a: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)

\(\Leftrightarrow x^3+8-x^3-2x=15\)

\(\Leftrightarrow2x=-7\)

hay \(x=-\dfrac{7}{2}\)

b: Ta có: \(\left(x-2\right)^3-\left(x-4\right)\left(x^2+4x+16\right)+6\left(x+1\right)^2=49\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+64+6\left(x+1\right)^2=49\)

\(\Leftrightarrow-6x^2+12x+56+6x^2+12x+6=49\)

\(\Leftrightarrow24x=-13\)

hay \(x=-\dfrac{13}{24}\)

b. (x + 4)² - (x + 1)(x - 1) = 16

<=> x² + 4x + 16 - (x² - 1) = 16

<=> x² + 4x + 16 - x² + 1 - 16 = 0

<=> x² - x² + 4x = 16 - 16 - 1

<=> 4x = -1

<=> x = \(\dfrac{-1}{4}\)

\(a,\Leftrightarrow-9x^2+30x-25+9x^2+18x+9=30\\ \Leftrightarrow48x=46\\ \Leftrightarrow x=\dfrac{23}{24}\\ b,\Leftrightarrow x^2+8x+16-x^2+1=16\\ \Leftrightarrow8x=-1\Leftrightarrow x=-\dfrac{1}{8}\)

Ta có: \(4\left(x+3\right)\left(3x-2\right)-3\left(x-1\right)\left(4x-1\right)=16\)

\(\Leftrightarrow4\left(3x^2-2x+9x-6\right)-3\left(4x^2-x-4x+1\right)=16\)

\(\Leftrightarrow12x^2+28x-24-12x^2+15x-3=16\)

\(\Leftrightarrow x=1\)

Lời giải:
$\frac{2}{3}x(x^2-16)=0$

$\Leftrightarrow x=0$ hoặc $x^2-16=0$

$\Leftrightarrow x=0$ hoặc $(x-4)(x+4)=0$

$\Leftrightarrow x=0$ hoặc $x-4=0$ hoặc $x+4=0$

$\Leftrightarrow x=0$ hoặc $x=\pm 4$

Không có đáp án nào đúng.

2: \(3x\left(x-4\right)+2x-8=0\)

=>\(3x\left(x-4\right)+2\left(x-4\right)=0\)

=>\(\left(x-4\right)\left(3x+2\right)=0\)

=>\(\left[{}\begin{matrix}x-4=0\\3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{2}{3}\end{matrix}\right.\)

3: 4x(x-3)+x²-9=0

=>\(4x\left(x-3\right)+\left(x+3\right)\left(x-3\right)=0\)

=>\(\left(x-3\right)\left(4x+x+3\right)=0\)

=>\(\left(x-3\right)\left(5x+3\right)=0\)

=>\(\left[{}\begin{matrix}x-3=0\\5x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{5}\end{matrix}\right.\)

4: \(x\left(x-1\right)-x^2+3x=0\)

=>\(x^2-x-x^2+3x=0\)

=>2x=0

=>x=0

5: \(x\left(2x-1\right)-2x^2+5x=16\)

=>\(2x^2-x-2x^2+5x=16\)

=>4x=16

=>x=4

a: Để A là số nguyên thì

x^3-2x^2+4 chia hết cho x-2

=>\(x-2\in\left\{1;-1;2;-2;4;-4\right\}\)

=>\(x\in\left\{3;1;4;0;6;-2\right\}\)

b: Để B là số nguyên thì

\(3x^3-x^2-6x^2+2x+9x-3+2⋮3x-1\)

=>\(3x-1\in\left\{1;-1;2;-2\right\}\)

=>\(x\in\left\{\dfrac{2}{3};0;1;-\dfrac{1}{3}\right\}\)

a:

ĐKXĐ: x<>-1/2

Để \(\dfrac{2x^3+x^2+2x+2}{2x+1}\in Z\) thì

\(2x^3+x^2+2x+1+1⋮2x+1\)

=>\(2x+1\inƯ\left(1\right)\)

=>2x+1 thuộc {1;-1}

=>x thuộc {0;-1}

b:

ĐKXĐ: x<>1/3

 \(\dfrac{3x^3-7x^2+11x-1}{3x-1}\in Z\)

=>3x^3-x^2-6x^2+2x+9x-3+2 chia hết cho 3x-1

=>2 chia hết cho 3x-1

=>3x-1 thuộc {1;-1;2;-2}

=>x thuộc {2/3;0;1;-1/3}

mà x nguyên

nên x thuộc {0;1}

c: 

ĐKXĐ: x<>2

\(\dfrac{x^4-16}{x^4-4x^3+8x^2-16x+16}\in Z\)

=>\(\left(x^2-4\right)\left(x^2+4\right)⋮\left(x-2\right)^2\left(x^2+4\right)\)

=>\(x+2⋮x-2\)

=>x-2+4 chia hết cho x-2

=>4 chia hết cho x-2

=>x-2 thuộc {1;-1;2;-2;4;-4}

=>x thuộc {3;1;4;0;6;-2}

a) \(\Rightarrow6x-2-4-12x=16\)

\(\Rightarrow-6x-6=16\)

\(\Rightarrow-6x=22\)

\(\Rightarrow x=\frac{-11}{3}\)

b) \(\Rightarrow2x^2-2x-3x^2+12x+x^2+2x=-3\)

\(\Rightarrow14x=-3\)

\(\Rightarrow x=\frac{-3}{14}.\)

a, \(\dfrac{x^3+27}{x^2-3x+9}=\dfrac{x+3}{M}\Leftrightarrow\dfrac{\left(x+3\right)\left(x^2-3x+9\right)}{x^2-3x+9}=\dfrac{x+3}{M}\)

\(\Rightarrow M=\dfrac{x+3}{x+3}=1\)

b, \(\dfrac{M}{x+4}=\dfrac{x^2-8x+16}{16-x^2}=\dfrac{\left(x-4\right)^2}{\left(4-x\right)\left(x+4\right)}=\dfrac{4-x}{x+4}\)

\(\Rightarrow M=\dfrac{\left(4-x\right)\left(x+4\right)}{x+4}=4-x\)

c, tương tự