Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
$\textbf{a)}$
\[\left(x+3-\frac1{x+3}\right)\cdot\frac{x+3}{x+4}\]
$\text{ĐKXĐ: }x\ne-3,\,-4.$
$=\dfrac{(x+3)^2-1}{x+3}\cdot\dfrac{x+3}{x+4}$
$=\dfrac{(x+2)(x+4)}{x+4}$
$=x+2.$
$\textbf{b)}$
$\left(2x-4-\frac{x-12}{3x+4}\right)\cdot\left(3x-2-\frac{10}{2x+1}\right)$
$\text{ĐKXĐ: }x\ne-\dfrac43,\,-\dfrac12.$
$=\dfrac{(2x-4)(3x+4)-(x-12)}{3x+4}\cdot\dfrac{(3x-2)(2x+1)-10}{2x+1}$
$=\dfrac{6x^2-5x-4}{3x+4}\cdot\dfrac{6x^2-x-12}{2x+1}.$
câu 1
a)\(ĐKXĐ:x^3-8\ne0=>x\ne2\)
b)\(\frac{3x^2+6x+12}{x^3-8}=\frac{3\left(x^2-2x+4\right)}{\left(x-2\right)\left(x^2-2x+4\right)}=\frac{3}{x-2}\left(#\right)\)
Thay \(x=\frac{4001}{2000}\)zô \(\left(#\right)\)ta được
\(\frac{3}{\frac{4001}{2000}-2}=\frac{3}{\frac{4001}{2000}-\frac{4000}{2000}}=\frac{3}{\frac{1}{2000}}=6000\)
a.=\(\frac{7x+2}{3xy^2}.\frac{x^2y}{14x+4}\)
=\(\frac{7x+2}{3y}.\frac{x^2y}{2\left(7x+2\right)}\)
=\(\frac{1}{3y}.\frac{x}{2}\)
=\(\frac{x}{6y}\)
b.=\(\frac{8xy}{3x-1}.\frac{5-15x}{12xy^3}\)
=\(\frac{2}{3x-1}.\frac{-15x+5}{3y^2}\)
=\(\frac{2}{3x-1}.\frac{-5\left(3x-1\right)}{3y^2}\)
=\(\frac{-10}{3y^2}\)
c.=\(\frac{3\left(x^3+1\right)}{x-1}.\frac{1}{x^2-x+1}\)
=\(\frac{3\left(x+1\right).\left(x^2-x+1\right)}{x-1}.\frac{1}{x^2-x+1}\)
=\(\frac{3x+3}{x-1}\)
d.=\(\frac{4\left(x+3\right)}{.\left(3x-1\right)}.\frac{1-3x}{x^2+3x}\)
=\(\frac{4\left(x+3\right)}{x.\left(3x-1\right)}.\frac{-\left(3x-1\right)}{x\left(x+3\right)}\)
=\(\frac{-4}{x^2}\)
e.=\(\frac{2\left(2x+3y\right)}{x-1}.\frac{1-x^3}{4x^2+12xy+9y^2}\)
=\(2.\frac{-\left(1+x+x^2\right)}{2x+3y}\)
=\(-\frac{2x^2+2x+2}{2x+3y}\)


