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a: \(=6x^4-9x^3+3x^2-4x^3+6x^2-2x+10x^2-15x+5\)
\(=6x^4-13x^3+19x^2-17x+5\)
b: \(=6x^4-\dfrac{9}{4}x^3-\dfrac{9}{2}x^2-\dfrac{8}{3}x^3+x^2+2x-\dfrac{20}{3}x^2+\dfrac{5}{2}x+5\)
\(=6x^4-\dfrac{59}{12}x^3-\dfrac{67}{6}x^2+\dfrac{9}{2}x+5\)
c: \(=3x^4-\dfrac{9}{8}x^3-\dfrac{3}{4}x^2+8x^3-3x^2-6x-\dfrac{4}{3}x^2+\dfrac{1}{2}x+1\)
\(=3x^4-\dfrac{55}{8}x^3-\dfrac{25}{12}x^2-\dfrac{11}{2}x+1\)
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: \(VT=\left[\dfrac{2}{\left(x+1\right)^3}\cdot\dfrac{x+1}{x}+\dfrac{1}{\left(x+1\right)^2}\cdot\dfrac{1+x^2}{x^2}\right]\cdot\dfrac{x^3}{x-1}\)
\(=\left(\dfrac{2}{x\left(x+1\right)^2}+\dfrac{x^2+1}{x^2\cdot\left(x+1\right)^2}\right)\cdot\dfrac{x^3}{x-1}\)
\(=\dfrac{2x+x^2+1}{x^2\cdot\left(x+1\right)^2}\cdot\dfrac{x^3}{x-1}\)
\(=\dfrac{\left(x+1\right)^2}{\left(x+1\right)^2}\cdot\dfrac{x}{x-1}=\dfrac{x}{x-1}\)
\(A=\dfrac{x^2+x-2+x^2-x-2-4}{x\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x\left(x-3\right)}{2\left(x+2\right)}=\dfrac{2\left(x-2\right)\left(x+2\right)\left(x-3\right)}{2\left(x-2\right)\left(x+2\right)^2}=\dfrac{x-3}{x+2}\\ A\le0\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-3\ge0\\x+2< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x-3\le0\\x+2>0\end{matrix}\right.\end{matrix}\right.\Rightarrow-2< x< 3;x\ne0\left(ĐKXD\right)\)
a) \(\dfrac{x^2-4}{2x^2-4x}\)
\(=\dfrac{\left(x-2\right)\left(x+2\right)}{2x\left(x-2\right)}\)
\(=\dfrac{x+2}{2x}\)
b) \(\dfrac{2x-x^2}{x^2-4x+4}\)
\(=\dfrac{x\left(2-x\right)}{\left(x-2\right)^2}\)
\(=\dfrac{x\left(2-x\right)}{\left(2-x\right)^2}\)
\(=\dfrac{x}{2-x}\)
c) \(\dfrac{x^2-xy-x+y}{x^2+xy-x-y}\)
\(=\dfrac{x\left(x-y\right)-\left(x-y\right)}{x\left(x+y\right)-\left(x+y\right)}\)
\(=\dfrac{\left(x-y\right)\left(x-1\right)}{\left(x+y\right)\left(x-1\right)}\)
\(=\dfrac{x-y}{x+y}\)
d) \(\dfrac{5x^2+10x+5}{x+x^2}\)
\(\dfrac{5\left(x^2+2x+1\right)}{x\left(1+x\right)}\)
\(=\dfrac{5\left(x+1\right)^2}{x\left(x+1\right)}\)
\(=\dfrac{5\left(x+1\right)}{x}\)
e) \(\dfrac{3x^2+3x}{\left(x+1\right)\left(2x+6\right)}\)
\(=\dfrac{3x\left(x+1\right)}{\left(x+1\right).2\left(x+3\right)}\)
\(=\dfrac{3x}{2\left(x+3\right)}\)
f) \(\dfrac{\left(2-3x\right)\left(x+1\right)}{x^2+2x+1}\)
\(=\dfrac{\left(2-3x\right)\left(x+1\right)}{\left(x+1\right)^2}\)
\(=\dfrac{2-3x}{x+1}\)
ĐKXĐ: \(x\notin\left\{-1;2;-2\right\}\)
a) Ta có: \(A=\left(\dfrac{\left(x+1\right)^2}{\left(x+1\right)^2-3x}-\dfrac{2x^2+4x-1}{x^3+1}-\dfrac{1}{x+1}\right):\dfrac{x^2-4}{3x^2+6x}\)
\(=\left(\dfrac{\left(x+1\right)^2}{x^2-x+1}-\dfrac{2x^2+4x-1}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{1}{x+1}\right):\dfrac{x^2-4}{3x^2+6x}\)
\(=\left(\dfrac{x^3+3x^2+3x+1}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{2x^2+4x-1}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}\right):\dfrac{\left(x-2\right)\left(x+2\right)}{3x\left(x+2\right)}\)
\(=\dfrac{x^3+3x^2+3x+1-2x^2-4x+1-x^2+x-1}{\left(x+1\right)\left(x^2-x+1\right)}:\dfrac{x-2}{3x}\)
\(=\dfrac{x^3+1}{\left(x+1\right)\left(x^2-x+1\right)}\cdot\dfrac{3x}{x-2}\)
\(=\dfrac{3x}{x-2}\)
b) Để A nguyên thì \(3x⋮x-2\)
\(\Leftrightarrow3x-6+6⋮x-2\)
mà \(3x-6⋮x-2\)
nên \(6⋮x-2\)
\(\Leftrightarrow x-2\inƯ\left(6\right)\)
\(\Leftrightarrow x-2\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)
hay \(x\in\left\{3;1;4;0;5;-1;8;-4\right\}\)
Kết hợp ĐKXĐ, ta được:
\(x\in\left\{3;1;4;0;5;8;-4\right\}\)
Vậy: Để A nguyên thì \(x\in\left\{3;1;4;0;5;8;-4\right\}\)
\(a,\dfrac{x^2-2x}{x^2-4}=\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{x}{x+2}\)
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 ?
\(A=\left(\dfrac{x-1}{x\left(x-2\right)}+\dfrac{x+1}{x\left(x+2\right)}-\dfrac{4}{x\left(x-2\right)\left(x+2\right)}\right)\cdot\dfrac{x\left(x-3\right)}{2\left(x+2\right)}\)
\(=\dfrac{x^2+x-2+x^2-x+2-4}{x\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x\left(x-3\right)}{2\left(x+2\right)}\)
\(=\dfrac{2x^2-4}{x\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x\left(x-3\right)}{2\left(x+2\right)}\)
\(=\dfrac{2x\left(x^2-2\right)\left(x-3\right)}{2x\left(x-2\right)\cdot\left(x+2\right)^2}=\dfrac{\left(x^2-2\right)\left(x-3\right)}{\left(x-2\right)\left(x+2\right)^2}\)