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\(a,=\dfrac{x^3-\left(x-1\right)\left(x^2+x+1\right)}{1-x}=\dfrac{x^3-x^3+1}{1-x}=\dfrac{1}{1-x}\\ b,=\dfrac{2x+x^2+3x+2+2-x}{\left(x+2\right)^2}=\dfrac{\left(x+2\right)^2}{\left(x+2\right)^2}=1\)
`a)3x(2x^2-3x+4)`
`=6x^3-9x^2+12x`
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`b)(x+3)^2+(3x-2)(x+4)`
`=x^2+6x+9+3x^2+12x-2x-8`
`=4x^2+16x+1`
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`c)[2x-4]/[x-1]+[2x+2]/[x^2-1]` `ĐK: x \ne +-1`
`=[(2x-4)(x+1)+2x+2]/[(x-1)(x+1)]`
`=[2x^2+2x-4x-4+2x+2]/[(x-1)(x+1)]`
`=[2x^2-2]/[x^2-1]`
`=2`
\(1.\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}.\Leftrightarrow\dfrac{x-1-3x}{3}=\dfrac{x-2}{2}.\Leftrightarrow\dfrac{-2x-1}{3}-\dfrac{x-2}{2}=0.\)
\(\Leftrightarrow\dfrac{-4x-2-3x+6}{6}=0.\Rightarrow-7x+4=0.\Leftrightarrow x=\dfrac{4}{7}.\)
\(2.\left(x-2\right)\left(2x-1\right)=x^2-2x.\Leftrightarrow\left(x-2\right)\left(2x-1\right)-x\left(x-2\right)=0.\)
\(\Leftrightarrow\left(x-2\right)\left(2x-1-x\right)=0.\Leftrightarrow\left(x-2\right)\left(x-1\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2.\\x=1.\end{matrix}\right.\)
\(3.3x^2-4x+1=0.\Leftrightarrow\left(x-1\right)\left(x-\dfrac{1}{3}\right)=0.\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=\dfrac{1}{3}.\end{matrix}\right.\)
\(4.\left|2x-4\right|=0.\Leftrightarrow2x-4=0.\Leftrightarrow x=2.\)
\(5.\left|3x+2\right|=4.\Leftrightarrow\left[{}\begin{matrix}3x+2=4.\\3x+2=-4.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}.\\x=-2.\end{matrix}\right.\)
\(1,\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}\\ \Leftrightarrow\dfrac{x-1}{3}-x=\dfrac{x-2}{2}\\ \Leftrightarrow\dfrac{2\left(x-1\right)-6x}{6}=\dfrac{3\left(x-2\right)}{6}\\ \Leftrightarrow2\left(x-1\right)-6x=3\left(x-2\right)\\ \Leftrightarrow2x-2-6x=3x-6\\ \Leftrightarrow-4x-2=3x-6\)
\(\Leftrightarrow3x-6+4x+2=0\\ \Leftrightarrow7x-4=0\\ \Leftrightarrow x=\dfrac{4}{7}\)
\(2,\left(x-2\right)\left(2x-1\right)=x^2-2x\\ \Leftrightarrow2x^2-4x-x+2=x^2-2x\\ \Leftrightarrow x^2-3x+2=0\\ \Leftrightarrow\left(x^2-2x\right)-\left(x-2\right)=0\\ \Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
\(3,3x^2-4x+1=0\\ \Leftrightarrow\left(3x^2-3x\right)-\left(x-1\right)=0\\ \Leftrightarrow3x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)
\(4,\left|2x-4\right|=0\\ \Leftrightarrow2x-4=0\\ \Leftrightarrow2x=4\\ \Leftrightarrow x=2\)
\(5,\left|3x+2\right|=4\\ \Leftrightarrow\left[{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)
\(6,\left|2x-5\right|=\left|-x+2\right|\\ \Leftrightarrow\left[{}\begin{matrix}2x-5=-x+2\\2x-5=x-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=7\\x=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=3\end{matrix}\right.\)
a.x-\(\dfrac{5x+2}{6}=\dfrac{7-3x}{4}\)
⇔\(x=\dfrac{7-3x}{4}+\dfrac{5x+2}{6}\)
⇔\(x=\dfrac{21-9x+10x+4}{12}\)
⇔x=\(\dfrac{x+25}{12}\)
⇔12x=x+25
⇔x=\(\dfrac{25}{11}\)
Vậy pt đã cho có n0 là S=\(\left\{\dfrac{25}{11}\right\}\)
b.ĐKXĐ:x≠-2;x≠2
\(\dfrac{x-2}{x+2}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{x^2-4}\)
⇔\(\dfrac{\left(x-2\right)\cdot\left(x-2\right)-3\cdot\left(x+2\right)}{\left(x-2\right)\cdot\left(x+2\right)}\)=\(\dfrac{2x-22}{\left(x-2\right)\cdot\left(x+2\right)}\)
⇔\(\dfrac{x^2-7x-2}{\left(x-2\right)\cdot\left(x+2\right)}=\dfrac{2x-22}{\left(x-2\right)\cdot\left(x+2\right)}\)
⇒\(\left(x^2-7x-2\right)\cdot\left(x-2\right)\cdot\left(x+2\right)=\left(2x-22\right)\cdot\left(x-2\right)\cdot\left(x+2\right)\)
⇔x2-7x-2=2x-22
⇔x2-9x+20=0
⇔(x-4)(x-5)=0
⇔\(\left\{{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)
Vậy pt đã cho có n0 là S={4;5}
b)
ĐKXĐ: \(x\notin\left\{2;3;\dfrac{1}{2}\right\}\)
Ta có: \(\dfrac{x+4}{2x^2-5x+2}+\dfrac{x+1}{2x^2-7x+3}=\dfrac{2x+5}{2x^2-7x+3}\)
\(\Leftrightarrow\dfrac{x+4}{\left(x-2\right)\left(2x-1\right)}+\dfrac{x+1}{\left(x-3\right)\left(2x-1\right)}=\dfrac{2x+5}{\left(2x-1\right)\left(x-3\right)}\)
\(\Leftrightarrow\dfrac{\left(x+4\right)\left(x-3\right)}{\left(x-2\right)\left(2x-1\right)\left(x-3\right)}+\dfrac{\left(x+1\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)\left(2x-1\right)}=\dfrac{\left(2x+5\right)\left(x-2\right)}{\left(2x-1\right)\left(x-3\right)\left(x-2\right)}\)
Suy ra: \(x^2-3x+4x-12+x^2-2x+x-2=2x^2-4x+5x-10\)
\(\Leftrightarrow2x^2-14=2x^2+x-10\)
\(\Leftrightarrow2x^2-14-2x^2-x+10=0\)
\(\Leftrightarrow-x-4=0\)
\(\Leftrightarrow-x=4\)
hay x=-4(nhận)
Vậy: S={-4}
a) (2x+12x−1−2x−12x+1):4x10x−5=(2x+1)2−(2x−1)2(2x−1)(2x+1).10x+54x(2x+12x−1−2x−12x+1):4x10x−5=(2x+1)2−(2x−1)2(2x−1)(2x+1).10x+54x
=4x2+4x+1−4x2+4x−1(2x−1)(2x+1).5(2x+1)4x4x2+4x+1−4x2+4x−1(2x−1)(2x+1).5(2x+1)4x
=8x.5(2x+1)(2x−1)(2
b) \(\left(\dfrac{1}{x^2+x}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)=\left(\dfrac{1}{x\left(x+1\right)}-\dfrac{x\left(2-x\right)}{x\left(x+1\right)}\right):\left(\dfrac{1}{x}+\dfrac{x^2}{x}-\dfrac{2x}{x}\right)=\left(\dfrac{1-2x+x^2}{x\left(x+1\right)}\right):\left(\dfrac{1+x^2-2x}{x}\right)=\left(\dfrac{\left(x-1\right)^2}{x\left(x+1\right)}\right)\cdot\left(\dfrac{x}{\left(x-1\right)^2}\right)=\dfrac{\left(x-1\right)^2\cdot x}{\left(x-1\right)^2\cdot x\cdot\left(x+1\right)}=\dfrac{1}{x+1}\)
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:
\(\dfrac{1}{\left(a-b\right)\left(b-c\right)}+\dfrac{1}{\left(b-c\right)\left(c-a\right)}+\dfrac{1}{\left(c-a\right)\left(a-b\right)}\)
\(=\dfrac{c-a+a-b+b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
a: \(=\dfrac{x^3-1}{x+2}\cdot\dfrac{x^2+x+1-x^2+1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x+2}{x+2}=1\)
b: \(=\dfrac{\left(x+2\right)\left(x-1\right)\left(x+1\right)}{2\left(x+5\right)}\cdot\left(\dfrac{x+1-2x+2}{\left(x-1\right)\left(x+1\right)}+\dfrac{1}{x+2}\right)\)
\(=\dfrac{\left(x+2\right)\left(x-1\right)\left(x+1\right)}{2\left(x+5\right)}\cdot\left(\dfrac{-\left(x-3\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{1}{x+2}\right)\)
\(=\dfrac{\left(x+2\right)\left(x-1\right)\left(x+1\right)}{2\left(x+5\right)}\cdot\dfrac{-\left(x^2-x-6\right)+x^2-1}{\left(x-1\right)\left(x+1\right)\left(x+2\right)}\)
\(=\dfrac{-x^2+x+6+x^2-1}{2\left(x+5\right)}=\dfrac{x+5}{2\left(x+5\right)}=\dfrac{1}{2}\)
\(=\dfrac{2x^2-x-x-1+2-x^2}{x-1}=\dfrac{x^2-2x+1}{x-1}=\dfrac{\left(x-1\right)^2}{x-1}=x-1\)
1) \(\dfrac{x^2-4}{x^2+2x+1}:\dfrac{4-2x}{2x+2}=\dfrac{\left(x-2\right)\left(x+2\right)2\left(x+1\right)}{\left(x+1\right)^22\left(2-x\right)}=\dfrac{2\left(x-2\right)\left(x+2\right)\left(x+1\right)}{-2\left(x-2\right)\left(x+1\right)\left(x+1\right)}=\dfrac{-\left(x+2\right)}{x+1}=\dfrac{-x-2}{x+1}\)
2) \(\dfrac{x+1}{x+2}:\left(\dfrac{x+2}{x+3}:\dfrac{x+3}{x+1}\right)=\dfrac{x+1}{x+2}:\dfrac{\left(x+2\right)\left(x+1\right)}{\left(x+3\right)\left(x+3\right)}=\dfrac{\left(x+1\right)\left(x+3\right)\left(x+3\right)}{\left(x+2\right)\left(x+2\right)\left(x+1\right)}=\dfrac{\left(x+3\right)^2}{\left(x+2\right)^2}=\dfrac{x^2+6x+9}{x^2+4x+4}\)