Lời giải:

a. $x(3x+1)+(x-1)^2-(2x+1)(2x-1)=0$

$\Leftrightarrow (3x^2+x)+(x^2-2x+1)-(4x^2-1)=0$

$\Leftrightarrow 3x^2+x+x^2-2x+1-4x^2+1=0$

$\Leftrightarrow (3x^2+x^2-4x^2)+(x-2x)+(1+1)=0$

$\Leftrightarrow -x+2=0$

$\Leftrightarrow x=2$

b.

$(x+1)^3+(2-x)^3-9(x-3)(x+3)=0$

$\Leftrightarrow [(x+1)+(2-x)][(x+1)^2-(x+1)(2-x)+(2-x)^2]-9(x-3)(x+3)=0$

$\Leftrightarrow 3[x^2+2x+1-(x-x^2+2)+(x^2-4x+4)]-9(x-3)(x+3)=0$

$\Leftrightarrow 3(3x^2-3x+3)-9(x^2-9)=0$

$\Leftrightarrow 9(x^2-x+1)-9(x^2-9)=0$

$\Leftrightarrow 9(x^2-x+1-x^2+9)=0$
$\Leftrightarrow 9(-x+10)=0$

$\Leftrightarrow -x+10=0\Leftrightarrow x=10$

c.

$(x-1)^3-(x+3)(x^2-3x+9)+3x^2=25$

$\Leftrightarrow (x^3-3x^2+3x-1)-(x^3+3^3)+3x^2=25$

$\Leftrightarrow x^3-3x^2+3x-1-x^3-27+3x^2=25$
$\Leftrightarrow (x^3-x^3)+(-3x^2+3x^2)+3x-28=25$

$\Leftrightarrow 3x-28=25$

$\Leftrightarrow x=\frac{53}{3}$

d.

$(x+2)^3-(x+1)(x^2-x+1)-6(x-1)^2=23$
$\Leftrightarrow (x^3+6x^2+12x+8)-(x^3+1)-6(x^2-2x+1)=23$

$\Leftrightarrow x^3+6x^2+12x+8-x^3-1-6x^2+12x-6=23$

$\Leftrightarrow (x^3-x^3)+(6x^2-6x^2)+(12x+12x)+(8-1-6)=23$
$\Leftrightarrow 24x+1=23$

$\Leftrgihtarrow 24x=22$

$\Leftrightarrow x=\frac{11}{12}$

e.

$(x+3)(x^2-3x+9)-x(x-2)(x+2)+11=0$

$\Leftrightarrow x^3+3^3-x(x^2-4)+11=0$

$\Leftrightarrow x^3+27-x^3+4x+11=0$

$\Leftrightarrow (x^3-x^3)+4x+(27+11)=0$

$\Leftrightarrow 4x+38=0$

$\Leftrightarrow x=\frac{-19}{2}$

f.

$x(x-3)-x+3=0$

$\Leftrightarrow x(x-3)-(x-3)=0$

$\Leftrightarrow (x-3)(x-1)=0$

$\Leftrightarrow x-3=0$ hoặc $x-1=0$

$\Leftrightarrow x=3$ hoặc $x=1$

mọi người ơi giúp với tui đang cần gấp .Ai làm nhanh nhất thì tui sẽ h cho

(\(3+\dfrac{x}{3-x}+\dfrac{2x}{3+x}-\dfrac{4x^2-3x-9}{x^2-9}\) ):\(\left(\dfrac{2}{3-x}-\dfrac{x-1}{3x-x^2}\right)\)\(=\left(\dfrac{3x^2-27}{\left(x-3\right)\left(x+3\right)}+\dfrac{-x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{2x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{4x^2-3x-9}{\left(x-3\right)\left(x+3\right)}\right)\)\(:\left(\dfrac{2x}{x\left(3-x\right)}-\dfrac{x-1}{x\left(3-x\right)}\right)\)

\(=\dfrac{3x^2-27-x^2-3x+2x^2-6x-4x^2+3x+9}{\left(x-3\right)\left(x+3\right)}:\dfrac{x+1}{x\left(3-x\right)}\) 

\(=\dfrac{-6x-18}{\left(x-3\right)\left(x+3\right)}:\dfrac{x+1}{x\left(3-x\right)}\) \(=\dfrac{-6\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}:\dfrac{x+1}{x\left(3-x\right)}\) 

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

a: Ta có: \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2+3x^2=-33\)

\(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+6x^2+12x+1+3x^2=-33\)

\(\Leftrightarrow39x=-34\)

hay \(x=-\dfrac{34}{39}\)

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

\(\Leftrightarrow x^3-27-x^3+4x=1\)

\(\Leftrightarrow4x=28\)

hay x=7

c: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x-3\right)\left(x+3\right)=26\)

\(\Leftrightarrow x^3+8-x^3+9x=26\)

\(\Leftrightarrow x=2\)

Em ơi em đăng câu hỏi thì dùng công thức toán học để viết chữ viết thế này ai hiểu đúng được em?

1: =>x^2+4x-21=0

=>(x+7)(x-3)=0

=>x=3 hoặc x=-7

2: =>(2x-5-4)(2x-5+4)=0

=>(2x-9)(2x-1)=0

=>x=9/2 hoặc x=1/2

3: =>x^3-9x^2+27x-27-x^3+27+9(x^2+2x+1)=15

=>-9x^2+27x+9x^2+18x+9=15

=>18x=15-9-27=-21

=>x=-7/6

6: =>4x^2+4x+1-4x^2-16x-16=9

=>-12x-15=9

=>-12x=24

=>x=-2

7: =>x^2+6x+9-x^2-4x+32=1

=>2x+41=1

=>2x=-40

=>x=-20

C

Đẳng thức nào đúng:

 A.(x-3)^2=x^2+6x+9                              B.(x-3)^2=9-x^2

C.(x-3)^2=9-6x+x^2                               D.(x-3)^2=x^2-3x+9

Câu C

\(d,\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=15\\ \Leftrightarrow24x=-10\Leftrightarrow x=-\dfrac{5}{12}\\ e,\Leftrightarrow x^3-3x^2+3x-1+8-x^3+3x^2+6x=17\\ \Leftrightarrow9x=10\Leftrightarrow x=\dfrac{10}{9}\\ f,\Leftrightarrow9x^2+18x+9-18x=36+x^3-27\\ \Leftrightarrow x^3-9x^2=0\Leftrightarrow x^2\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\)