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a)
2x-4=2(x-2)
2x+4=2(x+2)
x
Để P xác định thì
[2(x-2) => [2(x+2)
[2(x+2) =>[ 2(x-2)
[ (x-2)(x+2) => [(x+2)(x-2)
Vay 2(x+2) , 2(x-2), (x+2)(x-2) thi P xác định
$\textbf{a)}$
\[\left(x+3-\frac1{x+3}\right)\cdot\frac{x+3}{x+4}\]
$\text{ĐKXĐ: }x\ne-3,\,-4.$
$=\dfrac{(x+3)^2-1}{x+3}\cdot\dfrac{x+3}{x+4}$
$=\dfrac{(x+2)(x+4)}{x+4}$
$=x+2.$
$\textbf{b)}$
$\left(2x-4-\frac{x-12}{3x+4}\right)\cdot\left(3x-2-\frac{10}{2x+1}\right)$
$\text{ĐKXĐ: }x\ne-\dfrac43,\,-\dfrac12.$
$=\dfrac{(2x-4)(3x+4)-(x-12)}{3x+4}\cdot\dfrac{(3x-2)(2x+1)-10}{2x+1}$
$=\dfrac{6x^2-5x-4}{3x+4}\cdot\dfrac{6x^2-x-12}{2x+1}.$
\(\frac{x^4-y^4}{y^3-x^3}=\frac{\left(x^2+y^2\right)\left(x+y\right)\left(x-y\right)}{\left(y-x\right)\left(x^2+xy+y^2\right)}=-\frac{\left(x^2+y^2\right)\left(x+y\right)}{\left(x^2+xy+y^2\right)}\)
\(\frac{\left(2x-4\right)\left(x-3\right)}{\left(x-2\right)\left(3x^2-27\right)}=\frac{2\left(x-2\right)\left(x-3\right)}{\left(x-2\right)3\left(x-3\right)\left(x+3\right)}=\frac{2}{3\left(x+3\right)}\)
\(\frac{2x^3+x^2-2x-1}{x^3+2x^2-x-2}=\frac{\left(x-1\right)\left(x+1\right)\left(2x+1\right)}{\left(x-1\right)\left(x+1\right)\left(x+2\right)}=\frac{2x+1}{x+2}\)
\(\frac{x^4-y^4}{y^3-x^3}=\frac{\left(x^2+y^2\right)\left(x+y\right)\left(x-y\right)}{\left(y-x\right)\left(x^2+xy+y^2\right)}=-\frac{\left(x^2+y^2\right)\left(x+y\right)}{\left(x^2+xy+y^2\right)}\)
=\(\frac{x}{x+2}.\frac{x+2}{4}-\frac{\left(x-2\right)\left(x^2+2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}.\frac{x^2-2x+4}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{4}=\frac{x}{4}+\frac{x^2+2x+4}{4\left(x+2\right)}=\frac{x^2+2x+x^2+2x+4}{4\left(x+2\right)}=\frac{x^2+\left(x+2\right)^2}{4\left(x+2\right)}\)Có j ko biết thì hỏi nha...Chúc học dốt.....Dỏi thơn