a) \(\dfrac{1}{3x-2}-\dfrac{1}{3x+2}-\dfrac{3x-6}{9x^2-4}\)

\(=\dfrac{3x+2-3x+2-3x+6}{\left(3x-2\right)\left(3x+2\right)}\)

\(=\dfrac{-3x+10}{\left(3x-2\right)\left(3x+2\right)}\)

b) \(\dfrac{x+25}{2x^2-50}-\dfrac{x+5}{x^2-5x}-\dfrac{5-x}{2x^2+10x}\)

\(=\dfrac{x+25}{2\left(x-5\right)\left(x+5\right)}-\dfrac{x+5}{x\left(x-5\right)}+\dfrac{x-5}{2x\left(x+5\right)}\)

\(=\dfrac{x^2+25x-2\left(x+5\right)^2+\left(x-5\right)^2}{2x\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{x^2+25x-2x^2-20x-50+x^2-10x+25}{2x\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{-5x-25}{2x\left(x-5\right)\left(x+5\right)}\)

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

c) Ta có: \(\dfrac{1-2x}{2x}-\dfrac{4x}{2x-1}-\dfrac{3}{2x-4x^2}\)

\(=\dfrac{-\left(2x-1\right)^2-8x^2+3}{2x\left(2x-1\right)}\)

\(=\dfrac{-\left(4x^2-4x+1\right)-8x^2+3}{2x\left(2x-1\right)}\)

\(=\dfrac{-4x^2+4x-1-8x^2+3}{2x\left(2x-1\right)}\)

\(=\dfrac{-12x^2+4x+2}{2x\left(2x-1\right)}\)

a: \(\Leftrightarrow2x\left(x^2+2x+5\right)=0\)

=>x=0

b: \(\Leftrightarrow\dfrac{x}{x-1}-\dfrac{x+1}{x-3}=\dfrac{1}{2}\)

\(\Leftrightarrow x^2-4x+3=2x\left(x-3\right)-2\left(x^2-1\right)\)

\(\Leftrightarrow x^2-4x+3=2x^2-6x-2x^2+2=-6x+2\)

\(\Leftrightarrow x^2+2x+1=0\)

=>x=-1(nhận)

\(\Leftrightarrow2x\left(x^2+2x+5\right)=0\)

\(\Leftrightarrow x=0\) ( vì \(x^2+2x+5>0;\forall x\)

b.\(\Leftrightarrow\dfrac{x\left(x-4\right)}{\left(x-1\right)\left(x-4\right)}-\dfrac{1}{2}=\dfrac{x+1}{x-3}\)

\(ĐK:x\ne1;3;4\)

\(\Leftrightarrow\dfrac{x}{\left(x-1\right)}-\dfrac{1}{2}=\dfrac{x+1}{x-3}\)

\(\Leftrightarrow\dfrac{x\left(x-3\right)-\left(x-1\right)\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}=\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x-1\right)\left(x-3\right)}\)

\(\Leftrightarrow x\left(x-3\right)-\left(x-1\right)\left(x-3\right)=\left(x+1\right)\left(x-1\right)\)

\(\Leftrightarrow x^2-3x-\left(x^2-3x-x+3\right)=x^2-1\)

\(\Leftrightarrow x^2-3x-x^2+4x-3=x^2-1\)

\(\Leftrightarrow x^2+x-2=0\)

\(\Leftrightarrow x^2-x+2x-2=0\)

\(\Leftrightarrow x\left(x-1\right)+2\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(ktm\right)\\x=-2\left(tm\right)\end{matrix}\right.\)

Vậy \(S=\left\{-2\right\}\)

\(a,2x^3+4x^2+10x=0\\ \Leftrightarrow2x\left(x^2+2x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x=0\\x^2+2x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\\left(x^2+2x+1\right)+4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\\left(x+1\right)^2+4=0\left(vô..lí\right)\end{matrix}\right.\)

\(b,ĐKXĐ:\left\{{}\begin{matrix}x\ne1\\x\ne3\\x\ne4\end{matrix}\right.\\ \dfrac{x^2-4x}{x^2-5x+4}-\dfrac{1}{2}=\dfrac{x+1}{x-3}\\ \Leftrightarrow\dfrac{x\left(x-4\right)}{\left(x-1\right)\left(x-4\right)}-\dfrac{1}{2}=\dfrac{x+1}{x-3}\\ \Leftrightarrow\dfrac{x}{x-1}-\dfrac{1}{2}-\dfrac{x+1}{x-3}=0\\ \Leftrightarrow\dfrac{2x\left(x-3\right)}{2\left(x-1\right)\left(x-3\right)}-\dfrac{\left(x-1\right)\left(x-3\right)}{2\left(x-1\right)\left(x-3\right)}-\dfrac{2\left(x+1\right)\left(x-1\right)}{2\left(x-1\right)\left(x-3\right)}=0\)

\(\Leftrightarrow\dfrac{2x^2-6x}{2\left(x-1\right)\left(x-3\right)}-\dfrac{x^2-4x+3}{2\left(x-1\right)\left(x-3\right)}-\dfrac{2x^2-2}{\left(x-1\right)\left(x-3\right)}=0\)

\(\Leftrightarrow\dfrac{2x^2-6x-x^2+4x-3-2x^2+2}{2\left(x-1\right)\left(x-3\right)}=0\)

\(\Rightarrow-x^2-2x-1=0\)

\(\Leftrightarrow x^2+2x+1=0\\ \Leftrightarrow\left(x+1\right)^2=0\\ \Leftrightarrow x+1=0\\ \Leftrightarrow x=-1\left(tm\right)\)

a: \(=\dfrac{1}{x+2y}+\dfrac{1}{x-2y}-\dfrac{4y}{\left(x-2y\right)\left(x+2y\right)}\)

\(=\dfrac{x-2y+x+2y-4y}{\left(x+2y\right)\left(x-2y\right)}=\dfrac{2\left(x-2y\right)}{\left(x+2y\right)\left(x-2y\right)}=\dfrac{2}{x+2y}\)

b: \(=\dfrac{2x}{x-1}+\dfrac{5\left(x-1\right)\left(x+1\right)}{\left(x+1\right)^2}\cdot\dfrac{2\left(x+1\right)}{5\left(1-x\right)}\)

\(=\dfrac{2x}{x-1}-2=\dfrac{2x-2x+2}{x-1}=\dfrac{2}{x-1}\)

c: \(=\dfrac{5\left(x-1\right)}{2x}\cdot\dfrac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{5\cdot4x}{2x\cdot\left(x+1\right)}=\dfrac{10}{x+1}\)

\(a,\dfrac{x^2-2x}{x^2-4}=\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{x}{x+2}\)

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

c) \(\dfrac{x^4+4}{x\left(x^2+2\right)-2x^2-\left(x-1\right)^2-1}\)

\(=\dfrac{x^4+4x^2-4x^2+4}{x^3+2x-2x^2-x^2+2x-1-1}\)

\(=\dfrac{\left(x^2+2\right)^2-4x^2}{\left(x^3+2x-2x^2\right)-\left(x^2-2x+2\right)}\)

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

\(=\dfrac{x^2+2+2x}{x-1}\)

Bài 2:

a) \(\left(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}\right):\dfrac{4x}{10x-5}\)

\(=\dfrac{\left(2x+1\right)^2-\left(2x-1\right)^2}{\left(2x-1\right)\left(2x+1\right)}.\dfrac{5\left(2x-1\right)}{4x}\)

\(=\dfrac{8x}{\left(2x-1\right)\left(2x+1\right)}.\dfrac{5\left(2x-1\right)}{4x}\)

\(=\dfrac{10}{2x+1}\)

b) \(\left(\dfrac{1}{x^2+x}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)\)

\(=\dfrac{1-2x+x^2}{x\left(x+1\right)}:\dfrac{1+x^2-2x}{x}\)

\(=\dfrac{1}{x+1}\)

c) Trong ngoặc giữa hai phân số là dấu gì vậy ?

a: \(=\dfrac{3x}{5\left(x+y\right)}-\dfrac{x}{10\left(x-y\right)}\)

\(=\dfrac{6x\left(x-y\right)-x\left(x+y\right)}{10\left(x-y\right)\cdot\left(x+y\right)}\)

\(=\dfrac{6x^2-6xy-x^2-xy}{10\left(x-y\right)\left(x+y\right)}=\dfrac{5x^2-7xy}{10\left(x-y\right)\left(x+y\right)}\)

b: \(=\dfrac{7}{2\left(2x-3\right)\left(2x+3\right)}+\dfrac{1}{x\left(2x+3\right)}-\dfrac{1}{2\left(2x-3\right)}\)

\(=\dfrac{7x+2\left(2x-3\right)-x\left(2x+3\right)}{2x\left(2x+3\right)\left(2x-3\right)}\)

\(=\dfrac{7x+4x-6-2x^2-3x}{2x\left(2x+3\right)\left(2x-3\right)}\)

\(=\dfrac{-2x^2-6}{2x\left(2x+3\right)\left(2x-3\right)}=\dfrac{-x^2-3}{x\left(2x+3\right)\left(2x-3\right)}\)

c: \(=\dfrac{5}{x+1}+\dfrac{10}{x^2-x+1}-\dfrac{15}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(=\dfrac{5x^2-5x+5+10x+10-15}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(=\dfrac{5x^2+5x}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{5x}{x^2-x+1}\)

Đề có sai không bạn?

a) \(\left(x^2-4\right)-\left(x-2\right)\left(3-2x\right)\)

\(=\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(3-2x\right)\)

\(=\left(x-2\right)\left(x+2-3+2x\right)\)

\(=\left(x-2\right)\left(3x-1\right)\)

b) ĐKXĐ: x ≠ 5; x ≠ -5

Với điều kiện trên ta có:

\(\dfrac{x+5}{x^2-5x}-\dfrac{x-5}{2x^2+10x}=\dfrac{x+25}{2x^2-50}\)

\(\Leftrightarrow\dfrac{x+5}{x\left(x-5\right)}-\dfrac{x-5}{2x\left(x+5\right)}-\dfrac{x+25}{2\left(x^2-25\right)}=0\)

\(\Leftrightarrow\dfrac{x+5}{x\left(x-5\right)}-\dfrac{x-5}{2x\left(x+5\right)}-\dfrac{x+25}{2\left(x-5\right)\left(x+5\right)}=0\)

\(\Rightarrow2\left(x+5\right)^2-\left(x-5\right)^2-x\left(x+25\right)=0\)

\(\Leftrightarrow2x^2+20x+50-x^2+10x-25-x^2-25x=0\)

\(\Leftrightarrow5x-25=0\)

\(\Leftrightarrow5x=25\)

\(\Leftrightarrow x=5\)(Không thỏa mãn ĐKXĐ)

Vậy tập nghiệm của phương trình là S = ∅

c) ĐKXĐ: x ≠ 1

Với điều kiện trên ta có:

\(\dfrac{1}{x-1}-\dfrac{3x^2}{x^3-1}=\dfrac{2x}{x^2+x+1}\)

\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{2x}{x^2+x+1}=0\)

\(\Rightarrow x^2+x+1-3x^2-2x\left(x-1\right)=0\)

\(\Leftrightarrow x^2+x+1-3x^2-2x^2+2x=0\)

\(\Leftrightarrow-4x^2+3x+1=0\)

\(\Leftrightarrow-4x^2+4x-x+1=0\)

\(\Leftrightarrow-4x\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(-4x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\-4x-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(Khôngthoảman\right)\\x=-\dfrac{1}{4}\left(Thỏamãn\right)\end{matrix}\right.\)

Vậy tập nghiệm của phương trình là \(S=\left\{-\dfrac{1}{4}\right\}\)

a) Ta có: \(\dfrac{4}{5}-3\left|x\right|=\dfrac{1}{5}\)

\(\Leftrightarrow3\left|x\right|=\dfrac{4}{5}-\dfrac{1}{5}=\dfrac{3}{5}\)

\(\Leftrightarrow\left|x\right|=\dfrac{1}{5}\)

hay \(x\in\left\{\dfrac{1}{5};-\dfrac{1}{5}\right\}\)

b) Ta có: \(4x-\dfrac{1}{2}x+\dfrac{3}{5}x=\dfrac{4}{5}\)

nên \(\dfrac{41}{10}x=\dfrac{4}{5}\)

hay \(x=\dfrac{8}{41}\)

c) Ta có: \(\left(2x-8\right)\left(10-5x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-8=0\\10-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=8\\5x=10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)

d) Ta có: \(\dfrac{3}{4}+\dfrac{1}{4}\left|2x-1\right|=\dfrac{7}{2}\)

\(\Leftrightarrow\dfrac{1}{4}\left|2x-1\right|=\dfrac{7}{2}-\dfrac{3}{4}=\dfrac{14}{4}-\dfrac{3}{4}=\dfrac{11}{4}\)

\(\Leftrightarrow\left|2x-1\right|=11\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=11\\2x-1=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=12\\2x=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-5\end{matrix}\right.\)