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GV Nguyễn Trần Thành Đạt
Giáo viên
23 tháng 9 2023
\(a,\dfrac{x^2+2}{x^3+1}-\dfrac{1}{x+1}\left(ĐKXĐ:x\ne-1\right)\\ =\dfrac{x^2+2-\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\\ =\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{1}{x^2-x+1}\\ c,\dfrac{1}{2-2x}-\dfrac{3}{2+2x}+\dfrac{2x}{x^2-1}\\ =\dfrac{-1}{2\left(x-1\right)}-\dfrac{3}{2\left(x+1\right)}+\dfrac{2x}{\left(x-1\right)\left(x+1\right)}\left(ĐKXĐ:x\ne\pm1\right)\\ =\dfrac{-1\left(x+1\right)-3\left(x-1\right)+2x.2}{2\left(x+1\right)\left(x-1\right)}\\ =\dfrac{-x-1-3x+3+4x}{2\left(x+1\right)\left(x-1\right)}=\dfrac{2}{2\left(x+1\right)\left(x-1\right)}=\dfrac{1}{\left(x-1\right)\left(x+1\right)}\)
26 tháng 1 2023
m: \(=\left(\dfrac{2x}{\left(x-1\right)\left(x+1\right)}+\dfrac{x-1}{2\left(x+1\right)}\right)\cdot\dfrac{2x}{x+1}-\dfrac{3}{x-1}\)
\(=\dfrac{4x+x^2-2x+1}{2\left(x-1\right)\left(x+1\right)}\cdot\dfrac{2x}{x+1}-\dfrac{3}{x-1}\)
\(=\dfrac{\left(x+1\right)^2\cdot x}{\left(x-1\right)\left(x+1\right)^2}-\dfrac{3}{x-1}=\dfrac{x}{x-1}-\dfrac{3}{x-1}=\dfrac{x-3}{x-1}\)
p: \(=\left(\dfrac{-\left(x+2\right)}{x-2}+\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{x+2}\right)\cdot\dfrac{-x^2\left(x-2\right)}{x\left(x-3\right)}\)
\(=\dfrac{-x^2-4x-4+4x^2+x^2-4x+4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{-x\left(x-2\right)}{x-3}\)
\(=\dfrac{4x^2-8x}{\left(x+2\right)}\cdot\dfrac{-x}{x-3}=\dfrac{-4x^2\left(x-2\right)}{\left(x+2\right)\left(x-3\right)}\)
24 tháng 1 2021
Mk giải giúp bạn phần a thôi nha! (Dài lắm, lười :v)
a, 1 + \(\dfrac{x}{3-x}\) = \(\dfrac{5x}{\left(x+2\right)\left(x+3\right)}+\dfrac{2}{x+2}\) (x \(\ne\) -2; x \(\ne\) \(\pm\) 3)
\(\Leftrightarrow\) \(\dfrac{3}{3-x}=\dfrac{5x+2\left(x+3\right)}{\left(x+2\right)\left(x+3\right)}\)
\(\Leftrightarrow\) \(\dfrac{3}{3-x}=\dfrac{5x+2x+6}{\left(x+2\right)\left(x+3\right)}\)
\(\Leftrightarrow\) \(\dfrac{3}{3-x}=\dfrac{7x+6}{x^2+5x+6}\)
Vì 3 - x \(\ne\) 0; x2 + 5x + 6 \(\ne\) 0
\(\Rightarrow\) 3(x2 + 5x + 6) = (7x + 6)(3 - x)
\(\Leftrightarrow\) 3x2 + 15x + 18 = 21x - 7x2 + 18 - 6x
\(\Leftrightarrow\) 10x2 = 0
\(\Leftrightarrow\) x = 0 (TM)
Vậy S = {0}
Chúc bn học tốt! (Nếu bạn cần phần nào khác mk có thể giúp bn chứ đừng có đăng hết lên, ít người làm lắm :v)
NB
24 tháng 1 2021
b)\(\dfrac{x+2}{x-2}-\dfrac{2}{x^2-2x}=\dfrac{1}{x}\\ \Leftrightarrow\dfrac{x\left(x+2\right)}{x\left(x-2\right)}-\dfrac{2}{x\left(x-2\right)}=\dfrac{x-2}{x\left(x-2\right)}\Leftrightarrow x^2+2x-2=x-2\\ \Leftrightarrow x^2+2x-2-x+2=0\Leftrightarrow x^2-x=0\\ \Leftrightarrow x\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
vậy..
2 tháng 12 2021
\(a,=\dfrac{4x+8}{x^2+2x}=\dfrac{4\left(x+2\right)}{x\left(x+2\right)}=\dfrac{4}{x}\\ b,=\dfrac{\left(2x-3\right)-\left(2x-4\right)}{x-2}=\dfrac{2x-3-2x+4}{x-2}=\dfrac{1}{x-2}\\ c,=\dfrac{2x-1-3x-2}{x+3}=\dfrac{-x-3}{x+3}=\dfrac{-\left(x+3\right)}{x+3}=-1\\ d,=\dfrac{11x-18+x}{2x-3}=\dfrac{12x-18}{2x-3}=\dfrac{6\left(2x-3\right)}{2x-3}=6\)
\(e,=\dfrac{3x-6-9x+3}{2x+1}=\dfrac{-6x-3}{2x+1}=\dfrac{-3\left(2x+1\right)}{2x+1}=-3\)
6 tháng 12 2021
\(a,=\dfrac{x^2-2+2-x}{x\left(x-1\right)^2}=\dfrac{x\left(x-1\right)}{x\left(x-1\right)^2}=\dfrac{1}{x-1}\\ b,=\dfrac{6x-3+6x^2-6x+2x^2+1}{2x\left(2x-1\right)}=\dfrac{8x^2-2}{2x\left(2x-1\right)}\\ =\dfrac{2\left(2x-1\right)\left(2x+1\right)}{2x\left(2x-1\right)}=\dfrac{2x+1}{x}\\ c,=\dfrac{x^3+x^2+x+2x-2+4x^2-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x^3+5x^2+3x-3}{x^3-1}\)
14 tháng 5 2022
a: \(=\dfrac{2x-2x+y}{2\left(2x-y\right)}=\dfrac{y}{2\left(2x-y\right)}\)
b: \(=\dfrac{3x+1}{\left(x-1\right)\left(x+1\right)}-\dfrac{x}{2\left(x-1\right)}\)
\(=\dfrac{6x+2-x^2-x}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-x^2+5x+2}{2\left(x-1\right)\left(x+1\right)}\)
c: \(=\dfrac{1}{x+2}+\dfrac{x+8}{3x\left(x+2\right)}\)
\(=\dfrac{3x+x+8}{3x\left(x+2\right)}=\dfrac{4x+8}{3x\left(x+2\right)}=\dfrac{4}{3x}\)
d: \(=\dfrac{4x+6-2x^2+3x+2x+1}{\left(2x-3\right)\left(2x+3\right)}\)
\(=\dfrac{-2x^2+9x+7}{\left(2x-3\right)\left(2x+3\right)}\)
10 tháng 11 2023
a:
ĐKXĐ: x<>-1
\(\dfrac{x^2+2}{x^3+1}-\dfrac{1}{x+1}\)
\(=\dfrac{x^2+1}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{1}{x+1}\)
\(=\dfrac{x^2+1-x^2+x-1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{x}{\left(x+1\right)\left(x^2-x+1\right)}\)
b: \(\dfrac{x}{x^2-2x}-\dfrac{x^2+4x}{x^3-4x}-\dfrac{2}{x^2+2x}\)
\(=\dfrac{x}{x\left(x-2\right)}-\dfrac{x\left(x+4\right)}{x\left(x^2-4\right)}-\dfrac{2}{x\left(x+2\right)}\)
\(=\dfrac{1}{x-2}-\dfrac{x+4}{x^2-4}-\left(\dfrac{1}{x}-\dfrac{1}{x+2}\right)\)
\(=\dfrac{1}{x-2}-\dfrac{x+4}{x^2-4}-\dfrac{1}{x}+\dfrac{1}{x+2}\)
\(=\left(\dfrac{1}{x-2}-\dfrac{x+4}{x^2-4}+\dfrac{1}{x+2}\right)-\dfrac{1}{x}\)
\(=\dfrac{x+2-x-4+x-2}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x}\)
\(=\dfrac{x-4}{x^2-4}-\dfrac{1}{x}\)
\(=\dfrac{x^2-4x-x^2+4}{x\left(x^2-4\right)}=\dfrac{-4x+4}{x\left(x-2\right)\left(x+2\right)}\)
c: \(\dfrac{1}{2-2x}-\dfrac{3}{2+2x}+\dfrac{2x}{x^2-1}\)
\(=\dfrac{-1}{2\left(x-1\right)}-\dfrac{3}{2\left(x+1\right)}+\dfrac{2x}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-x-1-3x+3+4x}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{2}{2\left(x-1\right)\left(x+1\right)}=\dfrac{1}{x^2-1}\)
d:
\(\dfrac{1}{\left(a-b\right)\left(b-c\right)}+\dfrac{1}{\left(b-c\right)\left(c-a\right)}+\dfrac{1}{\left(c-a\right)\left(a-b\right)}\)
\(=\dfrac{c-a+a-b+b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
13 tháng 7 2022
b: \(=\left[\dfrac{2}{3x}-\dfrac{2}{x+1}\cdot\dfrac{x+1-3x^2-3x}{3x}\right]\cdot\dfrac{x}{x+1}\)
\(=\left(\dfrac{2}{3x}-\dfrac{2}{x+1}\cdot\dfrac{-3x^2-2x+1}{3x}\right)\cdot\dfrac{x}{x+1}\)
\(=\dfrac{2x+2+6x^2+4x-2}{3x\left(x+1\right)}\cdot\dfrac{x}{x+1}\)
\(=\dfrac{6x^2+6x}{3\left(x+1\right)}\cdot\dfrac{1}{x+1}\)
\(=\dfrac{6x\left(x+1\right)}{3\left(x+1\right)^2}=\dfrac{2x}{x+1}\)
c: \(VT=\left[\dfrac{2}{\left(x+1\right)^3}\cdot\dfrac{x+1}{x}+\dfrac{1}{\left(x+1\right)^2}\cdot\dfrac{1+x^2}{x^2}\right]\cdot\dfrac{x^3}{x-1}\)
\(=\left(\dfrac{2}{x\left(x+1\right)^2}+\dfrac{x^2+1}{x^2\cdot\left(x+1\right)^2}\right)\cdot\dfrac{x^3}{x-1}\)
\(=\dfrac{2x+x^2+1}{x^2\cdot\left(x+1\right)^2}\cdot\dfrac{x^3}{x-1}\)
\(=\dfrac{\left(x+1\right)^2}{\left(x+1\right)^2}\cdot\dfrac{x}{x-1}=\dfrac{x}{x-1}\)
20 tháng 1 2021
\(a,\dfrac{3}{2x-1}+1=\dfrac{2x-1}{2x+1};ĐKXĐ:x\ne\pm\dfrac{1}{2}\\ \Leftrightarrow\dfrac{3}{2x-1}-\dfrac{2x-1}{2x+1}+1=0\\ \Leftrightarrow\dfrac{3\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}-\dfrac{\left(2x-1\right)\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)}+\dfrac{\left(2x-1\right)\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}=0\\ \Rightarrow3\left(2x+1\right)-\left(2x-1\right)^2+\left(2x-1\right)\left(2x+1\right)=0\\ \Leftrightarrow6x+3-\left(4x^2-4x+1\right)+\left(4x^2-1\right)=0\\ \Leftrightarrow6x+3-4x^2+4x-1+4x^2-1=0\\ \Leftrightarrow10x+1=0\\ \Leftrightarrow10x=-1\\ \Leftrightarrow x=-\dfrac{1}{10}\)
Vậy \(x\in\left\{-\dfrac{1}{10}\right\}\)
\(đk:x\ne0;1;-1\\ \dfrac{x^2+x}{x^2-x}.\dfrac{2x^2-2x}{x+1}\\ =\dfrac{x\left(x+1\right)}{x\left(x-1\right)}.\dfrac{2x\left(x-1\right)}{x+1}\\ =2x\)