(=27x^3+8-27x^3-9x\)

=8-9x

\(a,P=\left[\dfrac{x+1}{3x\left(x+1\right)}-\dfrac{2x-1}{3x\left(2x-1\right)}-1\right]\cdot\dfrac{2x}{1-x}\left(x\ne1;x\ne-1;x\ne0\right)\\ P=\left(\dfrac{1}{3x}-\dfrac{1}{3x}-1\right)\cdot\dfrac{2x}{1-x}\\ P=-1\cdot\dfrac{2x}{1-x}=\dfrac{2x}{x-1}\\ b,P=2+\dfrac{2}{x-1}\in Z\\ \Leftrightarrow x-1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\\ \Leftrightarrow x\in\left\{2;3\right\}\left(x\ne-1;x\ne0\right)\\ c,P\le1\Leftrightarrow\dfrac{2x}{x-1}-1\le0\\ \Leftrightarrow\dfrac{x+1}{x-1}\le0\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1\le0\\x-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+1\ge0\\x-1< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow-1\le x< 1\)

a: \(P=\left(\dfrac{x+1}{3x\left(x+1\right)}-\dfrac{2x-1}{3x\left(2x-1\right)}-1\right)\cdot\dfrac{2x}{x-1}\)

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

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

(5x³ – 4x²) : 2x² + (3x⁴ + 6x) : 3x – x(x² – 1)

= 5x³ : 2x² + (-4x²): 2x² + 3x⁴ : 3x + 6x : 3x – [x. x² + x . (-1)]

= (5:2) . (x³ : x²) + [(-4) : 2] . (x² : x²) + (3 : 3) . (x⁴ : x) + (6 : 3). (x:x) – ( x³ – x)

= \(\dfrac{5}{2}\)x – 2 + x³ + 2 – x³ + x

= (x³ – x³) + (\(\dfrac{5}{2}\)x + x) + (-2 + 2)

= 0 + \(\dfrac{7}{2}\)x + 0

= \(\dfrac{7}{2}\)x

=9x²+24x+16-9x²+1

=24x+17

26:

A=12x^2+10x-6x-5-(12x^2-8x+3x-2)

=12x^2+4x-5-12x^2+5x+2

=9x-3

Khi x=-2 thì A=-18-3=-21

25:

b: \(\left(y-3\right)\left(y^2+y+1\right)-y\left(y^2-2\right)\)

=y^3+y^2+y-3y^2-3y-3-y^3+2y

=-2y^2-3

( 3 x   +   1 ) 2   –   2 ( 3 x   +   1 ) ( 3 x   +   5 )   +   ( 3 x   +   5 ) 2 =   ( ( 3 x   +   1 )   –   ( 3 x   +   5 ) ) 2 =   ( 3 x   +   1   –   3 x   –   5 ) 2     =   ( - 4 ) 2   =   16

Đáp án cần chọn là: B