Lời giải:

a. ĐKXĐ: $x^2-16\neq 0\Leftrightarrow (x-4)(x+4)\neq 0$

$\Leftrightarrow x\neq \pm 4$

b. $A=\frac{x^2+8x+16}{x^2-16}=\frac{(x+4)^2}{(x-4)(x+4)}=\frac{x+4}{x-4}$

c. $A=3\Leftrightarrow \frac{x+4}{x-4}=3$

$\Rightarrow x+4=3(x-4)$

$\Leftrightarrow -2x+16=0$

$\Leftrightarrow x=8$ (tm) 

d. 

$A=0\Leftrightarrow \frac{x+4}{x-4}=0\Leftrightarrow x+4=0\Leftrightarrow x=-4$

Mà theo ĐKXĐ thì $x\neq \pm 4$ nên không tồn tại $x$ để $A=0$

`a,ĐKXĐ:x-4 ne 0,2x+2 ne 0`

`<=>x ne 4,x me -1`

`b,ĐKXĐ:4x^2-25 ne 0`

`<=>(2x-5)(2x+5) ne 0`

`<=>x ne +-5/2`

`c,ĐKXĐ:8x^3+27 ne 0`

`<=>8x^3 ne -27`

`<=>2x ne -3`

`<=>x ne -3/2`

`d,2x+2 ne 0,4y^2-9 ne 0`

`<=>2x ne -2,(2y-3)(2y+3) ne 0`

`<=>x ne -1,y ne +-3/2`

b) ĐKXĐ: \(x\notin\left\{\dfrac{5}{2};-\dfrac{5}{2}\right\}\)

c) ĐKXĐ: \(x\ne-\dfrac{3}{2}\)

d) ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-1\\y\notin\left\{\dfrac{3}{2};-\dfrac{3}{2}\right\}\end{matrix}\right.\)

1) \(\frac{3}{x^2-4y^2}\)

\(=\frac{3}{\left(x-2y\right)\left(x+2y\right)}\)

Phân thức xác định khi \(\left(x-2y\right)\left(x+2y\right)\ne0\)

\(\Rightarrow\hept{\begin{cases}x-2y\ne0\\x+2y\ne0\end{cases}}\Rightarrow x\ne\pm2y\)

2) \(\frac{2x}{8x^3+12x^2+6x+1}\)

\(=\frac{2x}{\left(2x+1\right)^3}\)

Phân thức xác định khi \(\left(2x+1\right)^3\ne0\)

\(\Rightarrow2x+1\ne0\)

\(\Rightarrow x\ne-\frac{1}{2}\)

3) \(\frac{5}{2x-3x^2}\)

\(=\frac{5}{x\left(2-3x\right)}\)

Phân thức xác định khi : \(x\left(2-3x\right)\ne0\)

\(\Rightarrow\hept{\begin{cases}x\ne0\\2-3x\ne0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x\ne0\\x\ne\frac{2}{3}\end{cases}}\)

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

b: \(=\dfrac{2\left(4x^2-4x+1\right)}{4x-30+2x}=\dfrac{4\left(2x-1\right)^2}{6x-30}=\dfrac{2\left(2x-1\right)^2}{3\left(x-5\right)}\)

d: \(=\dfrac{x\left(x-6\right)}{2\left(x-6\right)\left(x+6\right)}=\dfrac{x}{2x+12}\)

Đcm học ngu k biết xài caskov

a) \(ĐKXĐ:\hept{\begin{cases}x\ne\pm2\\x\ne-3\end{cases}}\)

b) \(P=1+\frac{x+3}{x^2+5x+6}\div\left(\frac{8x^2}{4x^3-8x^2}-\frac{3x}{3x^2-12}-\frac{1}{x+2}\right)\)

\(\Leftrightarrow P=1+\frac{x+3}{\left(x+3\right)\left(x+2\right)}:\left(\frac{8x^2}{4x^2\left(x-2\right)}-\frac{3x}{3\left(x^2-4\right)}-\frac{1}{x+2}\right)\)

\(\Leftrightarrow P=1+\frac{1}{x+2}:\left(\frac{2}{x-2}-\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{1}{x+2}\right)\)

\(\Leftrightarrow P=1+\frac{1}{x+2}:\frac{2x+4-x-x+2}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow P=1+\frac{1}{x+2}:\frac{6}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow P=1+\frac{\left(x-2\right)\left(x+2\right)}{6\left(x+2\right)}\)

\(\Leftrightarrow P=1+\frac{x-2}{6}\)

\(\Leftrightarrow P=\frac{x+4}{6}\)

c) Để P = 0

\(\Leftrightarrow\frac{x+4}{6}=0\)

\(\Leftrightarrow x+4=0\)

\(\Leftrightarrow x=-4\)

Để P = 1

\(\Leftrightarrow\frac{x+4}{6}=1\)

\(\Leftrightarrow x+4=6\)

\(\Leftrightarrow x=2\)

d) Để P > 0

\(\Leftrightarrow\frac{x+4}{6}>0\)

\(\Leftrightarrow x+4>0\)(Vì 6>0)

\(\Leftrightarrow x>-4\)

1 x . x x + 1 . x + 1 x + 2 . x + 2 x + 3 . x + 3 x + 4 . x + 4 x + 5 . x + 5 x + 6 . x + 6 x + 7 . x + 7 x + 8 . x + 8 x + 9 . x + 9 x + 10 . x + 10 1 = 1

1b.=2((x+y)+(x+y)(x-y)+(x-y))=2(x²-y²+x+y+x-y)=2(x²-y²+2x)=2x²-2y²+4x

2a.=4xy+4xy+2y=8xy+2y=2y(4x+1)

b.=(3x)²+2.3x.y+y²-(2z)²=(3x+y)²-(2z)²=(3x+y-2z)(3x+y+2z)

c.=x²-x-7x+7=x(x-1)-7(x-1)=(x-1)(x-7)

\(\left(x+y\right)^2+2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\)

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

\(=\left(2x\right)^2\)

\(=4x^2\)

hk tốt

^^