ĐKXĐ: \(x\ne0;x\ne\pm1\)

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

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

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

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

\(=\dfrac{x+1}{2x\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{1}{2x\left(x-1\right)}\)

\(=\dfrac{1}{2x^2-2x}\)

Kết bạn với tôi đi ,tôi cô đơn quá 

Lời giải:
b. 

$(3x+2)(3x-2)-9x(1+x)=9x^2-4-(9x+9x^2)=9x^2-4-9x-9x^2=-4-9x$

c.

$\frac{1}{x-2}-\frac{1}{x+2}+\frac{2x}{x^2-4}$

$=\frac{x+2}{(x-2)(x+2)}-\frac{x-2}{(x+2)(x-2)}+\frac{2x}{(x-2)(x+2)}$

$=\frac{x+2-(x-2)+2x}{(x-2)(x+2)}=\frac{4+2x}{(x-2)(x+2)}$

$=\frac{2(x+2)}{(x-2)(x+2)}=\frac{2}{x-2}$

\(199^3-199\\=199(199^2-1)\\=199(199-1)(199+1)\\=199\cdot198\cdot200\)

Vì \(199\cdot198\cdot200⋮200\) nên \(199^3-199⋮200\)

Bắn tung tóe

\(x^2-3\\ =x^2-\left(\sqrt{3}\right)^2\\ =\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)\)

$x^2-3$

$=x^2-(\sqrt3)^2$

$=(x-\sqrt3)(x+\sqrt3)$

\(\dfrac{5x+y^2}{x^2y}-\dfrac{5y-x^2}{xy^2}\left(dkxd:x,y\ne0\right)\)

\(=\dfrac{y\left(5x+y^2\right)}{x^2y^2}-\dfrac{x\left(5y-x^2\right)}{x^2y^2}\)

\(=\dfrac{5xy+y^3-5xy+x^3}{x^2y^2}\)

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

`#3107.101107`

`x^3 - 3x^2 - 4x + 12 = 0`

`\Leftrightarrow (x^3 - 3x^2) - (4x - 12) = 0`

`\Leftrightarrow x^2(x - 3) - 4(x - 3) = 0`

`\Leftrightarrow (x^2 - 4)(x - 3) = 0`

`\Leftrightarrow (x - 2)(x + 2)(x - 3) = 0`

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

`\Leftrightarrow`\(\left[{}\begin{matrix}x=2\\x=-2\\x=3\end{matrix}\right.\)

Vậy, `x \in {-2; 2; 3}`

_______

Sử dụng HĐT: \(A^2-B^2=\left(A-B\right)\left(A+B\right).\)

`#3107.101107`

`(x - 1)(x + 2) - 2x - 4 = 0`

`\Leftrightarrow (x - 1)(x + 2) - (2x + 4) = 0`

`\Leftrightarrow (x - 1)(x + 2) - 2(x + 2) = 0`

`\Leftrightarrow (x + 2)(x - 1 - 2) = 0`

`\Leftrightarrow (x + 2)(x - 3) = 0`

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

`\Leftrightarrow`\(\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)

Vậy, `x \in {-2; 3}.`

\(\left(x-1\right)\left(x+2\right)-2x-4=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)-\left(2x+4\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)-2\left(x+2\right)=0\)

\(\Leftrightarrow\left(x-1-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\)

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

Vậy x=3 hoặc x=-2

Bài 1:

a. ĐKXĐ: $1-x\neq 0; 1+x\neq 0; 1-x^2\neq 0$

$\Leftrightarrow x\neq \pm 1$

b.

\(A=\frac{1+x}{(1-x)(1+x)}+\frac{2(1-x)}{(x+1)(1-x)}-\frac{5-x}{(1-x)(1+x)}\\ =\frac{1+x+(2-2x)-(5-x)}{(1-x)(1+x)}=\frac{-2}{(1-x)(1+x)}=\frac{-2}{1-x^2}\)