`9x^2 - 6x + 1 = 9`

`=> (3x)^2 - 2.3x.1 + 1^2 = 9`

`=> (3x - 1)^2 = 3^2`

`=> 3x - 1 = 3` hoặc `3x - 1 = -3`

`=> 3x = 4` hoặc `3x = -2`

`=> x =` \(\dfrac{4}{3}\) hoặc \(x=-\dfrac{2}{3}\)

\(-2\left(x-7\right)\left(x+3\right)+\left(5x-1\right)\left(x+4\right)-3x^2-27x\)

\(=-2\left(x^2-4x-21\right)+5x^2+20x-x-4-3x^2-27x\)

\(=-2x^2+8x+42+2x^2-8x-4\)

=38

|3x-7|=12

=>\(\left[{}\begin{matrix}3x-7=12\\3x-7=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=19\\3x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{19}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)

`|3x - 7| = 12`

`=> 3x - 7 = 12` hoặc `3x - 7 = -12`

`=> 3x = 19` hoặc `3x = -5`

`=> x =` \(\dfrac{19}{3}\) hoặc `x =` \(-\dfrac{5}{3}\)

`x^3 - 3x^2 + 3x - 1 - y^3`

`= x^3 - 3 . x^2 . 1 + 3 . x . 1^2 - 1^3 - y^3`

`= (x-1)^3 - y^3`

`= (x-1-y)[(x-1)^2 + (x-1)y + y^2]`

`= (x-1-y)(x^2 - 2x + 1 + xy-y + y^2)`

\(x^3-3x^2+3x-1-y^3\)

\(=\left(x^3-3x^2+3x-1\right)-y^3\)

\(=\left(x-1\right)^3-y^3\)

\(=\left(x-1-y\right)\left[\left(x-1\right)^2+y\left(x-1\right)+y^2\right]\)

\(=\left(x-y-1\right)\left(x^2-2x+1+xy-y+y^2\right)\)

`x(x+1)(x+6)-x^3=5x`

=> (𝑥²+𝑥)(𝑥+6)−𝑥³−5𝑥=0=

=> 𝑥³+𝑥²+6𝑥²+6𝑥−𝑥³−5𝑥=0

=> 7𝑥²+𝑥=0

=> 𝑥(7𝑥+1)=0

=> 𝑥=0 hoặc 𝑥 `=-1/7`
 

x(x+1)(x+6)-x³=5x

⇒x³+7x²+6x-x³=5x

⇒7x²+6x=5x

⇒7x²=-x

x²≥0∀x

7x²≥0∀x

⇒7x²=-x

⇔x=0

Lời giải:
Gọi đa thức thương và đa thức dư khi chia $f(x)$ cho $(x+1)(x^2+1)$ lần lượt là $Q(x)$ và $ax^2+bx+c$ với $a,b,c$ là số thực.

Ta có:

$f(x)=(x+1)(x^2+1)Q(x)+ax^2+bx+c$

$f(x)=(x+1)(x^2+1)Q(x)+a(x^2-1)+b(x-1)+(a+b+c)$

$=(x+1)[(x^2+1)Q(x)+a(x-1)+b]+(a+b+c)$

$\Rightarrow f(x)$ chia $x+1$ dư $a+b+c$

$\Rightarrow a+b+c=4(1)$

Lại có:

$f(x)=(x+1)(x^2+1)Q(x)+a(x^2+1)+bx+(c-a)$

$=(x^2+1)[(x+1)Q(x)+a]+bx+(c-a)$

$\Rightarrow f(x)$ chia $x^2+1$ dư $bx+(c-a)$

$\Rightarrow b=2; c-a=3(2)$

Từ $(1); (2)\Rightarrow b=2; c=2,5; a=-0,5$

\(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)

=>\(24x^2+16x-9x-6-\left(4x^2+16x+7x+28\right)=10x^2-2x+5x-1\)

=>\(24x^2+7x-6-4x^2-23x-28-10x^2-3x+1=0\)

=>\(10x^2-19x-33=0\)

=>\(10x^2-30x+11x-33=0\)

=>10x(x-3)+11(x-3)=0

=>(x-3)(10x+11)=0

=>\(\left[{}\begin{matrix}x-3=0\\10x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{11}{10}\end{matrix}\right.\)

`(x-2)(x-2) - (x-1)(x+1) `

`= (x-2)^2 - (x^2 - 1)`

`= x^2 - 4x + 4 - x^2 + 1`

`= -4x + 5`

`= -4 . 81 +5`

`= -319`

(x - 2)(x - 2) - (x - 1)(x + 1)

= (x^2 - 4) - (x^2 - 1)

= x^2 - 4 - x^2 + 1

= -3 

=> Biểu thức luôn có giá trị là -3 với mọi x 

\(\left(9x^3y^2+5x^2y-4xy\right):\left(2xy^2\right)\\ =9x^3y^2:2xy^2+5x^2y:2xy^2-4xy:2xy^2\\ =\dfrac{9}{2}x^2+\dfrac{5x}{2y}-\dfrac{2}{y}\)