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.

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

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

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

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

a: \(\frac{y}{4}-2+\frac{15}{4y}\)

\(=\frac{y-8}{4}+\frac{15}{4y}=\frac{y^2-8y+15}{4y}=\frac{\left(y-5\right)\left(y-3\right)}{4y}\)

\(\frac{y}{2}+\frac{6}{y}-\frac72=\frac{y-7}{2}+\frac{6}{y}=\frac{y\left(y-7\right)+12}{2y}\)

\(=\frac{y^2-7y+12}{2y}=\frac{\left(y-3\right)\left(y-4\right)}{2y}\)

Ta có: \(M=\frac{\frac{y}{4}-2+\frac{15}{4y}}{\frac{y}{2}+\frac{6}{y}-\frac72}\)

\(=\frac{\left(y-5\right)\left(y-3\right)}{4y}:\frac{\left(y-3\right)\left(y-4\right)}{2y}\)

\(=\frac{\left(y-5\right)\left(y-3\right)}{4y}\cdot\frac{2y}{\left(y-3\right)\left(y-4\right)}=\frac{y-5}{2\cdot\left(y-4\right)}\)

b: \(N=\frac{3b-\frac{1}{9b^2}}{1+\frac{1}{3b}+\frac{1}{9b^2}}\)

\(=\frac{27b^3-1}{9b^2}:\frac{9b^2+3b+1}{9b^2}\)

\(=\frac{\left(3b-1\right)\left(9b^2+3b+1\right)}{9b^2}\cdot\frac{9b^2}{9b^2+3b+1}\)

=3b-1

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

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

Bắn tung tóe

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

$x^2-3$

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

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

ΔAHB vuông tại H

=>\(AH^2+HB^2=AB^2\)

=>\(AH^2=6^2-3,6^2=\left(6-3,6\right)\left(6+3,6\right)=2,4\cdot9,6=4,8^2\)

=>AH=4,8(cm)

Xét ΔAHB vuông tại H có HE là đường cao

nên \(AE\cdot AB=AH^2\)

=>\(AE=4,8^2:6=3,84\left(\operatorname{cm}\right)\)

Xét tứ giác AEHF có \(\hat{AEH}=\hat{AFH}=\hat{FAE}=90^0\)

nên AEHF là hình chữ nhật

=>HF=AE

=>HF=3,84(cm)

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

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

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

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

`#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}`

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