a: \(\left(3y+1\right)\left(2y-3\right)-6y\left(y+2\right)=16\)

=>\(6y^2-9y+2y-3-6y^2-12y=16\)

=>-19y=19

=>y=-1

b: Để A và B đều chia hết cho C thì \(\left\{{}\begin{matrix}12x^{2n}y^{12-3n}⋮3x^3y^4\\3x^3y^7⋮3x^3y^4\left(đúng\right)\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2n>=3\\12-3n>=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}n>=1,5\\-3n>=-8\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}n>=1,5\\n< =\dfrac{8}{3}\end{matrix}\right.\Leftrightarrow n=2\)

a: \(x\left(2x^2-3\right)-x^2\left(5x+1\right)+x^2\)

\(=2x^3-3x-5x^3-x^2+x^2\)

\(=-3x^3-3x\)

b: \(3x\left(y-2\right)-5x\left(1-x\right)-8y\left(x+y\right)\)

\(=3xy-6x-5x+5x^2-8xy-8y^2\)

\(=5x^2-5xy-6x-5y-8y^2\)

c: \(\dfrac{1}{2}x^2\left(6y-3\right)-x\left(xy+\dfrac{1}{2}y\right)+y\left(x-2\right)\)

\(=3x^2y-\dfrac{3}{2}x^2-x^2y-\dfrac{1}{2}xy+xy-2y\)

\(=2x^2y-\dfrac{3}{2}x^2+\dfrac{1}{2}xy-2y\)

Lời giải:

Từ PT(1) $\Rightarrow x=7-2y$. Thay vào PT(2) ta có:

$2(7-2y)^2-y(7-2y)+3y^2+(7-2y)=27$

$\Leftrightarrow 13y^2-65y+105=27$

$\Leftrightarrow 13y^2-65y+78=0$

$\Leftrightarrow y^2-5y+6=0$

$\Leftrightarrow (y-2)(y-3)=0$

$\Leftrightarrow y=2$ hoặc $y=3$

Nếu $y=2$ thì $x=7-2y=7-2.2=3$

Nếu $y=3$ thì $x=7-2y=7-2.3=1$

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