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\(a,\dfrac{x^2-2x}{x^2-4}=\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{x}{x+2}\)
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 ?
\(a,\frac{x}{xy-y^2}+\frac{2x-y}{xy-x^2}:\left(\frac{1}{x}+\frac{1}{y}\right)\)
\(=\left(\frac{x}{y\left(x-y\right)}+\frac{y-2x}{x\left(x-y\right)}\right):\left(\frac{y}{xy}+\frac{x}{xy}\right)\)
\(=\left(\frac{x-y}{x\left(x-y\right)}\right):\left(\frac{x+y}{xy}\right)\)
\(=\frac{1}{x}.\frac{xy}{x+y}=\frac{y}{x+y}\)
\(A=\left(\dfrac{x^2-2x}{2\left(x^2+4\right)}+\dfrac{2x^2}{x^3-2x^2+4x-8}\right)\cdot\dfrac{x^2-x-2}{x^2}\)
\(=\left(\dfrac{x\left(x-2\right)}{2\left(x^2+4\right)}+\dfrac{2x^2}{\left(x-2\right)\left(x^2+4\right)}\right)\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\)
\(=\left(\dfrac{x\left(x^2-4x+4\right)+4x^2}{2\left(x-2\right)\left(x^2+4\right)}\right)\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\)
\(=\dfrac{x\left(x^2-4x+4+4x\right)}{2\left(x^2+4\right)}\cdot\dfrac{x+1}{x^2}=\dfrac{x\left(x^2+4\right)}{2\left(x^2+4\right)}\cdot\dfrac{x+1}{x^2}\)
\(=\dfrac{x+1}{2x}\)
( x2−2x / 2x2+8 − 2x2 / 8−4x+2x2−x3 ).(1− 1/x − 2/x2 )
=[ x2−2x / 2(x2+4) − 2x2 / 2(x2+4)−x(x2+4) ]. x2−x−2 / x2
=[x2−2x / 2(x2+4) − 2x2 / (2−x)(x2+3)] . x2−x−2 / x2
=(x2−2x)(2−x)−4x2 / 2(2−x)(x2+4) . x2+x−2x−2 / x2
= −x(x2+4) / 2(2−x)(x2+4). (x+1)(x−2) / x2
=x+1 / 2x
\(A=\left(\dfrac{x^2-2x}{2\left(x^2+4\right)}+\dfrac{2x^2}{x^3-2x^2+4x-8}\right)\cdot\dfrac{x^2-x-2}{x^2}\)
\(=\left(\dfrac{x^2-2x}{2\left(x^2+4\right)}+\dfrac{2x^2}{\left(x^2+4\right)\left(x-2\right)}\right)\cdot\dfrac{x^2-x-2}{x^2}\)
\(=\dfrac{\left(x^2-2x\right)\left(x-2\right)+4x^2}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\)
\(=\dfrac{x^3-2x^2-2x^2+4x+4x^2}{2\left(x^2+4\right)}\cdot\dfrac{x+1}{x^2}\)
\(=\dfrac{x\left(x^2+4\right)}{2\left(x^2+4\right)}\cdot\dfrac{x+1}{x^2}=\dfrac{x+1}{2x}\)
1) ĐKXĐ của \(x\):
\(\left\{{}\begin{matrix}2x-6\ne0\\2x^2+6x\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2\left(x-3\right)\ne0\\2x\left(x+3\right)\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x\ne0;x\ne-3\end{matrix}\right.\)
ĐKXĐ: \(x\ne0;x\ne\pm3\)
Ta có: \(\dfrac{3}{2x-6}-\dfrac{x-6}{2x^2+6x}\)
\(=\dfrac{3}{2\left(x-3\right)}-\dfrac{x-6}{2x\left(x+3\right)}\)
\(=\dfrac{3}{2\left(x-3\right)}+\dfrac{x-6}{2x\left(x-3\right)}\)
\(=\dfrac{3.2+x-6}{2x\left(x-3\right)}\)
\(=\dfrac{6+x-6}{2x\left(x-3\right)}\)
\(=\dfrac{x}{2x\left(x-3\right)}\)
\(=\dfrac{1}{2\left(x-3\right)}\)
2) ĐKXĐ của câu này bạn làm tương tự câu trên nhé, ở đây ngoặc không đủ
ĐKXĐ: \(x\ne0;x\ne\pm2;x\ne3\)
Ta có: \(A=\left(\dfrac{2+x}{2-x}+\dfrac{4x^2}{x^2-4}-\dfrac{2-x}{2+x}\right):\dfrac{x^2-3x}{2x^2-x^3}\)
\(A=\left(\dfrac{2+x}{2-x}+\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{2-x}{2+x}\right):\dfrac{x\left(x-3\right)}{x^2\left(2-x\right)}\)
\(A=\left(\dfrac{2+x}{2-x}-\dfrac{4x^2}{\left(2-x\right)\left(2+x\right)}-\dfrac{2-x}{2+x}\right).\dfrac{x^2\left(2-x\right)}{x\left(x-3\right)}\)
\(A=\dfrac{\left(2+x\right)\left(2+x\right)-4x^2-\left(2-x\right)\left(2-x\right)}{\left(2-x\right)\left(2+x\right)}.\dfrac{x^2\left(2-x\right)}{x\left(x-3\right)}\)
\(A=\dfrac{4+4x+x^2-4x^2-\left(4-4x+x^2\right)}{\left(2-x\right)\left(2+x\right)}.\dfrac{x^2\left(2-x\right)}{x\left(x-3\right)}\)
\(A=\dfrac{-4x^2+8x}{\left(2-x\right)\left(2+x\right)}.\dfrac{x^2\left(2-x\right)}{x\left(x-3\right)}\)
\(A=\dfrac{-4x\left(x-2\right)}{\left(2-x\right)\left(2+x\right)}.\dfrac{x^2\left(2-x\right)}{x\left(x-3\right)}\)
\(A=\dfrac{-4x^2\left(x-2\right)}{\left(2+x\right)\left(x-3\right)}\)
\(\left(\dfrac{2}{x+2}-\dfrac{4}{x^2+4x+4}\right):\left(\dfrac{2}{x^2-4}+\dfrac{1}{2-x}\right)\\ =\left(\dfrac{2\left(x+2\right)}{\left(x+2\right)^2}-\dfrac{4}{\left(x+2\right)^2}\right):\left(\dfrac{2}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x-2}\right)\\ =\left(\dfrac{2x}{\left(x+2\right)^2}\right):\left(\dfrac{2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x+2}{\left(x+2\right)\left(x-2\right)}\right)\\ =\dfrac{2x}{\left(x+2\right)^2}:\dfrac{-x}{x^2-4}\\ =\dfrac{2x}{\left(x+2\right)^2}.\dfrac{x^2-4}{-x}=-\dfrac{2x^2-8}{\left(x+2\right)^2}\)
ĐKXĐ: \(x\notin\left\{0;2;-2\right\}\)
\(P=\dfrac{x+2}{x^2-4x+4}:\left(\dfrac{6-x^2}{x^2-2x}-\dfrac{1}{2-x}+\dfrac{x+2}{x}\right)\)
\(=\dfrac{x+2}{\left(x-2\right)^2}:\left(\dfrac{6-x^2}{x\left(x-2\right)}+\dfrac{1}{x-2}+\dfrac{x+2}{x}\right)\)
\(=\dfrac{x+2}{\left(x-2\right)^2}:\dfrac{6-x^2+x+\left(x+2\right)\left(x-2\right)}{x\left(x-2\right)}\)
\(=\dfrac{x+2}{\left(x-2\right)^2}\cdot\dfrac{x\left(x-2\right)}{6-x^2+x+x^2-4}\)
\(=\dfrac{x+2}{x-2}\cdot\dfrac{x}{x+2}=\dfrac{x}{x-2}\)